confirmation and php

colake

<?php 
include "user.php";
include "../admin/head.php";
include "../inc.php";

if (empty($page)) $page = "";

if ($page == "update")
{
     $result = mysql_query("SELECT * FROM phpmeet WHERE user = '$username'");
     while ($i = mysql_fetch_array($result))
     {
     if ($rpassword <> $i[password])
      		{
      		echo "Het paswoord dat u hebt ingegeven is niet correct";
      		}
    else     

              {
      ?>
              <body>
              <form name="myfrm" > <input type="hidden" name="keuze"> </form>
               <script language="javascript"> 
               {
                if (confirm("Bent u zeker dat u deze acount wil verwijderen?") )
                 { myfrm.keuze = 0;}
                 else
                 { myfrm.keuze=1;}
              </script>
   <?php           

              if ($keuze == 1)
               {
               mysql_query("DELETE FROM phpmeet  WHERE user = '$username'");
        	 echo "De acount <b> $username</b> werd verwijderd";
               }  

               else 
               {  

               echo " De acount <b> $username</b> werd NIET verwijderd! ";
	        }
      }
  }
} 
else 
{

  echo "Geef het paswoord van de acount<b> $username </b> in en druk op Verwijder om deze acount te verwijderen ";  

?>
    <html>     

    <body>
    <form action=delete.php?username=<?php echo $username; ?>&page=update  method="post" >
    Paswoord: <input type=password name=rpassword><br>
			<br>
			<input type=hidden value=$user>  

			<input type="submit" name="submit" value="verwijder">
		</form></body></html>

<?php
}
?>

this is my code but it doesn't work correct.

What is the function of it?
A user can delete his acount whit it. First there is a control on the password and than there is a confirmation window so that the user can stop the deleting.

What is the probleme
The confirmation give the probleme. I have trade a few things and now I got this bad solution that doesn't work:

      ?>
              <body>
              <form name="myfrm" > <input type="hidden" name="keuze"> </form>
               <script language="javascript"> 
               {
                if (confirm("Bent u zeker dat u deze acount wil verwijderen?") )
                 { myfrm.keuze = 0;}
                 else
                 { myfrm.keuze=1;}
              </script>
   <?php           

              if ($keuze == 1)
               {
               mysql_query("DELETE FROM phpmeet  WHERE user = '$username'");
        	 echo "De acount <b> $username</b> werd verwijderd";
               }  

               else 
               {  

               echo " De acount <b> $username</b> werd NIET verwijderd! ";

Is there a person who can explain me the probleme and give a correct solution. And please I am a newbee so explain it so i can understand it. Many thx.

Frustrated CoLaKe

MrAlaska

Either I am reading your code wrong, or I see a bunch of problems. Your hidden input needs both a name and a value, I don't see anywhere you are actually submitting the form in the final example, and you cannot (to my knowledge) pass values directly from javascript to PHP. To do so, you need to actually have javascript write the value to the document before submitting the form.

Get your form working in PHP, first, then add the javascript. I think it would be much simpler to just have in your form tag
onsubmit return confirm()
or something like that ( I don't remember the exact syntax) but it will then prompt the user (if they have javascript enabled) and a negative answer will stop the form submission. You can 'force' the user to have javascript turned on by letting javascript submit the form, but the form will need to be complete and submitted for PHP to deal with it.

colake

I think there is a probleme with the understanding of my code. There are a few words in dutch in and I think when this where also in englische it was a little bit beter.

What do the code?

A user that would be deleted out of the database can this by given is password. And then clicking on submit.

The control on the password works correct. And the deleting also. But some persons do this and than think "Stop, I don't want this". So there is a confirmation window (javascript) so the deleting is not done.

The probleme is whit this but. Because there have to be first the test on the correct password and then if it's correct the confirmation.

In my first message you can see my code. And I still need help.

KoshNaranek

the dutch is not the problem 🙂

The problem is: the code does NOT stop when the popup appears. This is not VB 🙂

You have to make 1 more step.

