Problem, dont really know how to describe

[viper21634]

OK, I am having a problem, I have created a page, sql.php that has the following function

{
	$db="cvilleinfo";
	$server = "localhost";
	$uname = "sa";
	$pass = "jahugh";
	$conn = mssql_connect($server,$uname,$pass) or die("Couldnt open a connection to the MSSQL server at 	$server");
	$setdb=mssql_select_db($db,$conn);
	$query=mssql_query($sql,$conn);
	$sqlarray=mssql_fetch_assoc($query);
	return $sqlarray;
	return $conn;
}

then I have this code on my main page

//set db vars
$sql = "SELECT * FROM settings WHERE id = 1";
//include sql function page
include("sql.php");
//call to connectdb to open SQL connection	
connectdb($sql);
foreach($sqlarray as $key => $value)
{
	$varname = "SESSION_".strtoupper($key);
	$$varname = $value;
	echo $varname;
	echo "<br>\n";
}

now i keep getting a error, that says that I have passed a invalid argument to my foreach. What is the problem???

[devinemke]

a function can only return 1 value (but that value can ve compund like an array)

function connectdb($sql)
{
    $db="cvilleinfo";
    $server = "localhost";
    $uname = "sa";
    $pass = "jahugh";
    $conn = mssql_connect($server,$uname,$pass) or die("Couldnt open a connection to the MSSQL server at     $server");
    $setdb=mssql_select_db($db,$conn);
    $query=mssql_query($sql,$conn);
    $sqlarray=mssql_fetch_assoc($query);
    return $sqlarray;
}


//set db vars
$sql = "SELECT * FROM settings WHERE id = 1";
//include sql function page
include("sql.php");
//call to connectdb to open SQL connection    

$sqlarray = connectdb($sql);
foreach($sqlarray as $key => $value)
{
    $varname = "SESSION_".strtoupper($key);
    $$varname = $value;
    echo $varname;
    echo "<br>\n";
}

[viper21634]

Well, i set it to only return the $sqlarray var, but still get this message

Warning: Invalid argument supplied for foreach() in c:\inetpub\wwwroot\php\header.php on line 12

[viper21634]

well this is very odd, I set the function to just subtract 5 from 10, but I still dont get a value

//this is my function in a file sql.php
function fun1()
{
$conn = 10-5;
return $conn;
}


//then in header.php

include("sql.php");
fun1();
echo $conn;

I get a error saying that conn is not defined, what could the problem be?? have I missed something?

[planetsim]

maybe instead of return try echo outside the function

so its echo function_name(arguments);

that will get the whole function i used that when i couldnt get around mutliple returns when i started with that..

Now its all i use..

also give us 3lines about and 3lines below line with the error and of course the error line..

Otherwise we'd be looking everywhere

[viper21634]

well that was a pain, yeah I went back and started reading, forgot that you cant echo even a returned var in a function so I changed the code a little and got it to work like I want it.

Thanks a lot guys.

[Weedpacket]

Originally posted by viper21634
well this is very odd, I set the function to just subtract 5 from 10, but I still dont get a value

//this is my function in a file sql.php
function fun1()
{
$conn = 10-5;
return $conn;
}

//then in header.php

include("sql.php");
fun1();
echo $conn;

I get a error saying that conn is not defined, what could the problem be?? have I missed something? [/B]

Well, that's a simple scoping problem; the $conn inside fun1() is a different variable from the $conn outside the function. And when fun1() returns, you don't store the value anywhere so it just disappears down a bottomless pit.

include("sql.php");
$conn=fun1();
echo $conn;