trouble with mysql_fetch_array

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August 18th, 2003, 10:25 PM

DorkRawk

Join Date: Apr 2007

Posts: 173

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trouble with mysql_fetch_array

I'm either haveing trouble w/ mysql_fetch_array or mysql_query because I keep getting this error...

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/captains/www/www/bandonly/validate.php on line 6

Here is my code...

php Code:

Original - php Code
$result = mysql_query("SELECT * FROM login WHERE name=$name",$db); $myrow = mysql_fetch_array($result); echo $myrow["pw"];

 
$result = mysql_query("SELECT * FROM login WHERE name=$name",$db);
$myrow = mysql_fetch_array($result);
echo $myrow["pw"];

August 18th, 2003, 10:38 PM

sliver

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Location: WI, USA

Posts: 932

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RE: trouble with mysql_fetch_array

That means that 1 of 2 things happened:
1. An error occured in your query.
2. No results were returned from your query.

For error tracking add "or die(mysql_error())" after your query. (mysql_query("select...") or die(mysql_error());)

Throw an if statement in to see if any results were returned($row=@mysql_fetch_array($result); if($row) {...)

August 18th, 2003, 11:13 PM

DorkRawk

Join Date: Apr 2007

Posts: 173

Time spent in forums: < 1 sec
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RE: trouble with mysql_fetch_array

Ok, I here is my code now...

php Code:

Original - php Code
$result = mysql_query("SELECT * FROM login WHERE name=$name",$db) or die(mysql_error()); $array=@mysql_fetch_array($result); if($array) {echo "row exists";} echo $array["pw"];

 
$result = mysql_query("SELECT * FROM login WHERE name=$name",$db)
    or die(mysql_error());
$array=@mysql_fetch_array($result); 
if($array) {echo "row exists";}
echo $array["pw"];

and now I get this error...
Unknown column 'Dave' in 'where clause'

"Dave" is the value of $name.
and there is a row where the "name" field has a value of "Dave".

August 18th, 2003, 11:23 PM

sliver

Moderator

Join Date: Apr 2007

Location: WI, USA

Posts: 932

Time spent in forums: 1 Day 11 h
Reputation Power: 3

RE: trouble with mysql_fetch_array

In the query change name=$name to name='$name'

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