Issue with Like Query in MYSQL PHP

Question

Following problem:

If I want to search with 'Last Name' it works but if it is a variable in this form $results['22']['BuyerName2']

Here is what I got so far:

$rr=$results[22]['BuyerName2'];
echo $rr; //echos Last Name

$stmt = $db->prepare("Update base_1 SET UpdateStatus=2 WHERE BuyerName LIKE ?");
$stmt->bindValue(1, "%$rr%", PDO::PARAM_STR);
$stmt->execute();

If I put instead $rr the Name directly in the bind value part it works. But not with $rr.

And when you say it doesn't work, what happens exactly? Do you get an error? If so, what's the error? Or does it just not update anything? — jcaron
– jcaron, Commented Aug 16, 2014 at 0:11

Barmar · Accepted Answer · 2014-08-16 00:11:12Z

3

Maybe there are extra spaces in $rr. Try:

$rr = trim($results[22]['BuyerName2']);

answered Aug 16, 2014 at 0:11

Barmar

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halfer · Accepted Answer · 2014-08-16 00:34:36Z

1

In the line:

$stmt->bindValue(1, "%$rr%", PDO::PARAM_STR);

I am unsure whether the $ becomes escaped or not in the string you binded. As far as I know, only the underscore and the percent sign stay unescaped. I would suggest to try:

$rr = "%".$rr."%";

and edit the line to:

$stmt->bindValue(1, $rr, PDO::PARAM_STR);

edited Aug 16, 2014 at 0:34

halfer

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answered Aug 16, 2014 at 0:15

Drudge

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