How to create a tree-like structure from a JavaScript Object

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}

This is the Object sample I'm given:

Input :

{
"People": [
{
"id": "12",
"parentId": "0",
"text": "Man",
"level": "1",
"children": null
},
{
"id": "6",
"parentId": "12",
"text": "Boy",
"level": "2",
"children": null
},
{
"id": "7",
"parentId": "12",
"text": "Other",
"level": "2",
"children": null
},
{
"id": "9",
"parentId": "0",
"text": "Woman",
"level": "1",
"children": null
},
{
"id": "11",
"parentId": "9",
"text": "Girl",
"level": "2",
"children": null
}
] }

I want to transform it to a JSON format like this:

{

    "People": [

        {

            "id": "12",

            "parentId": "0",

            "text": "Man",

            "level": "1",

            "children": [

                {

                    "id": "6",

                    "parentId": "12",

                    "text": "Boy",

                    "level": "2",

                    "children": null

                },

                {

                    "id": "7",

                    "parentId": "12",

                    "text": "Other",

                    "level": "2",

                    "children": null

                }   

            ]

        }

}

Any Ideas/Help would be appreciated

edited Nov 17 '18 at 20:29

asked Nov 16 '18 at 11:46

Omar

1114

2

Can you show us what your attempt looked like?

– SpoonMeiser
Nov 16 '18 at 11:47

Wait a minute..

– Nurbol Alpysbayev
Nov 16 '18 at 11:57

I tried implementing the solution of this question: stackoverflow.com/questions/18017869/… but didn't work after editing :/ @SpoonMeiser

– Omar
Nov 16 '18 at 12:12

I almost finished it... it's taking too long.. :)

– Nurbol Alpysbayev
Nov 16 '18 at 12:12

I think that the first example is the best approach you can use. It's the more efficient way of representing the tree.

– Patrik Valkovič
Nov 16 '18 at 12:13

|
show 2 more comments

This is the Object sample I'm given:

Input :

{
"People": [
{
"id": "12",
"parentId": "0",
"text": "Man",
"level": "1",
"children": null
},
{
"id": "6",
"parentId": "12",
"text": "Boy",
"level": "2",
"children": null
},
{
"id": "7",
"parentId": "12",
"text": "Other",
"level": "2",
"children": null
},
{
"id": "9",
"parentId": "0",
"text": "Woman",
"level": "1",
"children": null
},
{
"id": "11",
"parentId": "9",
"text": "Girl",
"level": "2",
"children": null
}
] }

I want to transform it to a JSON format like this:

{

    "People": [

        {

            "id": "12",

            "parentId": "0",

            "text": "Man",

            "level": "1",

            "children": [

                {

                    "id": "6",

                    "parentId": "12",

                    "text": "Boy",

                    "level": "2",

                    "children": null

                },

                {

                    "id": "7",

                    "parentId": "12",

                    "text": "Other",

                    "level": "2",

                    "children": null

                }   

            ]

        }

}

Any Ideas/Help would be appreciated

edited Nov 17 '18 at 20:29

asked Nov 16 '18 at 11:46

Omar

1114

2

Can you show us what your attempt looked like?

– SpoonMeiser
Nov 16 '18 at 11:47

Wait a minute..

– Nurbol Alpysbayev
Nov 16 '18 at 11:57

I tried implementing the solution of this question: stackoverflow.com/questions/18017869/… but didn't work after editing :/ @SpoonMeiser

– Omar
Nov 16 '18 at 12:12

I almost finished it... it's taking too long.. :)

– Nurbol Alpysbayev
Nov 16 '18 at 12:12

I think that the first example is the best approach you can use. It's the more efficient way of representing the tree.

– Patrik Valkovič
Nov 16 '18 at 12:13

|
show 2 more comments

This is the Object sample I'm given:

Input :

{
"People": [
{
"id": "12",
"parentId": "0",
"text": "Man",
"level": "1",
"children": null
},
{
"id": "6",
"parentId": "12",
"text": "Boy",
"level": "2",
"children": null
},
{
"id": "7",
"parentId": "12",
"text": "Other",
"level": "2",
"children": null
},
{
"id": "9",
"parentId": "0",
"text": "Woman",
"level": "1",
"children": null
},
{
"id": "11",
"parentId": "9",
"text": "Girl",
"level": "2",
"children": null
}
] }

I want to transform it to a JSON format like this:

{

    "People": [

        {

            "id": "12",

            "parentId": "0",

            "text": "Man",

            "level": "1",

            "children": [

                {

                    "id": "6",

                    "parentId": "12",

                    "text": "Boy",

                    "level": "2",

                    "children": null

                },

                {

                    "id": "7",

                    "parentId": "12",

                    "text": "Other",

                    "level": "2",

                    "children": null

                }   

            ]

