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Display Array Contents

echo '<pre>';
print_r ($_POST);
echo '</pre>';

Comments

  1. Mark Cole
    Permalink to comment#

    Seriously? This is terrible lol.

  2. Permalink to comment#

    lmao!!!!!!!!!!!!!!!!!!!!!!!!! this is funny

  3. Permalink to comment#

    lollllll

    so, an attempt to correct this users coding..

    input:

    $array[0] = "CSS";
    $array[1] = "HTML";
    $array[2] = "PHP";
    $array[3] = "JavaScript";
    
    echo '';
    print_r ($array);
    echo '';

    output:

    Array
    (
        [0] => CSS
        [1] => HTML
        [2] => PHP
        [3] => JavaScript
    )
  4. Permalink to comment#

    haha this is too basic for a snippet

  5. Permalink to comment#

    Or this:


    echo '<pre>',print_r($array,1),'<pre>';

    • Gringer

      For beginners:

      The reason we use a comma instead of a period/dot between the ‘pre’ and the print_r() is because print_r() does an echo from itself.

      If you would use a dot instead then it PHP will echo print_r() and AFTER that a pre and a /pre.


      Gringer
      PS:

      As we can read here:

      http://nl3.php.net/manual/function.print-r.php

      print_r() takes a second argument to return instead of echo:

      echo '<pre>' . print_r($array,true) . '</pre>';

      Note: We use dots instead of comma’s, and give print_r() an extra ‘true’ (= return).

      Not as short, but just to explain to beginners how the spoon fits in the fork xD

  6. superman
    Permalink to comment#

    I’m sorry to say that not everybody is experienced who come here.It is a wonderful resource for the all levels of coder. Keep it up Chris.. Thanks

  7. Gringer. I agree with you. Thanks.

  8. Waterman
    Permalink to comment#

    You can as well use the foreach statement to get the content of an array out.. you use that when you need to perform something to every single one of the array element…

    $someArr = array(1,2,3,4,5…)
    foreach ($someArr as $newVar){
    $newVar*=9;
    //whatever you need to do to array members
    echo $newVar;
    }

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