php - return variable from function - php

Question

I am trying to return a variable from a function. I have the function below, which inserts a postid to the database. I need the postid's value returned to another page.
forum.php - Where the function is.
function newTopic(){
// Get the POST data
global $ir;
$postid = mysql_insert_id();
mysql_query("UPDATE forum_topics SET post_id='$postid' WHERE topic_id='$topicid'");
// No error found and the update was succesful - Return success!
return 100;
return $postid;
}
newtopic.php - Where I need the $postid variable.
if($_POST)
{
$newTopic = $forum->newTopic();
/*
* Return codes:
* 100: Success
*/
switch($newTopic)
{
//If no error = success.
case 100:
$success = 'You have successfully created the topic.';
$issuccess = 1;
$stop = true;
break;
}
$checkerror = $error;
$checksuccess = $success;
}
if($checksuccess){
$contents.="
".alert("success","$success")."";
refresh("3","/forum/t$id-$postid");
}
As you can see, I am trying to use $postid variable from the function newTopic(). Although, the $postid variable is empty.
How can I get the value from the function newTopic.php located in forum.php, to newtopic.php?

Answer 1

When you have used
return 100;
your code will never see
return $postid;
You can use this code to return
return array("code"=>"100","postid"=>$postid);
Now in new_topic.php use the code as shown
if($_POST)
{
$newTopic = $forum->newTopic();
/*
* Return codes:
* 100: Success
*/
switch($newTopic['code'])
{
//If no error = success.
case 100:
$success = 'You have successfully created the topic.';
$issuccess = 1;
$stop = true;
break;
}
$checkerror = $error;
$checksuccess = $success;
}
if($checksuccess){
$contents.="
".alert("success","$success")."";
refresh("3","/forum/t$id-$newTopic['postid']");
}

Answer 2

Your code looks that is returning value 100 before returning $postid variable.
That's wrong.Your code will exit the function at first return.
comment the line "//return 100;"
OR return an array. You cannot return two values like u do.
Instead of doing
return 100;
return $postid;
Do
//return 100;
return array("successs"=>100,"id"=>$postid);
Then use your $newTopic variable as folows:
In switch:
switch($newTopic['success'])
Use the postId anywhere else
$newTopic['id']

Answer 3

try with reference like in the below code
function newTopic(&$postid){
// Get the POST data
global $ir;
$postid = mysql_insert_id();
mysql_query("UPDATE forum_topics SET post_id='$postid' WHERE topic_id='$topicid'");
// No error found and the update was succesful - Return success!
return 100;
}
....... //some codes
$postid = null;
newTopic($postid);
$my_postId = $postid; //Now you have your post ID
OR in you exists code like
if($_POST)
{
$last_postid = null;
$newTopic = $forum->newTopic($last_postid );
/*
* Return codes:
* 100: Success
*/
switch($newTopic)
{
//If no error = success.
case 100:
$success = 'You have successfully created the topic.';
$issuccess = 1;
$stop = true;
$postid = $last_postid;
break;
}
$checkerror = $error;
$checksuccess = $success;
}
if($checksuccess){
$contents.="
".alert("success","$success")."";
refresh("3","/forum/t$id-$postid");
}
EDIT: Call-time pass-by-reference fixed.

Answer 4

If you want to return two values you can not just write two returns.
There are several options to return more values. One is using an array:
return array(100, $postId);
and
list($status, $postId) = $forum->newTopic();
You can also use an associatve array as s.d suggested.
However, as your one variable only contains the status you can also use exceptions in the case the operation fails.