How to get a key in a JavaScript Map by its value?

I have a js Map like this one

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');

what I want is some method to return a key by its value

let jhonKey = people.getKey('jhon'); // jhonKey should be '1'

edited Nov 6 '17 at 11:17

Cerbrus

49.7k1094116

asked Nov 6 '17 at 11:13

Engineer Passion

146111

13

Then why did you store it the wrong way round? :-P

– Bergi
Nov 6 '17 at 16:06

add a comment |

I have a js Map like this one

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');

what I want is some method to return a key by its value

let jhonKey = people.getKey('jhon'); // jhonKey should be '1'

edited Nov 6 '17 at 11:17

Cerbrus

49.7k1094116

asked Nov 6 '17 at 11:13

Engineer Passion

146111

13

Then why did you store it the wrong way round? :-P

– Bergi
Nov 6 '17 at 16:06

add a comment |

I have a js Map like this one

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');

what I want is some method to return a key by its value

let jhonKey = people.getKey('jhon'); // jhonKey should be '1'

edited Nov 6 '17 at 11:17

Cerbrus

49.7k1094116

asked Nov 6 '17 at 11:13

Engineer Passion

146111

I have a js Map like this one

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');

what I want is some method to return a key by its value

let jhonKey = people.getKey('jhon'); // jhonKey should be '1'

javascript dictionary

edited Nov 6 '17 at 11:17

Cerbrus

49.7k1094116

asked Nov 6 '17 at 11:13

Engineer Passion

146111

edited Nov 6 '17 at 11:17

Cerbrus

49.7k1094116

asked Nov 6 '17 at 11:13

Engineer Passion

146111

edited Nov 6 '17 at 11:17

Cerbrus

49.7k1094116

edited Nov 6 '17 at 11:17

Cerbrus

49.7k1094116

edited Nov 6 '17 at 11:17

Cerbrus

49.7k1094116

asked Nov 6 '17 at 11:13

Engineer Passion

146111

asked Nov 6 '17 at 11:13

Engineer Passion

146111

asked Nov 6 '17 at 11:13

Engineer Passion

146111

13

Then why did you store it the wrong way round? :-P

– Bergi
Nov 6 '17 at 16:06

add a comment |

13

Then why did you store it the wrong way round? :-P

– Bergi
Nov 6 '17 at 16:06

Then why did you store it the wrong way round? :-P

– Bergi
Nov 6 '17 at 16:06

add a comment |

4 Answers
4

active

oldest

votes

You could seach for it in an array of entries.

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



let jhonKeys = [...people.entries()]

        .filter(({ 1: v }) => v === 'jhon')

        .map(([k]) => k);



console.log(jhonKeys); // if empty, no key foudn otherwise all found keys.

edited Oct 9 '18 at 6:07

answered Nov 6 '17 at 11:17

Nina Scholz

189k1598172

@Rajesh, actually, we need the key, not the index.

– Nina Scholz
Nov 6 '17 at 11:26

2

Actually we need the key not the index or value.. :)

– Keith
Nov 6 '17 at 11:26

@NinaScholz I am curious, are you really using this kind of one-liners in your job, and whether you are debugging your code in any way?

– Engineer
Nov 9 '17 at 12:58

1

@NinaScholz You don't need the array access and || logic if you destructure the map values directly: [...people.values()]

– philraj
Sep 12 '18 at 20:40

1

@philraj, i changed the answer. anothher solution could be the use of Array.from with a mapping of the value.

– Nina Scholz
Sep 12 '18 at 20:49

|
show 5 more comments

You can use for..of loop to loop directly over the map.entries and get the keys.

function getByValue(map, searchValue) {

  for (let [key, value] of map.entries()) {

    if (value === searchValue)

      return key;

  }

}



let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



console.log(getByValue(people, 'jhon'))

console.log(getByValue(people, 'abdo'))

edited Nov 6 '17 at 13:21

answered Nov 6 '17 at 11:33

Rajesh

16.4k52253

4

One advantage to this solution is possibly speed, as there is no array transform. To really get into the spirit of ES6 though, might be nice if you change the var to a const..

