mysql query when you have coma seperated string both sides
-3
hello I have this problem where I have a list of ids like
ID= 1,2,3,4,5
and the field in the database has values like
3,4,56,34,1,2,3 // field name can be users
and now I want to select all the tables in the database which has any of the values in ID
I tried this with
FIND_IN_SET
but it's not working properly
FIND_IN_SET (users, $ID)
can anyone help me with this, please??
php mysql mysqli
edited Nov 12 at 7:58
Gufran Hasan
3,43441326
asked Nov 12 at 7:55
Amani
279
|
show 3 more comments
-3
hello I have this problem where I have a list of ids like
ID= 1,2,3,4,5
and the field in the database has values like
3,4,56,34,1,2,3 // field name can be users
and now I want to select all the tables in the database which has any of the values in ID
I tried this with
FIND_IN_SET
but it's not working properly
FIND_IN_SET (users, $ID)
can anyone help me with this, please??
php mysql mysqli
edited Nov 12 at 7:58
Gufran Hasan
3,43441326
asked Nov 12 at 7:55
Amani
279
Redesign you DB
– Jens
Nov 12 at 7:57
Please your query here which you tried?
– Gufran Hasan
Nov 12 at 8:00
Do you want to select Tables or Rows??
– Mᴏʀᴀᴅɴᴇᴊᴀᴅ
Nov 12 at 8:00
You might want to use the IN operator.
– Jean-Marc Zimmer
Nov 12 at 8:01
1
Possible duplicate of MySql: FIND_IN_SET does not work properly
– Gufran Hasan
Nov 12 at 8:02
|
show 3 more comments
-3
-3
-3
hello I have this problem where I have a list of ids like
ID= 1,2,3,4,5
and the field in the database has values like
3,4,56,34,1,2,3 // field name can be users
and now I want to select all the tables in the database which has any of the values in ID
I tried this with
FIND_IN_SET
but it's not working properly
FIND_IN_SET (users, $ID)
can anyone help me with this, please??
php mysql mysqli
edited Nov 12 at 7:58
Gufran Hasan
3,43441326
asked Nov 12 at 7:55
Amani
279
hello I have this problem where I have a list of ids like
ID= 1,2,3,4,5
and the field in the database has values like
3,4,56,34,1,2,3 // field name can be users
and now I want to select all the tables in the database which has any of the values in ID
I tried this with
FIND_IN_SET
but it's not working properly
FIND_IN_SET (users, $ID)
can anyone help me with this, please??
php mysql mysqli
php mysql mysqli
edited Nov 12 at 7:58
Gufran Hasan
3,43441326
asked Nov 12 at 7:55
Amani
279
edited Nov 12 at 7:58
Gufran Hasan
3,43441326
asked Nov 12 at 7:55
Amani
279
edited Nov 12 at 7:58
Gufran Hasan
3,43441326
edited Nov 12 at 7:58
Gufran Hasan
3,43441326
edited Nov 12 at 7:58
Gufran Hasan
3,43441326
3,43441326
asked Nov 12 at 7:55
Amani
279
asked Nov 12 at 7:55
Amani
279
asked Nov 12 at 7:55
Amani
279
279
Redesign you DB
– Jens
Nov 12 at 7:57
Please your query here which you tried?
– Gufran Hasan
Nov 12 at 8:00
Do you want to select Tables or Rows??
– Mᴏʀᴀᴅɴᴇᴊᴀᴅ
Nov 12 at 8:00
You might want to use the IN operator.
– Jean-Marc Zimmer
Nov 12 at 8:01
1
Possible duplicate of MySql: FIND_IN_SET does not work properly
– Gufran Hasan
Nov 12 at 8:02
|
show 3 more comments
Redesign you DB
– Jens
Nov 12 at 7:57
Please your query here which you tried?
– Gufran Hasan
Nov 12 at 8:00
Do you want to select Tables or Rows??
– Mᴏʀᴀᴅɴᴇᴊᴀᴅ
Nov 12 at 8:00
You might want to use the IN operator.
– Jean-Marc Zimmer
Nov 12 at 8:01
1
Possible duplicate of MySql: FIND_IN_SET does not work properly
– Gufran Hasan
Nov 12 at 8:02
Redesign you DB
– Jens
Nov 12 at 7:57
Redesign you DB
– Jens
Nov 12 at 7:57
Please your query here which you tried?
– Gufran Hasan
Nov 12 at 8:00
Please your query here which you tried?
– Gufran Hasan
Nov 12 at 8:00
Do you want to select Tables or Rows??
– Mᴏʀᴀᴅɴᴇᴊᴀᴅ
Nov 12 at 8:00
Do you want to select Tables or Rows??
– Mᴏʀᴀᴅɴᴇᴊᴀᴅ
Nov 12 at 8:00
You might want to use the IN operator.
– Jean-Marc Zimmer
Nov 12 at 8:01
You might want to use the IN operator.
– Jean-Marc Zimmer
Nov 12 at 8:01
1
1
Possible duplicate of MySql: FIND_IN_SET does not work properly
– Gufran Hasan
Nov 12 at 8:02
Possible duplicate of MySql: FIND_IN_SET does not work properly
– Gufran Hasan
Nov 12 at 8:02
|
show 3 more comments
2 Answers
2
active
oldest
votes
1
like this?
