How to add a PHP variable inside REGEXP in mySQL

Question

My code is like this

SELECT count(product_id) as totalCount FROM ref_products WHERE misc REGEXP '(.*\"level_id\":\'".$course_level_id."'\.*)'

And what happens then.?Is there any error.?

웃웃웃웃웃
– 웃웃웃웃웃

2013-12-11 12:35:42 +00:00
Commented Dec 11, 2013 at 12:35 — 웃웃웃웃웃
– 웃웃웃웃웃, Commented Dec 11, 2013 at 12:35

Ronald Swets · Accepted Answer · 2013-12-11 13:13:14Z

1

Assuming you use a double quoted string ("):

$sql = "SELECT count(product_id) as totalCount FROM ref_products WHERE misc REGEXP '(.*\"level_id\":\'".$course_level_id."'.*)'";

Assuming you use a single quoted string ('):

$sql = 'SELECT count(product_id) as totalCount FROM ref_products WHERE misc REGEXP \'(.*\"level_id\":'.$course_level_id.'.*)\'';

edited Dec 11, 2013 at 13:13

answered Dec 11, 2013 at 12:46

Ronald Swets

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4 Comments

UI Dev Over a year ago

This is working perfect in database. so i will use a dynamic variable in php as $course_level_id. Instead of "20" i have to give the variable. SELECT * FROM ref_products WHERE misc REGEXP '(.*\"level_id\":\20\.*)'

UI Dev Over a year ago

Its not a double quote string

Ronald Swets Over a year ago

Added the single quote sql as well

Ronald Swets Over a year ago

Might be needed to escape the double quotes in there. Updated answer

UI Dev · Accepted Answer · 2013-12-11 13:31:44Z

1

Thanks to all guys for their time & effort. Here is the correct answer i found. Please check

SELECT count(product_id) as totalCount FROM ref_products WHERE misc REGEXP '(.*\"level_id\":$course_level_id.*)'

answered Dec 11, 2013 at 13:31

UI Dev

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Comments

Bohdan Yurov · Accepted Answer · 2013-12-11 12:35:38Z

0

SELECT count(product_id) as totalCount FROM ref_products WHERE misc REGEXP '(.*"level_id":"'.$course_level_id.'".*)'

answered Dec 11, 2013 at 12:35

Bohdan Yurov

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5 Comments

Amal Over a year ago

@RIADev: not working is not very descriptive. We don't know what you're thinking or the output you're getting -- we only know what you're telling us. So, please be descriptive and explain the issue.

UI Dev Over a year ago

Sorry..This is working perfect in database. so i will use a dynamic variable in php as $course_level_id. Instead of "20" i have to give the variable. The query is like this SELECT * FROM ref_products WHERE misc REGEXP '(.*\"level_id\":\20\.*)'

UI Dev Over a year ago

@Amal Any upadtes from you ??

Bohdan Yurov Over a year ago

You are using quotes wrongly. Correct format: '(.*"level_id":'. $course_level_id.'.*)'

UI Dev Over a year ago

@nick In php it showing error near '(.*"level_id":' --- Undefined level_id. So we need to escape the double quote. Refer my answer its working perfect. If you have any quotes then refer ronald answer

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