        }

}

Any Ideas/Help would be appreciated

edited Nov 17 '18 at 20:29

asked Nov 16 '18 at 11:46

Omar

1114

This is the Object sample I'm given:

Input :

{
"People": [
{
"id": "12",
"parentId": "0",
"text": "Man",
"level": "1",
"children": null
},
{
"id": "6",
"parentId": "12",
"text": "Boy",
"level": "2",
"children": null
},
{
"id": "7",
"parentId": "12",
"text": "Other",
"level": "2",
"children": null
},
{
"id": "9",
"parentId": "0",
"text": "Woman",
"level": "1",
"children": null
},
{
"id": "11",
"parentId": "9",
"text": "Girl",
"level": "2",
"children": null
}
] }

I want to transform it to a JSON format like this:

{

    "People": [

        {

            "id": "12",

            "parentId": "0",

            "text": "Man",

            "level": "1",

            "children": [

                {

                    "id": "6",

                    "parentId": "12",

                    "text": "Boy",

                    "level": "2",

                    "children": null

                },

                {

                    "id": "7",

                    "parentId": "12",

                    "text": "Other",

                    "level": "2",

                    "children": null

                }   

            ]

        }

}

Any Ideas/Help would be appreciated

javascript json object

edited Nov 17 '18 at 20:29

asked Nov 16 '18 at 11:46

Omar

1114

edited Nov 17 '18 at 20:29

asked Nov 16 '18 at 11:46

Omar

1114

edited Nov 17 '18 at 20:29

asked Nov 16 '18 at 11:46

Omar

1114

asked Nov 16 '18 at 11:46

Omar

1114

asked Nov 16 '18 at 11:46

Omar

1114

2

Can you show us what your attempt looked like?

– SpoonMeiser
Nov 16 '18 at 11:47

Wait a minute..

– Nurbol Alpysbayev
Nov 16 '18 at 11:57

I tried implementing the solution of this question: stackoverflow.com/questions/18017869/… but didn't work after editing :/ @SpoonMeiser

– Omar
Nov 16 '18 at 12:12

I almost finished it... it's taking too long.. :)

– Nurbol Alpysbayev
Nov 16 '18 at 12:12

I think that the first example is the best approach you can use. It's the more efficient way of representing the tree.

– Patrik Valkovič
Nov 16 '18 at 12:13

|
show 2 more comments

2

Can you show us what your attempt looked like?

– SpoonMeiser
Nov 16 '18 at 11:47

Wait a minute..

– Nurbol Alpysbayev
Nov 16 '18 at 11:57

I tried implementing the solution of this question: stackoverflow.com/questions/18017869/… but didn't work after editing :/ @SpoonMeiser

– Omar
Nov 16 '18 at 12:12

I almost finished it... it's taking too long.. :)

– Nurbol Alpysbayev
Nov 16 '18 at 12:12

I think that the first example is the best approach you can use. It's the more efficient way of representing the tree.

– Patrik Valkovič
Nov 16 '18 at 12:13

Can you show us what your attempt looked like?

– SpoonMeiser
Nov 16 '18 at 11:47

Wait a minute..

– Nurbol Alpysbayev
Nov 16 '18 at 11:57

I tried implementing the solution of this question: stackoverflow.com/questions/18017869/… but didn't work after editing :/ @SpoonMeiser

– Omar
Nov 16 '18 at 12:12

I almost finished it... it's taking too long.. :)

– Nurbol Alpysbayev
Nov 16 '18 at 12:12

I think that the first example is the best approach you can use. It's the more efficient way of representing the tree.

– Patrik Valkovič
Nov 16 '18 at 12:13

|
show 2 more comments

2 Answers
2

active

oldest

votes

You could use an object for the reference to either child or parent and collect the children and parents to get only the person for the root.

var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Nov 16 '18 at 12:22

Nina Scholz

196k15108179

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

add a comment |

OMG this seemingly simple stuff took so long

const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)

edited Nov 16 '18 at 13:30

answered Nov 16 '18 at 12:21

Nurbol Alpysbayev

4,8341533

you could add your answer as a code snipet

– Dhaval Pankhaniya
Nov 16 '18 at 12:27

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

@Hussler you could at least upvote it if it works...

