How to correct Php Query for LIKE data?

Question

I am getting a error in my PHP query. I am attempting to do the below query. However, when I put the the code in PHP. The data is not pulling the "like" data. The search I want:

"SELECT * FROM `TABLE` WHERE `ID` like \'NAME%\'";

My php code:

$TABLE= $_POST['cptCode1'];
$data = mysql_query("SELECT * FROM TABLE WHERE ID like'".$searchName."'") or die(mysql_error());

What am I doing wrong?

**mysql_*** functions are depreciated. Use **PDO** Or MySQLi : http://www.php.net/manual/en/book.pdo.php http://in2.php.net/manual/en/book.mysqli.php — Subin, Apr 05 '14 at 05:33

score 1 · Accepted Answer · answered Apr 05 '14 at 05:34

1

Try with this.

$data = mysql_query("SELECT * FROM TABLE WHERE ID like '".$searchName."%'") or die(mysql_error());

answered Apr 05 '14 at 05:34

Jenz

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score 0 · Answer 2 · answered Apr 05 '14 at 05:33

0

I guess a space and % missing:

like '".$searchName."%'"

answered Apr 05 '14 at 05:33

Sahil Mittal

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score 0 · Answer 3 · edited May 23 '17 at 12:11

0

    $data = mysql_query("SELECT * FROM TABLE WHERE ID LIKE '$searchName%'");

Or Check here how-to-use-php-string-in-mysql-like-query

edited May 23 '17 at 12:11

Community

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answered Apr 05 '14 at 05:36

Himanshu

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score 0 · Answer 4 · answered Apr 05 '14 at 05:38

0

Follow this, it works

Replace

$data = mysql_query("SELECT * FROM TABLE WHERE ID like'".$searchName."'") or die(mysql_error());

with

$data = mysql_query("SELECT * FROM TABLE WHERE ID like '".$searchName."%'") or die(mysql_error());

answered Apr 05 '14 at 05:38

Birlla

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How to correct Php Query for LIKE data?

4 Answers4