– Keith
Nov 6 '17 at 13:15

@Keith Thanks for pointing it out. Have updated it to let.

– Rajesh
Nov 6 '17 at 13:22

1

Nice one,. just a heads up const works here too in for of. It's doesn't work like a normal for in this regard.

– Keith
Nov 6 '17 at 13:28

1

Just my personal preference to use let over const.

– Rajesh
Nov 6 '17 at 13:38

You don't need the map.entries(), the map itself acts as an iterator. I submitted an edit.

– philraj
Sep 12 '18 at 20:44

add a comment |

There is no direct method for picking out information in this direction, so if all you have is the map you need to loop through the set as suggested by others.

If the map/array/other is large enough that such a loop would be a performance issue and the requirement for a reverse lookup is common within the project, you could implement your own structure using a pair of maps/arrays/other with one as per the current object and the other with the key and value reversed. That way the reverse lookup is as efficient as the normal one. Of course, you have more work to do as you need to implement each method that you need as a pass-through to one or both of the underlying objects so if the map is small and/or the reverse lookup is not needed often the scan-via-loop option is likely to be preferable due to being simpler to maintain and possible simpler for the JiT compiler to optimise.

In any case one thing to be wary of is the possibility that multiple keys could have the same value. If this is possible then when looping the through your map you need to decide if you are fine to return one of the possible keys arbitrarily (probably the first one) or if you want to return an array of keys, and if implementing a reverse index for data that could have duplicate values the same issue also needs to be accounted for.

answered Nov 6 '17 at 15:36

David Spillett

1,050917

add a comment |

Though late and other great answers already exist, still you can give below "..." and "Array.find" a try

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



function getKey(value) {

  return [...people].find(([key, val]) => val == value)[0]

}



console.log('Jasmein - ',getKey('jasmein'))



console.log('Jhon - ',getKey('jhon'))

answered Nov 15 '18 at 6:39

Nitish Narang

2,9601815

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

You could seach for it in an array of entries.

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



let jhonKeys = [...people.entries()]

        .filter(({ 1: v }) => v === 'jhon')

        .map(([k]) => k);



console.log(jhonKeys); // if empty, no key foudn otherwise all found keys.

edited Oct 9 '18 at 6:07

answered Nov 6 '17 at 11:17

Nina Scholz

189k1598172

@Rajesh, actually, we need the key, not the index.

– Nina Scholz
Nov 6 '17 at 11:26

2

Actually we need the key not the index or value.. :)

– Keith
Nov 6 '17 at 11:26

@NinaScholz I am curious, are you really using this kind of one-liners in your job, and whether you are debugging your code in any way?

– Engineer
Nov 9 '17 at 12:58

1

@NinaScholz You don't need the array access and || logic if you destructure the map values directly: [...people.values()]

– philraj
Sep 12 '18 at 20:40

1

@philraj, i changed the answer. anothher solution could be the use of Array.from with a mapping of the value.

– Nina Scholz
Sep 12 '18 at 20:49

|
show 5 more comments

You could seach for it in an array of entries.

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



let jhonKeys = [...people.entries()]

        .filter(({ 1: v }) => v === 'jhon')

        .map(([k]) => k);



console.log(jhonKeys); // if empty, no key foudn otherwise all found keys.

edited Oct 9 '18 at 6:07

answered Nov 6 '17 at 11:17

Nina Scholz

189k1598172

@Rajesh, actually, we need the key, not the index.

– Nina Scholz
Nov 6 '17 at 11:26

2

Actually we need the key not the index or value.. :)

– Keith
Nov 6 '17 at 11:26

@NinaScholz I am curious, are you really using this kind of one-liners in your job, and whether you are debugging your code in any way?

– Engineer
Nov 9 '17 at 12:58

1

@NinaScholz You don't need the array access and || logic if you destructure the map values directly: [...people.values()]

– philraj
Sep 12 '18 at 20:40

1

@philraj, i changed the answer. anothher solution could be the use of Array.from with a mapping of the value.

– Nina Scholz
Sep 12 '18 at 20:49

|
show 5 more comments

You could seach for it in an array of entries.