– Nurbol Alpysbayev
Nov 16 '18 at 13:26

@NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

– Omar
Nov 16 '18 at 13:50

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53337241%2fhow-to-create-a-tree-like-structure-from-a-javascript-object%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

You could use an object for the reference to either child or parent and collect the children and parents to get only the person for the root.

var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Nov 16 '18 at 12:22

Nina Scholz

196k15108179

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

add a comment |

You could use an object for the reference to either child or parent and collect the children and parents to get only the person for the root.

var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Nov 16 '18 at 12:22

Nina Scholz

196k15108179

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

add a comment |

You could use an object for the reference to either child or parent and collect the children and parents to get only the person for the root.

var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Nov 16 '18 at 12:22

Nina Scholz

196k15108179

You could use an object for the reference to either child or parent and collect the children and parents to get only the person for the root.

var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

var data = { John: "James", Samar: "Michel", Albert: "Michel", Michel: "James", James: "Sarah" },

    parents = new Set,

    children = new Set,

    references = {},

    result;



Object

    .entries(data)

    .forEach(([child, parent]) => {

        references[child] = references[child] || ;

        references[parent] = references[parent] || ;

        references[parent].push({ [child]: references[child] });

        parents.add(parent);

        children.add(child);

    });



result = [...parents]

    .filter(p => !children.has(p))

    .map(p => ({ [p]: references[p] }));



console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Nov 16 '18 at 12:22

Nina Scholz

196k15108179

answered Nov 16 '18 at 12:22

Nina Scholz

196k15108179

answered Nov 16 '18 at 12:22

Nina Scholz

196k15108179

answered Nov 16 '18 at 12:22

Nina Scholz

196k15108179

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

add a comment |

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

add a comment |

OMG this seemingly simple stuff took so long

const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)

edited Nov 16 '18 at 13:30

answered Nov 16 '18 at 12:21

Nurbol Alpysbayev

4,8341533

you could add your answer as a code snipet

– Dhaval Pankhaniya
Nov 16 '18 at 12:27

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

@Hussler you could at least upvote it if it works...

– Nurbol Alpysbayev
Nov 16 '18 at 13:26

@NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

– Omar
Nov 16 '18 at 13:50

add a comment |

OMG this seemingly simple stuff took so long

const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)

edited Nov 16 '18 at 13:30

answered Nov 16 '18 at 12:21

Nurbol Alpysbayev

4,8341533

you could add your answer as a code snipet

– Dhaval Pankhaniya
Nov 16 '18 at 12:27

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

@Hussler you could at least upvote it if it works...

– Nurbol Alpysbayev
Nov 16 '18 at 13:26

@NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

– Omar
Nov 16 '18 at 13:50

add a comment |

OMG this seemingly simple stuff took so long

const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)

edited Nov 16 '18 at 13:30

answered Nov 16 '18 at 12:21

Nurbol Alpysbayev

4,8341533

OMG this seemingly simple stuff took so long

const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)

const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)

const users = {

        "John": "James",

        "Samar": "Michel",

        "Albert": "Michel",

        "Michel": "James",

        "James": "Sarah"

    }

    

    const findRoots = () => Object.keys(users).filter(k => !(users[k] in users)).map(k => users[k])

    

    const findSubordinates = (boss) => Object.keys(users).filter(k => users[k] === boss)

    

    const traverseBoss = (boss) => {

        let subs = findSubordinates(boss)

    

        let subsCollection = 

    

        subs.forEach(s => {

            subsCollection.push({

                [s]: traverseBoss(s)

            })

        })

    

        return subsCollection

    }

    

    

    const result = {}

    findRoots().forEach(root => result[root] = traverseBoss(root))

    

    console.log(result)

edited Nov 16 '18 at 13:30

answered Nov 16 '18 at 12:21

Nurbol Alpysbayev

4,8341533

edited Nov 16 '18 at 13:30

answered Nov 16 '18 at 12:21

Nurbol Alpysbayev

4,8341533

answered Nov 16 '18 at 12:21

Nurbol Alpysbayev

4,8341533

answered Nov 16 '18 at 12:21

Nurbol Alpysbayev

4,8341533

you could add your answer as a code snipet

– Dhaval Pankhaniya
Nov 16 '18 at 12:27

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

@Hussler you could at least upvote it if it works...

– Nurbol Alpysbayev
Nov 16 '18 at 13:26

@NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

– Omar
Nov 16 '18 at 13:50

add a comment |

you could add your answer as a code snipet

– Dhaval Pankhaniya
Nov 16 '18 at 12:27

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

@Hussler you could at least upvote it if it works...

– Nurbol Alpysbayev
Nov 16 '18 at 13:26

@NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

– Omar
Nov 16 '18 at 13:50

you could add your answer as a code snipet

– Dhaval Pankhaniya
Nov 16 '18 at 12:27

This works perfectly, Thanks

– Omar
Nov 16 '18 at 12:36

@Hussler you could at least upvote it if it works...

– Nurbol Alpysbayev
Nov 16 '18 at 13:26

@NurbolAlpysbayev I'm sorry but I have a new account and I cannot upvote :/

– Omar
Nov 16 '18 at 13:50

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Ndtyjky