Reputation: 1
//( ! ) Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given
No idea getting this problem. Please guys help me out. Here is my code:
$query = "SELECT t1.employeecode, t1.employeename, t2.v, t2.w, t3.total, t3.totals
FROM invoice t1,salaries t2,table1 t3
WHERE t1.employeecode = salaries.employeecode AND
t1.employeecode = t3.employeecode
ORDER BY t1.employeecode ASC";
$result = mysql_query($query,$con);
if(mysql_num_rows($result)>0){
echo '<table><tr><th>Article title</th> </tr>';
while($row=mysql_fetch_array($result)){
//$postedon = strftime("%A %e %b %Y",strtotime($row['postedon']));
echo '<h1><tr><td><a href="3.php?employeecode='.$row["employeecode"].'">'.$row["deparment"].'</a></td></tr></h1>';
}
Upvotes: 0
Views: 115
Reputation: 26451
$result is boolean (false) cause there must be problem with your query. Help yourself by getting error in your query using mysql_error()
,
$result = mysql_query($query,$con) or die(mysql_error());
Note: Please, don't use mysql_*
functions in new code. They are no longer maintained and are officially deprecated.
So use either PDO
or MySQLi
(IMO PDO is way to go)
Upvotes: 2
Reputation: 951
Try changing your salaries to be t2..
WHERE t1.employeecode = salaries.employeecode AND
like
WHERE t1.employeecode = t2.employeecode AND
I don't know, but what I know when we done give some alias to a table, our dbms would read that alias, not that native table name.
Upvotes: 0
Reputation: 2540
Since you aliased your salaries
table to t2
you have to reference it as t2
in your where clause:
$query = "SELECT t1.employeecode, t1.employeename, t2.v, t2.w, t3.total,
t3.totals
FROM invoice t1,salaries t2,table1 t3
WHERE t1.employeecode = t2.employeecode AND
t2.employeecode = t3.employeecode
ORDER BY t1.employeecode ASC";
Upvotes: 2
Reputation: 179
it seems your query is not working and fails returning FALSE bool value. To check the reason you need to check your query after dynamically building at run time follow below steps
$result = mysql_query($query) or die($query."
".mysql_error());
check the error message. seems some wrong column must be included
Upvotes: 0