PHP Approval/Deny help

Status
Not open for further replies.

Jayar

Active Member
629
2009
15
40
Hey erybody! I'm making a CMS and right now I'm working on the avatars.
When they upload their avatars, it automatically sets their approved status to false, because i want to be able to approve/deny them in the acp.
This is what I have for the Approve/deny portion, since i already got the uploading script all taken care of.

PHP:
$picture = mysql_query("SELECT * FROM images WHERE approved='false'");
echo "<form action='approve.php?do=submit' method='post'>"
while($result = mysql_fetch_array($picture)){
echo "<img src='../".$result['path']."'>";
echo "<input type='checkbox' id='".$result['id']."' />
}
echo "</form>";

Approve.php?do=submit :
PHP:
$do = $_GET['do'];
//this says whats going to be in approve.php?do=submit
if($do == submit){
//code here
}

I've been stuck on this for a few hours trying to figure and google a way on how to get more than one $_POST thingies and update them in the mysql.
Im not sure if thats a good description of my problem or not lol
 
7 comments
something like this? i wrote this for streamscene.me :)

PHP:
    if (isset($_GET['do']) && isset($_POST['id']) && is_array($_POST['id']))
    {
        foreach ($_POST['id'] as $id)
        {
            if (!mysql_query("UPDATE movies SET top=1 WHERE id = '$id'"))
            {
                echo error_message(sql_error());
            }
        }
    }
 
So,

PHP:
$do = $_GET['do']; 
//this says whats going to be in approve.php?do=submit 
if($do == submit){ 
 if (isset($_GET['do']) && isset($_POST['id']) && is_array($_POST['id'])) 
    { 
        foreach ($_POST['id'] as $id) 
        {
            if (!mysql_query("UPDATE images SET approve=true WHERE id = '$id'")) 
            { 
                echo error_message(sql_error()); 
            } 
        } 
    }  
}

ill try it:)
 
I don't really get what you're trying to do?
From what I can see, isn't essentially just you manually approving avatars? So you initially have the avatar approved='false' and you want it on submit to make approved='true'?
I don't get what you mean by 'multiple post thingies'
 
Notice I changed your checkbox.
I added name=" and value=""

I'm fairly tired atm and can't focus. So could be entirely wrong :|

PHP:
<?php
if($_GET['do'] == 'submit' && !empty($_POST['id'])){
  foreach($_POST['id'] as $ID){
    if(!is_numeric($ID)){
      echo ' Whaaaat, the ID '.$ID.' isn\'t a number? ';
      exit;
   }
   mysql_query('UPDATE `images` SET `approved` = "true" WHERE `id` = '.$ID);
  }
}else{
  // Show form
  $picture = mysql_query("SELECT path, id FROM images WHERE approved='false'"); 
  echo "<form action='approve.php?do=submit' method='post'>" ;
  while($result = mysql_fetch_array($picture)){ 
   echo "<img src='../".$result['path']."'>"; 
   echo '<input type="checkbox" name="id[]" value="'.$result['id'].'" id="'.$result['id'].'" />';
  } 
  echo "</form>";
}
?>
 
what your going to want to do is a switch statement heres a the base structure from a login i had minus the code between as an example.

PHP:
<?php
if (isset ( $_GET ['do'] )) {
	switch ($_GET ['do']) {
		
		case 'logout' :
		//code for ?do=logout
		break;
			
		case 'view' :
	   //code for ?do=view
	   break;
		
		case 'register' :
        //code for ?do=register
        break;
        
		default :
		die ( 'Invalid Command' );
		break;
	}
 }
?>


That will allow you to do page.php?do=submit and it will execute whatever code in in the submit code area.
 
Notice I changed your checkbox.
I added name=" and value=""

I'm fairly tired atm and can't focus. So could be entirely wrong :|

PHP:
<?php
if($_GET['do'] == 'submit' && !empty($_POST['id'])){
  foreach($_POST['id'] as $ID){
    if(!is_numeric($ID)){
      echo ' Whaaaat, the ID '.$ID.' isn\'t a number? ';
      exit;
   }
   mysql_query('UPDATE `images` SET `approved` = "true" WHERE `id` = '.$ID);
  }
}else{
  // Show form
  $picture = mysql_query("SELECT path, id FROM images WHERE approved='false'"); 
  echo "<form action='approve.php?do=submit' method='post'>" ;
  while($result = mysql_fetch_array($picture)){ 
   echo "<img src='../".$result['path']."'>"; 
   echo '<input type="checkbox" name="id[]" value="'.$result['id'].'" id="'.$result['id'].'" />';
  } 
  echo "</form>";
}
?>

Worked perfect thanks:D
 
Status
Not open for further replies.
Back
Top