Post the data to a "hidden" form on another "function".

NOW popup your question. The answer is the value of another hidden field.

Now "post" the hidden form to the final step.

MrAlaska

PHP does not understand javascript any more than I understand Dutch, but HTML understands it (if the browser accent is not too bad). Try something like this, which is a self posting form which should do what you are attempting:

<?php
if ($page == "update"){
	$result = mysql_query("SELECT * FROM phpmeet WHERE user = '$username'");
	while ($i = mysql_fetch_array($result)){
		if ($rpassword <> $i[password]){
			  echo "Het paswoord dat u hebt ingegeven is niet correct";
		}
		else {
			mysql_query("DELETE FROM phpmeet  WHERE user = '$username'");
			echo "De acount <b> $username</b> werd verwijderd";
		}
	}
}
else{

  echo "Geef het paswoord van de acount<b> $username </b> ".
  		"in en druk op Verwijder om deze acount te verwijderen ";
?>
    <html>
    <body>
    <form method="post"
    	onsubmit="return confirm('Bent u zeker dat u deze acount wil verwijderen?')";>
    Paswoord: <input type=password name=rpassword><br>
            <br>
            <input type=hidden name=page value=update>
            <input type=hidden name=username value='<?php echo $username ?>'>
            <input type="submit" name="submit" value="verwijder">
        </form></body></html>
<?php
}
?>

colake

Now is every thing ok. But there is one thing. When a user click on Cancel there is no message that the acount isn't deleted. Can I put something in it so that it give a message like when the acount is deleted?

For the last time me code :-). Thx for the help I've got.

<?php
if (empty($page)) $page = "";


if ($page == "update")
{
    $result = mysql_query("SELECT * FROM phpmeet WHERE user = '$username'");
    while ($i = mysql_fetch_array($result)){
        if ($rpassword <> $i[password]){
              echo "Het paswoord dat u hebt ingegeven is niet correct";
        }
        else {
            mysql_query("DELETE FROM phpmeet  WHERE user = '$username'");
            echo "De acount <b> $username</b> werd verwijderd";
        }
    }
}
else{

  echo "Geef het paswoord van de acount<b> $username </b> ".
          "in en druk op Verwijder om deze acount te verwijderen ";
?>
    <html>
    <body>
    <form method="post" action=delete.php?username=<?php echo $username; ?>&page=update
        onsubmit="return confirm('Bent u zeker dat u deze acount wil verwijderen?')";>
    Paswoord: <input type=password name=rpassword><br>
            <br>
            <input type=hidden name=page value=update>
            <input type=hidden name=username value='<?php echo $username ?>'>
            <input type="submit" name="submit" value="verwijder">
        </form></body></html>
<?php
}
?>

MrAlaska

Since you are using javascript to stop the form from being submitted, you would put any confirmation message within a javascript function and only need to change the HTML. If you want the form to actually go to another page with a confirmation message on an aborted delete you can have the javascript redirect you to another page inside the function. The code below will give an alert, but stays on the same page. You might need to fix my translation attempt 😃

Since this form is constructed to be self posting, you can leave the action attribute out of the form and it will post to itself. I think it is legal to do it this way, it always seems to work. By putting '$username' and '$page' variables into hidden inputs, you do not need to include them in the URL as part of the $GET bacause those values are now being passed in the $POST.


<html>
<body>
<script language="javascript">
function verify(){
	var check;
	check=confirm('Bent u zeker dat u deze acount wil verwijderen?');
	if (check==0){
		alert("nein acount wil verwijderen");
		return false;
	}
	else {
		return true;
	}
}
</script>

<form method="post" onsubmit="return verify()";>
Paswoord: <input type=password name=rpassword><br>
	<br>
	<input type=hidden name=page value=update>
	<input type=hidden name=username value='<?php echo $username ?>'>
	<input type="submit" name="submit" value="verwijder">
</form></body></html>

colake

Thx,

it's working. Not so elegant has I would first but that's a thing that I maybee can change when I use php a few years :-)