mysql_fetch_assoc() expects parameter 1 to be resource array given

Question

i am having a problem with my code i dont understand whats really wrong with it. i want to get the user id against the LoginID from a table in database but its giving an error of

mysql_fetch_assoc() expects parameter 1 to be resource array given.....

calling to function...

$login_id= $_SESSION['a_id'];
$result=$add_task -> get_id_from_login_id($login_id);
while($id=mysql_fetch_assoc($result)) {
 echo $id;
}

and in the function i have this code

$SQL="SELECT id FROM admin WHERE LoginID='$login_id'";
   return $this->execute_sql($SQL);

What part of the error message is not clear? If you wrote `mysql_fetch_assoc('hello world')` would you expect that to work? — Jon, Jul 25 '13 at 08:56
I suggest you read [this](http://stackoverflow.com/questions/60174/how-to-prevent-sql-injection-in-php). — Hristo Valkanov, Jul 25 '13 at 08:57
that why its not returning the integer value which is my requirment — Qadeer Khan, Jul 25 '13 at 08:57
what function is `get_id_from_login_id`? `$this->execute_sql()` in what class is, what does it? Explain better.. your function `get_id_from_login_id` is returning directly the result in array — Alessandro, Jul 25 '13 at 08:58
You are calling `get_id_from_login_id($login_id)`. **What do you think that returns???** I 'd say it returns the id you want if you asked me. All the rest is useless copy/pasted code. — Jon, Jul 25 '13 at 08:58
You have to show us the code inside the functions you are using if you want help — Alessandro, Jul 25 '13 at 09:01
@jon i want to know that whats wrong with the code that its not returning the id — Qadeer Khan, Jul 25 '13 at 09:01
on function i have this`$SQL="SELECT id FROM admin WHERE LoginID='$login_id'"; return $this->execute_sql($SQL);` — Qadeer Khan, Jul 25 '13 at 09:02
@QadeerKhan : what does this function do : execute_sql ?, mysql_query ???? — Prasanth Bendra, Jul 25 '13 at 09:05
WHAT IS $this->execute_sql($SQL)??? What's the code inside? You're using $this so it's an object? What object is? — Alessandro, Jul 25 '13 at 09:05
$SQL="SELECT id FROM admin WHERE LoginID = '$login_id' "; `$SQL="SELECT id FROM admin WHERE LoginID = '$login_id' ";` — Qadeer Khan, Jul 25 '13 at 09:09

score 0 · Answer 1 · answered Jul 25 '13 at 08:58

0

The error says that there are no results in your query, its may be because of invalid parameters or invalid query. check your connection first then check you query.

answered Jul 25 '13 at 08:58

Rakitha Fernando

81
1
8

score 0 · Answer 2 · answered Jul 25 '13 at 08:59

0

It seems that $result doesn't get data. Check your query in terminal or phpmyadmin. I often use this one:

$result = mysql_query($sql) or die(mysql_error());

answered Jul 25 '13 at 08:59

Dima Savva

86
3

score -1 · Answer 3 · answered Jul 25 '13 at 08:58

-1

Mysqlfetch_assoc returns an array with the columns of your database.

You should use :

while($id=mysql_fetch_assoc($result)) {
 echo $id['LoginID'];
}

answered Jul 25 '13 at 08:58

rachids

331
1
3
10

thanks all of u but i got the error inside the query there should be space as this `$SQL="SELECT id FROM admin WHERE LoginID = '$login_id' ";` – Qadeer Khan Jul 25 '13 at 09:08

mysql_fetch_assoc() expects parameter 1 to be resource array given

3 Answers3