Oh dear - I forgot the rights in the new table:
select command denied to user: 'henning@localhost' for table 'postnummer'
Thanks for helping, John!
Henning
John Wards wrote:
> not having the time to look at your code I am not sure but do this to see if
> its mysql causeing the errors
>
> $result = mysql_query($query) or die(mysql_error());
> ----- Original Message -----
> From: "Henning" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Wednesday, September 18, 2002 5:12 PM
> Subject: [PHP] Join - problem
>
>
>
>>Hello
>>I'm using PHP4 and MySQL on Apache webserver.
>>I have a table valled "liste" with names of some people, and a table
>>called "postnummer" with zip-codes and citynames.
>>My select should join the adress and zip from one table with the
>>cityname from the other table.
>>But my join-line does not work. =:(
>>
>>My code results in this line:
>>
>>Warning: Supplied argument is not a valid MySQL result resource in
>>/var/www/html/find/resultat.php on line 27
>>
>>The code is:
>>mysql_select_db("adresser");
>>$query = ("SELECT fornavn, efternavn, gade, liste.postnummer,
>>postnummer.postbynavn FROM liste
>>left join postnummer on liste.postnummer = postnummer.postnummer
>>WHERE MATCH (fornavn, efternavn) AGAINST ('$search');");
>>$result = mysql_query($query);
>>while(list( $fornavn, $efternavn, $gade, $postnummer, $postbynavn) =
>>mysql_fetch_row($result))
>>
>
> print("<TR><TD>$fornavn</TD><TD>$efternavn</TD><TD>$gade</TD><TD>$postnummer
> </TD><TD>$postbynavn</TD></TR>\n");
>
>>
>>What is going on?
>>The select is ok if I do not join, but then I'm not able to get the
>>field "postbynavn" from the other table.
>>
>>PLEASE HELP!
>>
>>Henning
>>
>>
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>
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