Assume query looks OK when you echo it? You're not seeing a literal "$id",
are you?
Does an id exist for $id? Is it numeric or char?
Check mysql_affected_rows() - what does it return?
Miles Thompson
At 02:07 PM 1/31/2002 +0000, Alex Francis wrote:
>I have an empty field in a mysql database table which needs to have text
>inserted when my client answers an email. The code is as follows
>
>
>$query = 'UPDATE submissions SET answers = "$array[answer]" where id = "$id"
>';
> // $query = $query . " where id = $id ";
>
> // echo ("The query is: <BR>$query<P>\n");
>
>if (mysql_db_query ($dbname, $query, $link)){
> echo ("The answer was successfully inserted!<BR>\n");
> } else {
> echo ("The answer could not be inserted!" . mysql_error () . "<BR>\n");
> }
>
>
> mysql_close ($link);
>
>The query seems to work alright, no errors showing up and "The answer was
>successfully inserted!" displayed. However, the table is not updated and I
>cannot see why. The same happens on another page when I try to insert a 'Y'.
>This script was copied from a third page which works alright.
>
>I hope someone can help. I am going mad!!!!!!!.
>--
>Alex Francis
>Cameron Design
>35, Drumillan Hill
>Greenock PA16 0XD
>
>Tel 01475 798106
>[EMAIL PROTECTED]
>http://www.camerondesign.co.uk
>
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>
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