Question

TABLE bcompany

companyID | cName | ...

I have the input field, where user searches for the records including the “input” characters:

<input type="text" class="bigblack" name="srch" />

PHP:

$req="%".mysql_real_escape_string($_POST['srch'])."%";

$query = mysql_query("SELECT companyID, cName FROM bcompany WHERE
companyID OR cName LIKE $req ORDER BY companyID LIMIT 10");

OR

$query = mysql_query("SELECT companyID, cName FROM bcompany WHERE
companyID,cName LIKE $req ORDER BY companyID LIMIT 10");

Both queries return an error:

mysql_fetch_array() expects parameter 1 to be resource….

There is probably something wrong with the MYSQL SELECT.

Can you please help me solve this out?

Thanks in advance 🙂

Rest of the code:

while($res = mysql_fetch_array($query)) {


    echo $res["companyID"];
    echo $res["cName"]."<br>";



}

Answer 1

You query should be:

$query = mysql_query("SELECT companyID, cName FROM bcompany WHERE
companyID LIKE '$req' OR cName LIKE '$req' ORDER BY companyID LIMIT 10");

Here are the docs for MySQL LIKE:

http://dev.mysql.com/doc/refman/5.0/en/string-comparison-functions.html

Alternatively you could use the MySQL REGEXP in place of LIKE in your queries.

It looks like mysql_fetch_array is griping when you call it because the query is incorrect and not producing a result which mysql_fetch_array can use.