[SOLVED] Foreach argument problem

[Sebby123]

Hello there,

 

I'm currently having problems with the foreach statement. I'm new to using foreach so I don't understand it very well.

 

Could someone explain to me this problem and/or what is wrong with my code so that it produces this error?

 

Warning: Invalid argument supplied for foreach() in /DirHere/AdditionCtrl.php on line 32

 

Lines 30 - 44:

var_dump($Addon);
/*
array(2) {
  ["Hangman"]=>
  string(7) "Hangman"
  ["TicTakToe"]=>
  string(11) "Tic Tak Toe"
}
*/
function GetAddonInfo($AddonName) {
foreach ($Addon as $Folder => $Name) {
	if ($AddonName == $Name) {
		echo "Addon Information:\n";
		echo "Name: ".GetAddonINI($Folder."/Info.ini", "Name");
		echo "Description: ".GetAddonINI($Folder."/Info.ini", "Description");
		echo "Author: ".GetAddonINI($Folder."/Info.ini", "Author");
		echo "Email: ".GetAddonINI($Folder."/Info.ini", "Email");
		echo "Website: ".GetAddonINI($Folder."/Info.ini", "Website");
	} else {
		echo "Cannot find the addon '$AddonName'.";
	}
}
}

 

Thank you in advanced.

 

Seb.

[WolfRage]

In the foreach I think you meant that "$Addon" should be "$AddonName".

<?php
var_dump($Addon);
/*
array(2) {
  ["Hangman"]=>
  string(7) "Hangman"
  ["TicTakToe"]=>
  string(11) "Tic Tak Toe"
}
*/
function GetAddonInfo($AddonName) {
  foreach ($AddonName as $Folder => $Name) {
    if ($AddonName == $Name) {
      echo "Addon Information:\n";
      echo "Name: ".GetAddonINI($Folder."/Info.ini", "Name");
      echo "Description: ".GetAddonINI($Folder."/Info.ini", "Description");
      echo "Author: ".GetAddonINI($Folder."/Info.ini", "Author");
      echo "Email: ".GetAddonINI($Folder."/Info.ini", "Email");
      echo "Website: ".GetAddonINI($Folder."/Info.ini", "Website");
    } else {
      echo "Cannot find the addon '$AddonName'.";
    }
  }
}
?>

[MatthewJ]

I would think something like:

 

var_dump($Addon);
/*
array(2) {
  ["Hangman"]=>
  string(7) "Hangman"
  ["TicTakToe"]=>
  string(11) "Tic Tak Toe"
}
*/
function GetAddonInfo($AddonName, $Addon) {
   foreach ($Addon as $Folder => $Name) {
      if ($AddonName == $Name) {
         echo "Addon Information:\n";
         echo "Name: ".GetAddonINI($Folder."/Info.ini", "Name");
         echo "Description: ".GetAddonINI($Folder."/Info.ini", "Description");
         echo "Author: ".GetAddonINI($Folder."/Info.ini", "Author");
         echo "Email: ".GetAddonINI($Folder."/Info.ini", "Email");
         echo "Website: ".GetAddonINI($Folder."/Info.ini", "Website");
      } else {
         echo "Cannot find the addon '$AddonName'.";
      }
   }
}

GetAddonInfo('addon name', $Addon);

 

It looks like you were just trying to work on an array without passing it in to the function.

[Sebby123]

It looks like you were just trying to work on an array without passing it in to the function.

 

Ah, yes, I presumed that it would have the variable $Addon because it was defined before the function. Didn't realise I had to give it to the function. Well now I know for next time as well.

 

Thank you very much. =]