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up vote 1 down vote favorite

i am posting an array of checkboxes. and i cant get it to work. i didnt include the proper syntax in the foreach loop to keep it simple. but it is working. i tested in by trying to do the same thing with a text field instead of a checkbox and it worked with the textfield.

<?php
<form method="post">
 echo'
foreach{

<input id="'.$userid.'" value="'.$userid.'"  name="invite[]" type="checkbox">
<input type="submit">';

}
 </form>
?>

here is the part that is not working. it is echoing 'invite' instead of array. 

<?php
    if(isset($_POST['invite'])){
$invite = $_POST['invite'];
echo $invite;
}
share|improve this question
$invite is an array. Try print_r($invite); instead of echo – Tom Pietrosanti May 18 '12 at 15:22
2  
Do a print_r($_POST); and see what is actually set. – Nick May 18 '12 at 15:24
it says 'invite' when i do print_r – arboles May 18 '12 at 15:24
Try adding multiple checkbox named 'invite[]'. You'll get the output as an array. And you shouldn't be using <Form ** inside the loop as there'll always be single instance of invite in each form. – J A May 18 '12 at 15:25
I think your PHP code is still wrong. Your foreach shouldn't be inside your echo and your HTML is inside of you <?PHP ?> tags. – Tom Pietrosanti May 18 '12 at 15:40
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4 Answers

up vote 6 down vote accepted

Your $_POST array contains the invite array, so reading it out as

<?php
if(isset($_POST['invite'])){
  $invite = $_POST['invite'];
  echo $invite;
}
?>

won't work since it's an array. You have to loop through the array to get all of the values.

<?php
if(isset($_POST['invite'])){
  if (is_array($_POST['invite'])) {
    foreach($_POST['invite'] as $value){
      echo $value;
    }
  } else {
    echo $value;
  }
}
?>
share|improve this answer
when i tried this. i got Warning: Invalid argument supplied for foreach() in C:\xampp\htdocs\social_learning\site_pages\groups_page.php on line 95 – arboles May 18 '12 at 15:34
Can you copy/paste the full results of an echo '<pre>'; print_r($_POST); echo '</pre>'; – Ryan May 18 '12 at 15:40
I edited my answer to add a check of whether the invite value is an array or string. If you only pass one checked invite checkbox it will be a string value. That may fix the error you are receiving. – Sean May 18 '12 at 15:43
here is the <pre><pre> results Array ( [invite] => Invite [rememberme] => undefine ) – arboles May 18 '12 at 15:48
i tried the edited for loop. and i still got an undefined variable. Notice: Undefined variable: value in C:\xampp\htdocs\social_learning\site_pages\groups_page.php on line 99 i think the issue has to be with the data before it is being POSTED. but when i inspect the element. the data in the html is all correct. so the problem is somewhere from before the html is sent. – arboles May 18 '12 at 15:52
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up vote 1 down vote

I just used the following code:

<form method="post">
    <input id="user1" value="user1"  name="invite[]" type="checkbox">
    <input id="user2" value="user2"  name="invite[]" type="checkbox">
    <input type="submit">
</form>

<?php
    if(isset($_POST['invite'])){
        $invite = $_POST['invite'];
        print_r($invite);
    }
?>

When I checked both boxes, the output was:

Array ( [0] => user1 [1] => user2 )

I know this doesn't directly answer your question, but it gives you a working example to reference and hopefully helps you solve the problem.

share|improve this answer
thanks. my code must have some bug, type or something i am not seeing. the weird part is that i tried this exact same thing with a text field, and it worked. so the issue is related specifically to the checkbox. – arboles May 18 '12 at 15:45
up vote 0 down vote 
// if you do the input like this
<input id="'.$userid.'" value="'.$userid.'"  name="invite['.$userid.']" type="checkbox">

// you can access the value directly like this:
$invite = $_POST['invite'][$userid];
share|improve this answer
up vote 0 down vote

Because your <form> element is inside the foreach loop, you are generating multiple forms. I assume you want multiple checkboxes in one form.

Try this...

<form method="post">
foreach{
<?php echo'
<input id="'.$userid.'" value="'.$userid.'"  name="invite[]" type="checkbox">
<input type="submit">';
?>
}
</form>
share|improve this answer
1  
Nice catch. I would even move the submit button outside the loop as well. – Tom Pietrosanti May 18 '12 at 15:27
sorry, that was a mistake i made when i typed it in the question. the code isnt really like that. – arboles May 18 '12 at 15:28

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