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



let jhonKeys = [...people.entries()]

        .filter(({ 1: v }) => v === 'jhon')

        .map(([k]) => k);



console.log(jhonKeys); // if empty, no key foudn otherwise all found keys.

edited Oct 9 '18 at 6:07

answered Nov 6 '17 at 11:17

Nina Scholz

189k1598172

You could seach for it in an array of entries.

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



let jhonKeys = [...people.entries()]

        .filter(({ 1: v }) => v === 'jhon')

        .map(([k]) => k);



console.log(jhonKeys); // if empty, no key foudn otherwise all found keys.

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



let jhonKeys = [...people.entries()]

        .filter(({ 1: v }) => v === 'jhon')

        .map(([k]) => k);



console.log(jhonKeys); // if empty, no key foudn otherwise all found keys.

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



let jhonKeys = [...people.entries()]

        .filter(({ 1: v }) => v === 'jhon')

        .map(([k]) => k);



console.log(jhonKeys); // if empty, no key foudn otherwise all found keys.

edited Oct 9 '18 at 6:07

answered Nov 6 '17 at 11:17

Nina Scholz

189k1598172

edited Oct 9 '18 at 6:07

answered Nov 6 '17 at 11:17

Nina Scholz

189k1598172

answered Nov 6 '17 at 11:17

Nina Scholz

189k1598172

answered Nov 6 '17 at 11:17

Nina Scholz

189k1598172

@Rajesh, actually, we need the key, not the index.

– Nina Scholz
Nov 6 '17 at 11:26

2

Actually we need the key not the index or value.. :)

– Keith
Nov 6 '17 at 11:26

@NinaScholz I am curious, are you really using this kind of one-liners in your job, and whether you are debugging your code in any way?

– Engineer
Nov 9 '17 at 12:58

1

@NinaScholz You don't need the array access and || logic if you destructure the map values directly: [...people.values()]

– philraj
Sep 12 '18 at 20:40

1

@philraj, i changed the answer. anothher solution could be the use of Array.from with a mapping of the value.

– Nina Scholz
Sep 12 '18 at 20:49

|
show 5 more comments

@Rajesh, actually, we need the key, not the index.

– Nina Scholz
Nov 6 '17 at 11:26

2

Actually we need the key not the index or value.. :)

– Keith
Nov 6 '17 at 11:26

@NinaScholz I am curious, are you really using this kind of one-liners in your job, and whether you are debugging your code in any way?

– Engineer
Nov 9 '17 at 12:58

1

@NinaScholz You don't need the array access and || logic if you destructure the map values directly: [...people.values()]

– philraj
Sep 12 '18 at 20:40

1

@philraj, i changed the answer. anothher solution could be the use of Array.from with a mapping of the value.

– Nina Scholz
Sep 12 '18 at 20:49

@Rajesh, actually, we need the key, not the index.

– Nina Scholz
Nov 6 '17 at 11:26

Actually we need the key not the index or value.. :)

– Keith
Nov 6 '17 at 11:26

@NinaScholz I am curious, are you really using this kind of one-liners in your job, and whether you are debugging your code in any way?

– Engineer
Nov 9 '17 at 12:58

@NinaScholz You don't need the array access and || logic if you destructure the map values directly: [...people.values()]

– philraj
Sep 12 '18 at 20:40

@philraj, i changed the answer. anothher solution could be the use of Array.from with a mapping of the value.

– Nina Scholz
Sep 12 '18 at 20:49

|
show 5 more comments

You can use for..of loop to loop directly over the map.entries and get the keys.

function getByValue(map, searchValue) {

  for (let [key, value] of map.entries()) {

    if (value === searchValue)

      return key;

  }

}



let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



console.log(getByValue(people, 'jhon'))

console.log(getByValue(people, 'abdo'))

edited Nov 6 '17 at 13:21

answered Nov 6 '17 at 11:33

Rajesh

16.4k52253

4

One advantage to this solution is possibly speed, as there is no array transform. To really get into the spirit of ES6 though, might be nice if you change the var to a const..

– Keith
Nov 6 '17 at 13:15

@Keith Thanks for pointing it out. Have updated it to let.

– Rajesh
Nov 6 '17 at 13:22

1

Nice one,. just a heads up const works here too in for of. It's doesn't work like a normal for in this regard.

