使用PHP我发出一个MySQL数据库的简单查询-当在phpMyAdmin中发出相同查询时，我得到一个错误，我得到信息行

Question

I've run this on my local windows machine and on an ubuntu server with the same results.

Query to run in PHP:

$job_sql="SELECT * FROM job WHERE job_title = 'SIP Opportunities where sipsubmitted EQ 1 and still in BAB PROCESS'";

$job_ret = mysql_query($job_sql);

$job_row = mysql_fetch_array($job_ret,MYSQL_ASSOC);

Error from PHP Script:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\\wamp\\www\\tracker\\inc\\common.php on line 489

Other similar queries work fine in same script (different job_titles).

phpMyAdmin:

SELECT * 
FROM job
WHERE job_title = 'SIP Opportunities where sipsubmitted EQ 1 and still in BAB PROCESS'
LIMIT 0 , 30

Showing rows 0 - 0 (1 total, Query took 0.0004 sec) - successfully found the row!!!

Answer 1

Looks like you may have misplace a single quote in your example code. With the code as it is now, you're looking for the following job title:

'SIP Opportunities where sipsubmitted EQ 1 and still in BAB PROCESS'

Did you mean for that entire string to be the job_title ?