Undefined variable When values are empty

Question

I have 2 blocks of code....

// 1st block

<div id="a1">
<?php

if (is_array($new_array) || is_object($new_array))
{
  foreach ($new_array as $name => $val)
   {
     echo $name . " : " . $val[0] . " , " . $val[1]. " , " . $val[2];
   }
}
unset($new_array);
?>
</div>

2nd block

<div id="a2">
    <?php

    if (is_array($new_array) || is_object($new_array))
    {
      foreach ($new_array as $name => $val)
       {
         echo $name . " : " . $val[0] . " , " . $val[1]. " , " . $val[2];
       }
    }
    unset($new_array);
    ?>
    </div>

Either 1st or 2nd Block will give empty results in a day. Means if Today , 1st block will give empty result & tomorrow 2nd Block will give empty result.... Alternatively....

Issue :

Today, Value is empty for 2nd block , it gave Notice: Undefined variable: new_array error , so I initialized this before 2nd block of code :

$new_array='';

it worked fine.... but tomorrow 2nd block code will give this result :

Warning: Illegal string offset , Fatal error: Uncaught Error: Cannot use string offset as an array

So i need to remove this code : $new_array=''; before 2nd block & i need to place before 1st block.....

MorganFreeFarm · Accepted Answer · 2019-05-30 09:25:33Z

7

What about to use if (isset($new_array)) {...} or to initialize it like array, $new_array = [];

You can add isset to other checks, like this:

if (isset($new_array) && (is_array($new_array) || is_object($new_array))) { .... }

More info about isset()

edited May 30, 2019 at 9:25

answered Sep 12, 2018 at 6:55

MorganFreeFarm

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6 Comments

user8444404 Over a year ago

i tried $new_array = []; but it give this error : Notice: Undefined variable: new_array

MorganFreeFarm Over a year ago

Where you place $new_array = []; ?

Krishanu Over a year ago

use isset($new_array), bro! @MorganFreeFarm give a simple solution.

user8444404 Over a year ago

i am already using this : if (is_array($new_array) || is_object($new_array)) , how i can use if (isset) ? i tried if (isset(if (is_array($new_array) || is_object($new_array))) , but it gave error....

user8444404 Over a year ago

@MorganFreeFarm sorry , i can't unset() , otherwise i for today , i will get same result as yesterday.....

|

Krishna Joshi · Accepted Answer · 2018-09-12 07:04:20Z

0

You can use isset() function to check if $new_array is set or not. Edit your code as below

// 1st Block

<div id="a1">
<?php
  if(isset($new_array)){
    if (is_array($new_array) || is_object($new_array))
    {
       foreach ($new_array as $name => $val)
       {
          echo $name . " : " . $val[0] . " , " . $val[1]. " , " . $val[2];
       }
    }
  }
  unset($new_array);
?>
</div>

// 2nd Block

<div id="a2">
<?php
  if(isset($new_array)){
    if (is_array($new_array) || is_object($new_array))
    {
       foreach ($new_array as $name => $val)
       {
          echo $name . " : " . $val[0] . " , " . $val[1]. " , " . $val[2];
       }
    }
  }
  unset($new_array);
?>
</div>

answered Sep 12, 2018 at 7:04

Krishna Joshi

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Comments

Ronnie Oosting · Accepted Answer · 2018-09-12 07:05:09Z

You are looking for isset()/empty() and perhaps a short if/else.

Example:

<div id="a1">
<?php

if (is_array($new_array) || is_object($new_array))
{
  foreach ($new_array as $name => $val)
   {
     echo $name . " : " . $val[0] . " , " . $val[1]. " , " . $val[2];
   }
}
unset($new_array);
?>
</div>

Let's say $new_array is not set/not an array/not and object. Your if statement will not continue, because it will receive false.

In case you wish to always execute your code you could do something like this: (see the notes for explanation).

<div id="a1">
<?php

$new_array = []; // [] = short for array();. $new_array will now always be an array.

// You could also do something like this:
$new_array = (! is_array($new_array)) ? $new_array = [] : $new_array; // Here you say $new_array is always set with his code and else it will be an empty array.


if (is_array($new_array)) // 
{
  foreach ($new_array as $name => $val)
   {
     echo $name . " : " . $val[0] . " , " . $val[1]. " , " . $val[2];
   }
}
unset($new_array);
?>
</div>

http://php.net/manual/en/control-structures.if.php // Shorthand if/else
http://www.php.net/manual/en/function.is-array.php // is_array

Exterminator · Accepted Answer · 2018-09-12 07:16:21Z

0

You can check first if array exists or not. If it exists then check if it is empty(if it is an array) or count the number of elements in it is greater than zero (in case of object array)

<div id="a1">
<?php
if (is_array($new_array) || is_object($new_array)) {
    if (isset($new_array)) {
        if (empty($new_array) || count($new_array) > 0) {
            foreach ($new_array as $name => $val) {
                echo $name . " : " . $val[0] . " , " . $val[1] . " , " . $val[2];
            }
        }
    }
}
unset($new_array);
?>
</div>

answered Sep 12, 2018 at 7:16

Exterminator

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Comments

prasanna puttaswamy · Accepted Answer · 2018-09-12 07:00:07Z

-1

data type mismatch in php 7 result in this error. instead of

$new_array='';

Initialize like

$new_array=array();

Once you unset it destroys that array. So intialize that before first block code

answered Sep 12, 2018 at 7:00

prasanna puttaswamy

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1 Comment

user8444404 Over a year ago

Thanks for answer, i also tried that...... but still it dont work for me as ` new_array` acts as variable. not array.....

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