[Resolved] Small function problem

Sayian

Hello , i am receiving an error stating
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 'where userid='2165'' at line 1

when i try to use this php function

function view_visit_list($user_id) {
global $tb_visits,$base_url;

$sql = "select * from $tb_visits where userid='$user_id'";
$res = mysql_query($sql) or die(mysql_error());

$out = "";  

if(mysql_num_rows($res) == 0 ) {$out.= "";}
else {

while($rs = mysql_fetch_array($res)) {
$fav_name = get_username($rs['visitid']);
$fuserid=$rs['visitid'];
$out.= "<a class=\"link01\" href=\"/acct_profile.php?acctid=$fuserid\">$fav_name</a> &nbsp; ";
}
return $out;}
}

Any idea why this is happening?

kburger

Are you sure that's the code? It looks correct to me, unless, for some strange reason $user_id is 2165'.

Sayian

Yes , thats the correct code , and this works for the exact same function , except i change the tables to grab info from another table.

the table names are correct

it's getting the userid from the profile thats currently being viewed

this function works

function view_friends_list($user_id) {
global $tb_favourites,$base_url;

$sql = "select * from $tb_favourites where user_id='$user_id'";
$res = mysql_query($sql) or die(mysql_error());

$out = "";  

if(mysql_num_rows($res) == 0 ) {$out.= "This user Hasnt Added Any Freinds.";}
else {

while($rs = mysql_fetch_array($res)) {
$fav_name = get_username($rs['fav_user_id']);
$fuserid=$rs['fav_user_id'];
$out.= "<a class=\"link01\" href=\"/acct_profile.php?acctid=$fuserid\">$fav_name</a> &nbsp; ";
}
return $out;}
}

This one doesnt


function view_visit_list($user_id) {
global $tb_visits,$base_url;

$sql = "select * from $tb_visits where user_id='$user_id'";
$res = mysql_query($sql) or die(mysql_error());

$out = "";  

if(mysql_num_rows($res) == 0 ) {$out.= "No one has viewed this profile yet.";}
else {

while($rs = mysql_fetch_array($res)) {
$fav_name = get_username($rs['view_id']);
$fuserid=$rs['view_id'];
$out.= "<a class=\"link01\" href=\"/acct_profile.php?acctid=$fuserid\">$fav_name</a> &nbsp; ";
}
return $out;}
}

Im totally lost

halojoy

in first post
you use userid instead of user_id

Sayian

b/c i changed it to see if that was the problem , that wasnt the problem.

halojoy

so it will work with both userid and user_id ??
okay, thats good
problem has to be elsewhere

BuzzLY

Check to make sure that $tb_visits is populated with the name of the table, and that the table name is the proper name and case.

Once you verify that, copy/paste the SQL statement and test it within your database to make sure it works correctly (with the proper userid of course).

bradgrafelman

Also consider visting the Debugging 101 thread (link sould be in my sig I believe). A suggestion you'll find there is probably echo out the $sql, visibly checking it for errors, and (as BuzzLY said) copy it directly from the PHP output into a SQL GUI/CLI.