query not returning

[Robert Morgan]

Can anyone tell me whats wrong with this code? The variables are being past
from other page and the $connection is ok as is $result. this query should
return 6 rows.
the error message I get is "Warning: mysql_fetch_row (): supplied argument is
not a valid MySQL result resource in /var/www/html"

The user selects a field from a drop down then enters the parameter to
search the field with in a input type text field then clicks a button to run
the query. I'm new at this so any help appreciated.

Bob

<?php
$SField=($_POST["SField"]);
$Parameter = ($_POST["Parameter"]);
$connection = mysql_connect(" localhost", "connection ", "pword");
print_r($Parame ter);
print_r(connect ion);
mysql_select_db ("HelpData", $connection);
$FLDquery= "select * from tblWorkstation where '".$SField." ' =
".$Parameter."" ;
$result = mysql_query($FL Dquery, $connection) ;


while ($row = mysql_fetch_row ($result))
{

for($i=0;$i<mys ql_numFields($r esult);$i++)
echo $row[$i]." ";
echo "\n";
}



?>

[Jochen Daum]

Re: query not returning

Hi Robert!

Robert Morgan wrote:
[color=blue]
> Can anyone tell me whats wrong with this code? The variables are being
> past from other page and the $connection is ok as is $result.[/color]

Use print_r($result );

I doubt it is ok -- should return "resource(. .."
[color=blue]
> this query
> should return 6 rows.
> the error message I get is "Warning: mysql_fetch_row (): supplied argument
> is not a valid MySQL result resource in /var/www/html"[/color]

Alternatively, paste the query into phpmyadmin.[color=blue]
>
> The user selects a field from a drop down then enters the parameter to
> search the field with in a input type text field then clicks a button to
> run the query. I'm new at this so any help appreciated.
>
> Bob
>
> <?php
> $SField=($_POST["SField"]);
> $Parameter = ($_POST["Parameter"]);
> $connection = mysql_connect(" localhost", "connection ", "pword");
> print_r($Parame ter);
> print_r(connect ion);[/color]

where is the $ in $connection. Also, use error_reporting (E_ALL);

HTH, Jochen
--
PHPDBEditTk: Toolkit for nice and easy
administration interfaces. http://phpdbedittk.sourceforge.net

[Rainer Herbst]

Re: query not returning

Robert Morgan schrieb:[color=blue]
> Can anyone tell me whats wrong with this code? The variables are being past
> from other page and the $connection is ok as is $result. this query should
> return 6 rows.
> the error message I get is "Warning: mysql_fetch_row (): supplied argument is
> not a valid MySQL result resource in /var/www/html"
>
> The user selects a field from a drop down then enters the parameter to
> search the field with in a input type text field then clicks a button to run
> the query. I'm new at this so any help appreciated.
>
> Bob
>
> <?php
> $SField=($_POST["SField"]);
> $Parameter = ($_POST["Parameter"]);
> $connection = mysql_connect(" localhost", "connection ", "pword");
> print_r($Parame ter);
> print_r(connect ion);
> mysql_select_db ("HelpData", $connection);
> $FLDquery= "select * from tblWorkstation where '".$SField." ' =
> ".$Parameter."" ;
> $result = mysql_query($FL Dquery, $connection) ;
>
>
> while ($row = mysql_fetch_row ($result))
> {
>
> for($i=0;$i<mys ql_numFields($r esult);$i++)
> echo $row[$i]." ";
> echo "\n";
> }
>
>
>
> ?>
>
>[/color]

Your query will result in an SQL statement like:
select * from tblWorkstation where 'myField' = xyz;

but the statement should look like
select * from tblWorkstation where myField = 'xyz';

The quota converts the fieldname into a string, which will normally not
be equal to the parameter!

HTH!
Rainer

--
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ZEIK User #319157
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