mysql_fetch_array (MYSQL)

Question

I'm a beginner and I have a error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\VertrigoServ\www\index.php on line 35

and the code...

$sresult = mysql_query("SELECT code, location FROM banners");
while ($row_s = mysql_fetch_array($sresult))
{
    $banner[$row_s["location"]]=$row_s["code"];
}

Nanne · Accepted Answer · 2011-03-16 20:09:07Z

1

Try this

$sresult = mysql_query("SELECT code, location FROM banners");
if (!$result) {
    die('Invalid query: ' . mysql_error());
}
while ($row_s = mysql_fetch_array($sresult))
{
    $banner[$row_s["location"]]=$row_s["code"];
}

And check what the error is.

answered Mar 16, 2011 at 20:09

Nanne

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Comments

Mārtiņš Briedis · Accepted Answer · 2011-03-16 20:32:29Z

0

Something is wrong with the query. try:

$result = mysql_query("..");
if(!$result){
  echo "Query error: " . mysql_error();
}

edited Mar 16, 2011 at 20:32

answered Mar 16, 2011 at 20:08

Mārtiņš Briedis

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2 Comments

Martin Tournoij Over a year ago

or die() :-( :-( phpfreaks.com/blog/or-die-must-die -- I never understood why people keep using this ... Or even started using it in the first place...

Mārtiņš Briedis Over a year ago

Agreed on die(), I, personally, have NEVER used it in my own code, but it was the first thing that popped in mind when writing here, strange :)

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