Return Array from Function? How?

[JSProgramIA]

This simple example does not work:

function getProjectInfo($projectId){

	global $row;

	// Retrieve info for supplied project ID
	// Returns $row
	$page_sql = "SELECT * from myTable";
	
	$result = mysql_query($page_sql);
        $row = mysql_fetch_array($result);
        return $row;

}

and if I call it:

getProjectInfo("999");

I get nothing ( and I Know there are records ).

Any advice?

 

[sleipnir214]

What is the $projectID parameter supposed to do? You don't use it.

Why are you performing "global" on $row? You should do either a return or just set the value to a global varible, not both.

Where is your database connection handle? It's not global, so you should be using "global" on that variable and explicitly passing it to your mysql_query() invocation.

What are your PHP installation's error outputs set to? Are you getting any errors?


Want the best answers? Ask the best questions!

TANSTAAFL!!

 

[lgarner]

I was wondering about the db connection handle also. Since the default for mysql_query() is to use the last-opened handle, does the handle need to be specified and made global when it's used within a function?

I'd suggest print_r($row) inside the function to see what's there and determine if it's a query problem or a return problem.

 

[JSProgramIA]

I was assuimg a lot in my last post. I will be more specific.

1. My connection is included via:

<? include ("includes/config.php") ?>

Why are you performing "global" on $row? You should do either a return or just set the value to a global varible, not both.

I don't know. I was just trying it.

What is the $projectID parameter supposed to do? You don't use it.

In reality, my SQL is a lot longer and does return rows (as confirmed by Navicat). So for the sake of this example, let's use:

$page_sql = "SELECT * from myTable where projectid = '$project_id' ";

Lastly, I am not getting any errors.

Could this be something in my $conn string not being global or something? Or do I need to pass the returned row by reference?

Thanks again. Any simple example of returning an array (mysql rows) would be helpful.

Regards.

 

[JSProgramIA]

print_r within the function works.

print_r outside of it returns NULL.

 

[JSProgramIA]

NEVER MIND I AM AN IDIOT!!!!!!!!

Can you see it?

function getProjectInfo($projectId){


	// Retrieve info for supplied project ID
	// Returns $row
	$page_sql = "SELECT * , total_hours-hours_paid AS hours_owed,  DATE_FORMAT(last_updated, '%m-%d-%Y') AS last_update FROM  `project`  INNER JOIN `hours` ON (`project`.`project_id` = `hours`.`project_id`) ";
	
	$result = mysql_query($page_sql);
	if(!$result){
	
		die("Error 1");
	
	} else {
	
		$returned_row = mysql_fetch_array($result);
		
		return $row;
		
	}
}

THANKS SO MUCH. I APPRECIATE YOUR HELP.

 

[lgarner]

I just tried it and it works fine. And, I've confirmed that the connection handle doesn't need to be specified and made global; omitting it in the function works. I can't see anything in your code to cause the problem (except for the query string, I cut & pasted it to test).

For what it's worth, here's what I tried:

<?php
$conn = mysql_connect("localhost","user","password");
if (!$conn) die;

function getinfo($name)
{
$result = mysql_query("select * from assets.assets where name='$name'");
$row = mysql_fetch_array($result);
return $row;
}

$r = getinfo("JDP01WW0917");
print_r($r);
?>

And the results:
Array
(
[0] => 0917
[tag] => 0917
[1] => JDP01WW0917
[name] => JDP01WW0917
[2] => Server
[type] => Server
...
)

With or without using a global for $row made no difference.

 

[lgarner]

OK, never mind.... Variable names matter