SELECT * FROM `tableName` WHERE CONCAT(",", `users`, ",") REGEXP ",(1|2|3|4|5),"
answered Nov 12 at 8:10
suresh bambhaniya
861113
This is actually working, but i want to get results only if all the values of ID matches with users . i tried using & insted of | but dosent works
– Amani
Nov 12 at 9:35
you can store users in ascending order(1,2,3,4,34,56) and then use SELECT * FROMtableNameWHEREusersREGEXP "1,2,3,4,5"
– suresh bambhaniya
Nov 12 at 9:56
add a comment |
1
This should work :
$ID = "(1, 2, 3, 4, 5)";
$q = "SELECT * FROM `users` WHERE `ID` IN " . $ID;
//Execute $q here
answered Nov 12 at 8:05
Jean-Marc Zimmer
37414
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');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53257888%2fmysql-query-when-you-have-coma-seperated-string-both-sides%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
like this?
SELECT * FROM `tableName` WHERE CONCAT(",", `users`, ",") REGEXP ",(1|2|3|4|5),"
answered Nov 12 at 8:10
suresh bambhaniya
861113
This is actually working, but i want to get results only if all the values of ID matches with users . i tried using & insted of | but dosent works
– Amani
Nov 12 at 9:35
you can store users in ascending order(1,2,3,4,34,56) and then use SELECT * FROMtableNameWHEREusersREGEXP "1,2,3,4,5"
– suresh bambhaniya
Nov 12 at 9:56
add a comment |
1
like this?
SELECT * FROM `tableName` WHERE CONCAT(",", `users`, ",") REGEXP ",(1|2|3|4|5),"
answered Nov 12 at 8:10
suresh bambhaniya
861113
This is actually working, but i want to get results only if all the values of ID matches with users . i tried using & insted of | but dosent works
– Amani
Nov 12 at 9:35
you can store users in ascending order(1,2,3,4,34,56) and then use SELECT * FROMtableNameWHEREusersREGEXP "1,2,3,4,5"
– suresh bambhaniya
Nov 12 at 9:56
add a comment |
1
1
1
like this?
SELECT * FROM `tableName` WHERE CONCAT(",", `users`, ",") REGEXP ",(1|2|3|4|5),"
answered Nov 12 at 8:10
suresh bambhaniya
861113
like this?
SELECT * FROM `tableName` WHERE CONCAT(",", `users`, ",") REGEXP ",(1|2|3|4|5),"
answered Nov 12 at 8:10
suresh bambhaniya
861113
answered Nov 12 at 8:10
suresh bambhaniya
861113
answered Nov 12 at 8:10
suresh bambhaniya
861113
answered Nov 12 at 8:10
suresh bambhaniya
861113
861113
This is actually working, but i want to get results only if all the values of ID matches with users . i tried using & insted of | but dosent works
– Amani
Nov 12 at 9:35
you can store users in ascending order(1,2,3,4,34,56) and then use SELECT * FROMtableNameWHEREusersREGEXP "1,2,3,4,5"
– suresh bambhaniya
Nov 12 at 9:56
add a comment |
This is actually working, but i want to get results only if all the values of ID matches with users . i tried using & insted of | but dosent works
– Amani
Nov 12 at 9:35
you can store users in ascending order(1,2,3,4,34,56) and then use SELECT * FROMtableNameWHEREusersREGEXP "1,2,3,4,5"
– suresh bambhaniya
Nov 12 at 9:56
This is actually working, but i want to get results only if all the values of ID matches with users . i tried using & insted of | but dosent works
– Amani
Nov 12 at 9:35
This is actually working, but i want to get results only if all the values of ID matches with users . i tried using & insted of | but dosent works
– Amani
Nov 12 at 9:35
you can store users in ascending order(1,2,3,4,34,56) and then use SELECT * FROM
tableName WHERE users REGEXP "1,2,3,4,5"– suresh bambhaniya
Nov 12 at 9:56
you can store users in ascending order(1,2,3,4,34,56) and then use SELECT * FROM
tableName WHERE users REGEXP "1,2,3,4,5"– suresh bambhaniya
Nov 12 at 9:56
add a comment |
1
This should work :
$ID = "(1, 2, 3, 4, 5)";
$q = "SELECT * FROM `users` WHERE `ID` IN " . $ID;
//Execute $q here
answered Nov 12 at 8:05
Jean-Marc Zimmer
37414
add a comment |
1
This should work :
$ID = "(1, 2, 3, 4, 5)";
$q = "SELECT * FROM `users` WHERE `ID` IN " . $ID;
//Execute $q here
answered Nov 12 at 8:05
Jean-Marc Zimmer
37414
add a comment |
1
1
1
This should work :
$ID = "(1, 2, 3, 4, 5)";
$q = "SELECT * FROM `users` WHERE `ID` IN " . $ID;
//Execute $q here
answered Nov 12 at 8:05
Jean-Marc Zimmer
37414
This should work :
$ID = "(1, 2, 3, 4, 5)";
$q = "SELECT * FROM `users` WHERE `ID` IN " . $ID;
//Execute $q here
answered Nov 12 at 8:05
Jean-Marc Zimmer
37414
answered Nov 12 at 8:05
Jean-Marc Zimmer
37414
answered Nov 12 at 8:05
Jean-Marc Zimmer
37414
answered Nov 12 at 8:05
Jean-Marc Zimmer
37414
37414
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53257888%2fmysql-query-when-you-have-coma-seperated-string-both-sides%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Redesign you DB
– Jens
Nov 12 at 7:57
Please your query here which you tried?
– Gufran Hasan
Nov 12 at 8:00
Do you want to select Tables or Rows??
– Mᴏʀᴀᴅɴᴇᴊᴀᴅ
Nov 12 at 8:00
You might want to use the IN operator.
– Jean-Marc Zimmer
Nov 12 at 8:01
1
Possible duplicate of MySql: FIND_IN_SET does not work properly
– Gufran Hasan
Nov 12 at 8:02