– Keith
Nov 6 '17 at 13:28

1

Just my personal preference to use let over const.

– Rajesh
Nov 6 '17 at 13:38

You don't need the map.entries(), the map itself acts as an iterator. I submitted an edit.

– philraj
Sep 12 '18 at 20:44

add a comment |

You can use for..of loop to loop directly over the map.entries and get the keys.

function getByValue(map, searchValue) {

  for (let [key, value] of map.entries()) {

    if (value === searchValue)

      return key;

  }

}



let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



console.log(getByValue(people, 'jhon'))

console.log(getByValue(people, 'abdo'))

edited Nov 6 '17 at 13:21

answered Nov 6 '17 at 11:33

Rajesh

16.4k52253

4

One advantage to this solution is possibly speed, as there is no array transform. To really get into the spirit of ES6 though, might be nice if you change the var to a const..

– Keith
Nov 6 '17 at 13:15

@Keith Thanks for pointing it out. Have updated it to let.

– Rajesh
Nov 6 '17 at 13:22

1

Nice one,. just a heads up const works here too in for of. It's doesn't work like a normal for in this regard.

– Keith
Nov 6 '17 at 13:28

1

Just my personal preference to use let over const.

– Rajesh
Nov 6 '17 at 13:38

You don't need the map.entries(), the map itself acts as an iterator. I submitted an edit.

– philraj
Sep 12 '18 at 20:44

add a comment |

You can use for..of loop to loop directly over the map.entries and get the keys.

function getByValue(map, searchValue) {

  for (let [key, value] of map.entries()) {

    if (value === searchValue)

      return key;

  }

}



let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



console.log(getByValue(people, 'jhon'))

console.log(getByValue(people, 'abdo'))

edited Nov 6 '17 at 13:21

answered Nov 6 '17 at 11:33

Rajesh

16.4k52253

You can use for..of loop to loop directly over the map.entries and get the keys.

function getByValue(map, searchValue) {

  for (let [key, value] of map.entries()) {

    if (value === searchValue)

      return key;

  }

}



let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



console.log(getByValue(people, 'jhon'))

console.log(getByValue(people, 'abdo'))

function getByValue(map, searchValue) {

  for (let [key, value] of map.entries()) {

    if (value === searchValue)

      return key;

  }

}



let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



console.log(getByValue(people, 'jhon'))

console.log(getByValue(people, 'abdo'))

function getByValue(map, searchValue) {

  for (let [key, value] of map.entries()) {

    if (value === searchValue)

      return key;

  }

}



let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



console.log(getByValue(people, 'jhon'))

console.log(getByValue(people, 'abdo'))

edited Nov 6 '17 at 13:21

answered Nov 6 '17 at 11:33

Rajesh

16.4k52253

edited Nov 6 '17 at 13:21

answered Nov 6 '17 at 11:33

Rajesh

16.4k52253

answered Nov 6 '17 at 11:33

Rajesh

16.4k52253

answered Nov 6 '17 at 11:33

Rajesh

16.4k52253

4

One advantage to this solution is possibly speed, as there is no array transform. To really get into the spirit of ES6 though, might be nice if you change the var to a const..

– Keith
Nov 6 '17 at 13:15

@Keith Thanks for pointing it out. Have updated it to let.

– Rajesh
Nov 6 '17 at 13:22

1

Nice one,. just a heads up const works here too in for of. It's doesn't work like a normal for in this regard.

– Keith
Nov 6 '17 at 13:28

1

Just my personal preference to use let over const.

– Rajesh
Nov 6 '17 at 13:38

You don't need the map.entries(), the map itself acts as an iterator. I submitted an edit.

– philraj
Sep 12 '18 at 20:44

add a comment |

4

One advantage to this solution is possibly speed, as there is no array transform. To really get into the spirit of ES6 though, might be nice if you change the var to a const..

– Keith
Nov 6 '17 at 13:15

@Keith Thanks for pointing it out. Have updated it to let.

– Rajesh
Nov 6 '17 at 13:22

1

Nice one,. just a heads up const works here too in for of. It's doesn't work like a normal for in this regard.

