Cant access variable value in php

Question

I'm working on simple app that add array items on button click and display the values on click.

My problem is after adding the items to array I cant display the values.

Hope you help me.

HTML

<form action="" method="POST">
    <button name="save">save</button>
    <button name="display">display</button>
</form>

PHP

$arr = array();

    if (isset($_POST['save'])) {

        $items = array('Mark', '12', 'Japan');

        array_push($arr, $items);

    }

    if (isset($_POST['display'])) {

        print_r($arr);

    }

The reason this isn't working is because when clicking on the save button you filled the $arr variable with the items and never saved them, but than when you clicked on the display button it recreated the $arr array as an empty list. To do this, on the save button you need to store that data somewhere like a database or a file, and than on the display button retrieve that data first and than display it. — Igor Ilic
– Igor Ilic, Commented Dec 17, 2018 at 7:38
You don't actually "save" your array after adding values to it — user8034901
– user8034901, Commented Dec 17, 2018 at 7:51

Zhorov · Accepted Answer · 2018-12-17 08:19:42Z

1

You may try to use PHP sessions. This example is very basic, but may help you to understand how to preserve your data across subsequent accesses:

<?php
session_start();

if (isset($_POST['save'])) {
    echo 'Save'.'<br>';
    $items = array('Mark', '12', 'Japan');
    $_SESSION['SimpleArray'] = array();
    array_push($_SESSION['SimpleArray'], $items);
}

if (isset($_POST['display'])) {
    echo 'Display'.'<br>';
    print_r($_SESSION['SimpleArray']);
    echo '<br>';
}
?>

answered Dec 17, 2018 at 8:19

Zhorov

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Comments

Joseph_J · Accepted Answer · 2018-12-17 08:36:58Z

Here is a way that you can use a hidden input element to keep updating your POST array with the added information.

Use this:

<?php

//Check to make sure there is a value set for the hidden field in the post array.
  if(isset($_POST['myArray']) && $_POST['myArray']){

  //If there is we are going to decode and unserialize it and pass it to the $arr varaible.
  $arr = unserialize(base64_decode($_POST['myArray'])); 

  }

  //Check to see if you hit the save button.
  if (isset($_POST['save'])) {

    //We did so we are adding an element to the $arr array.
    $arr[] = array('Mark', '12', 'Japan');

  }

  //Check to see if we want to display all the data from hitting the save
  //button multiple times.
  if (isset($_POST['display'])) {

    //We did so lets print out the array of what we have "saved so far."
    print_r($arr);

  }

  //Checking to make sure we have an array.  If we do we are going to 
  //serialize and encode the array so we can pass it to the hidden input element.
  if(isset($arr)){

    //serialize and encode.
    $arr = base64_encode(serialize($arr));

  } else {

    //Nothing happened.  Pass nothing to the hidden input element.
    $arr = '';

  }


?>


<form action="" method="POST">
    <button name="save">save</button>
    <button name="display">display</button>
    <input type="hidden" name="myArray" value="<?php echo $arr; ?>">
</form>

Hope it helps.

alexey-novikov · Accepted Answer · 2018-12-17 07:47:42Z

0

It is two different conditions.

So when you are clicking on the save button, everything is working correctly and $arr is contain needed value but now, when you are clicking to the display button, you are re-submitting this form again and the first condition isn't working correctly because server forgets about your $arr. It is completely different requests.

Use this code:

$arr = array();

if (isset($_POST['save'])) {

    $items = array('Mark', '12', 'Japan');

    array_push($arr, $items);
    echo 'one <br>';

}

if (isset($_POST['display'])) {
    echo 'two <br>';
    print_r($arr);

}

And you will see that one isn't showing once you click the display button.

You can use session as the simplest solution for this.

edited Dec 17, 2018 at 7:47

answered Dec 17, 2018 at 7:41

alexey-novikov

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1 Comment

Rotimi Over a year ago

This is more of a debugging help than an actual answer. Maybe add a sample example using sessions. It seems OP is a newbie to PHP

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