Use mysql_error() can give a hint of what is going wrong.
$result = mysql_query($query) or die( mysql_error());
HTH
Jayme.
-----Mensagem Original-----
De: Oson, Chris M. <[EMAIL PROTECTED]>
Para: Jim Ray <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Enviada em: sábado, 10 de março de 2001 16:57
Assunto: RE: [PHP-DB] PHP and MYSQL query
> For starters, your statement is wrong...
>
> $result = mysql_query($strSQL, $db);
>
> where $db is your connection string to the database you already have
opened.
>
> If you have problems with your sql statement, try something like this and
> see
> what your results are:
>
> $strSQL = "SELECT * FROM $TA WHERE $SortField = '$search%' ORDER BY
> $SortField";
> print $strSQL;
>
> $result = mysql_query($strSQL);
>
> Once you have the results of the query printed out, copy and paste it into
> the
> command line in mySQL. You'll know if you're doing something wrong or
not.
>
> -----Original Message-----
> From: Jim Ray [mailto:[EMAIL PROTECTED]]
> Sent: Friday, March 09, 2001 8:03 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] PHP and MYSQL query
>
>
> I have a simpe query that I can not seem to get to work.
>
> Here is the HTML side:
> <td width="27%" align="center"><form
> action="luquery.php?DB=csjoa&TA=associates&SortField=company"
> method="post"><input type="text" name="search"><br><input
> type="submit"></form></td>
>
>
>
> Here is the PHP side:
>
> The fields are being past, but I get 0 in the results?
>
> $result=mysql_query("select * from $TA where $SortField='$search%'
order
> by $SortField");
>
> Am I missing something here?
>
> Thanks for the help.
>
> Jim
>
>
>
>
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