– Keith
Nov 6 '17 at 13:28

1

Just my personal preference to use let over const.

– Rajesh
Nov 6 '17 at 13:38

You don't need the map.entries(), the map itself acts as an iterator. I submitted an edit.

– philraj
Sep 12 '18 at 20:44

One advantage to this solution is possibly speed, as there is no array transform. To really get into the spirit of ES6 though, might be nice if you change the var to a const..

– Keith
Nov 6 '17 at 13:15

@Keith Thanks for pointing it out. Have updated it to let.

– Rajesh
Nov 6 '17 at 13:22

Nice one,. just a heads up const works here too in for of. It's doesn't work like a normal for in this regard.

– Keith
Nov 6 '17 at 13:28

Just my personal preference to use let over const.

– Rajesh
Nov 6 '17 at 13:38

You don't need the map.entries(), the map itself acts as an iterator. I submitted an edit.

– philraj
Sep 12 '18 at 20:44

add a comment |

There is no direct method for picking out information in this direction, so if all you have is the map you need to loop through the set as suggested by others.

answered Nov 6 '17 at 15:36

David Spillett

1,050917

add a comment |

There is no direct method for picking out information in this direction, so if all you have is the map you need to loop through the set as suggested by others.

answered Nov 6 '17 at 15:36

David Spillett

1,050917

add a comment |

There is no direct method for picking out information in this direction, so if all you have is the map you need to loop through the set as suggested by others.

answered Nov 6 '17 at 15:36

David Spillett

1,050917

There is no direct method for picking out information in this direction, so if all you have is the map you need to loop through the set as suggested by others.

answered Nov 6 '17 at 15:36

David Spillett

1,050917

answered Nov 6 '17 at 15:36

David Spillett

1,050917

answered Nov 6 '17 at 15:36

David Spillett

1,050917

answered Nov 6 '17 at 15:36

David Spillett

1,050917

add a comment |

Though late and other great answers already exist, still you can give below "..." and "Array.find" a try

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



function getKey(value) {

  return [...people].find(([key, val]) => val == value)[0]

}



console.log('Jasmein - ',getKey('jasmein'))



console.log('Jhon - ',getKey('jhon'))

answered Nov 15 '18 at 6:39

Nitish Narang

2,9601815

add a comment |

Though late and other great answers already exist, still you can give below "..." and "Array.find" a try

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



function getKey(value) {

  return [...people].find(([key, val]) => val == value)[0]

}



console.log('Jasmein - ',getKey('jasmein'))



console.log('Jhon - ',getKey('jhon'))

answered Nov 15 '18 at 6:39

Nitish Narang

2,9601815

add a comment |

Though late and other great answers already exist, still you can give below "..." and "Array.find" a try

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



function getKey(value) {

  return [...people].find(([key, val]) => val == value)[0]

}



console.log('Jasmein - ',getKey('jasmein'))



console.log('Jhon - ',getKey('jhon'))

answered Nov 15 '18 at 6:39

Nitish Narang

2,9601815

Though late and other great answers already exist, still you can give below "..." and "Array.find" a try

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



function getKey(value) {

  return [...people].find(([key, val]) => val == value)[0]

}



console.log('Jasmein - ',getKey('jasmein'))



console.log('Jhon - ',getKey('jhon'))

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



function getKey(value) {

  return [...people].find(([key, val]) => val == value)[0]

}



console.log('Jasmein - ',getKey('jasmein'))



console.log('Jhon - ',getKey('jhon'))

let people = new Map();

people.set('1', 'jhon');

people.set('2', 'jasmein');

people.set('3', 'abdo');



function getKey(value) {

  return [...people].find(([key, val]) => val == value)[0]

}



console.log('Jasmein - ',getKey('jasmein'))



console.log('Jhon - ',getKey('jhon'))

answered Nov 15 '18 at 6:39

Nitish Narang

2,9601815

answered Nov 15 '18 at 6:39

Nitish Narang

2,9601815

answered Nov 15 '18 at 6:39

Nitish Narang

2,9601815

answered Nov 15 '18 at 6:39

Nitish Narang

2,9601815

add a comment |

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