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mysql trouble
Posted: Fri Apr 09, 2010 8:49 am
by darin
Excuse me for the silly (I guess) topic, but I'm stuck. I'm trying to pull some data from a table with the following code and it keeps telling me "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in W:\www\elis\includes\userrights.php on line 8". PLEASE help!
Code: Select all
$fieldid = array(1,2,3,4,7);
$userid = 1;
foreach($fieldid as $id){
$query ="SELECT right ";
$query .="FROM userrights ";
$query .="WHERE (userid = '{$userid}') AND (fieldid = '{$id}')";
$queryresult = mysql_query($query, $connection);
$j = mysql_fetch_array($queryresult);
$right[$id] = $j['right'];
}
My table userrights looks like this:
userrightid userid fieldid right
1 1 1 1
2 1 2 1
3 1 3 1
4 1 4 1
5 1 5 1
6 1 6 1
7 1 7 1
Once again I apologise for being such a newbie...
Darin
Re: mysql trouble
Posted: Fri Apr 09, 2010 10:30 am
by pickle
Welcome to the boards, but...
Thread moved.
Re: mysql trouble
Posted: Fri Apr 09, 2010 10:42 am
by AbraCadaver
mysql_connect() mysql_select_db() ???
Re: mysql trouble
Posted: Fri Apr 09, 2010 10:54 am
by darin
Sorry for the wrong place pickle
AbraCadaver thisi is not it. Please find bellow the whole code, the trouble is that the first select works, but not the second
Code: Select all
<?php
function readrights($userid, $screenid){
global $connection;
$query1 ="SELECT fieldid ";
$query1 .="FROM fields ";
$query1 .="WHERE screenid = '{$screenid}'";
$quryresult1 = mysql_query($query1, $connection);
$i = 0;
while($field = mysql_fetch_array($quryresult1)){
$fieldid[$i] = $field['fieldid'];
$i = $i+1;
}
foreach($fieldid as $id){
$query ="SELECT right ";
$query .="FROM userrights ";
$query .="WHERE (userid = '{$userid}') AND (fieldid = '{$id}')";
$queryresult = mysql_query($query, $connection);
$j = mysql_fetch_array($queryresult);
$right[$id] = $j['right'];
}
return $right;
}
?>
Re: mysql trouble
Posted: Fri Apr 09, 2010 11:10 am
by AbraCadaver
Code: Select all
$queryresult = mysql_query($query, $connection) or die(mysql_error());
Re: mysql trouble
Posted: Fri Apr 09, 2010 11:21 am
by darin
AbraCadaver wrote:Code: Select all
$queryresult = mysql_query($query, $connection) or die(mysql_error());
Thanks for the support, here is what i get:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM userrights WHERE (userid = '1') AND (fieldid = '1')' at line 1
The names of the DB fields and table are ten times checked

....
Re: mysql trouble
Posted: Fri Apr 09, 2010 11:26 am
by AbraCadaver
Re: mysql trouble
Posted: Fri Apr 09, 2010 11:30 am
by darin
Yes! That did solve my problem, thank you verry much, but how in the world the first select works

?
Kind regards

,
Darin
oops you gave a link, i'll read it....