php - if ANY $_POST equals another $_POST die('error');? - Stack Overflow

admin管理员组

文章数量:1340292

I am trying to get this form to:

if any $_POST vars equals any other of the $_POST vars throw an error.

if it was just a few it wouldnt be an issue but I have about 20 or so so if i wanted to do it I would have to go like

    <?php 
    if($_POST['input1']==$_POST['input2'] || $_POST['input1']==$_POST['input3']){
die('whatever');
}

    ?>

But that's not good coding (and it would take forever) I thought about arrays and different ways...

but I am brain dead atm so I thought I could get some help.. thanks in advance

ps it would be nice to do it in PHP (server side) but Jquery is always an option.

I am trying to get this form to:

if any $_POST vars equals any other of the $_POST vars throw an error.

if it was just a few it wouldnt be an issue but I have about 20 or so so if i wanted to do it I would have to go like

    <?php 
    if($_POST['input1']==$_POST['input2'] || $_POST['input1']==$_POST['input3']){
die('whatever');
}

    ?>

But that's not good coding (and it would take forever) I thought about arrays and different ways...

but I am brain dead atm so I thought I could get some help.. thanks in advance

ps it would be nice to do it in PHP (server side) but Jquery is always an option.

• php
• javascript
• jquery

Share Improve this question Follow asked Aug 26, 2010 at 18:33 Luke3butlerLuke3butler 21022 silver badges1313 bronze badges 4

• Btw, die('some screwed up message') is NOT the way to handle errors. – hanse Commented Aug 26, 2010 at 18:40
• @Hanse still better than //TODO: fix me – Ishtar Commented Aug 26, 2010 at 18:43
• I was only using die as an example... I rarely use die. – Luke3butler Commented Aug 26, 2010 at 18:52
• @Hanse: What's your opinion on handling error messages? – Orson Commented Aug 26, 2010 at 20:44

Add a ment  | 

8 Answers 8

Sorted by: Reset to default 16

Delete duplicate values with array_unique() and check if it still equals to your array:

if($_POST != array_unique($_POST))
    die("...");

if ($_POST == array_unique($_POST)) {}

function testPost(){
foreach ($_POST as $keya=>$vala){
    foreach($_POST as $keyb=>$valb){
        if ($keya==$keyb){
            continue;
        } else {
            if ($vala == $valb){
                 return FALSE;
            }
        }
    }
}
return TRUE;
}

This is only an answer to zebediah49's post. A more effective implementation would be:

$post = array_value($_POST);
$count = count($post);
for ($i = 0; $i < $count; ++$i) {
    for ($j = $i + 1; $j < $count; ++$j) {
        if ($post[$i] == $post[$j]) {
            die();
        }
    }
}

This saves all the multiple checks. So it results in O(2*n) instead of O(n^2) (if I got that O stuff right). Though I don't know how much this is slowed down by the additional array_values.

$postValues = array_values($_POST);

if (array_unique($postValues)) != $postValues) {
    die('error!');
}

not so good, but

I'd say you have to do it in a double-loop; while it's O(n^2), it'll not be problematic in practice

foreach($_POST as $keya=>$vala) {
    foreach($_POST as $keyb => $valb) {
        if($vala == $valb && $keya != $keyb) {
            die('whatever');
        }
    }
}

You can do:

$count_array = array_count_values($_POST);
foreach($_POST as $post) {
  if($count_array[$post] >1 ) {
    die();
 }
}

This is my final code... so basically i'm putting the ones I want into an array

credits will go to @akellehe because his was the closest to my end result....

works great

$titles=array();
        $num=1;
        while($num!=15){
        $set1='title'.$num;
        $set2=$_REQUEST["title$num"];
        $titles[$set1]=$set2;
        unset($set1);
        unset($set2);
        $num+=1;
        }
        foreach($titles as $d => $p){
            foreach($titles as $e =>$q){
            if($p==$q && $d!=$e){
            if(!empty($p)){
            $_SESSION['error']='Duplicates not allowed!';
            }
            }
            }
        }

本文标签： phpif ANY POST equals another POST die(39error39)Stack Overflow