Home » Php » Why do I get "Resource id #4" when I apply print_r() to an array in PHP? [duplicate]

Why do I get "Resource id #4" when I apply print_r() to an array in PHP? [duplicate]

Posted by: admin December 1, 2021 Leave a comment


Possible Duplicate:
How do i “echo” a “Resource id #6” from a MySql response in PHP?

Below is the code:

$result=mysql_query("select * from choices where a_id='$taskid'")or die(mysql_error());

I get “Resource id #4”, any idea?

After I added

{ print_r($row); }

I just got []

What’s wrong?


You are trying to print a mysql resource variable instead of the values contained within the resource it references. You must first try to extract the values you have gotten by using a function such as mysql_fetch_assoc().

You might also try mysql_fetch_array() or mysql_fetch_row(), but I find associative arrays quite nice as they allow you to access their values by the field name as in Mike’s example.


mysql_query() does not return an array as explained in the manual. Use mysql_fetch_array(), mysql_fetch_assoc(), or mysql_fetch_row() with your $result. See the link above for more info on how to manipulate query results.

$result = mysql_query('SELECT * FROM table');
while ($row = mysql_fetch_assoc($result)) {
    echo $row["userid"];
    echo $row["fullname"];
    echo $row["userstatus"];


$result is a resource variable returned by mysql_query. More about resource variables: http://php.net/manual/en/language.types.resource.php

You must use other functions such as mysql_fetch_array() or mysql_fetch_assoc() to get the array of the query resultset.

$resultset = array();
$result=mysql_query("select * from choices where a_id='$taskid'") or die(mysql_error());
while($row = mysql_fetch_assoc($result)){
  $resultset[] = $row; // fetch each row...
mysql_free_result($result); // optional though...





Resources are special variable types used by PHP to track external resources like database connections, file handles, sockets, etc.