How to find the not null values in php and display in if condition [duplicate]

Question

Here i have one array in that array i want to find not null values and display in frond end ,suppose in this array all key values in null means i want to display all values are null,suppose any one of the key is not null means i want to display what is that value

<?php
$array = array('a' => '','b' => 'Kani' , 'c' => '', 'd' => 'Raja');

 if (in_array(null, $array)) {

     echo "There are null values.";
 }else{
  echo "Not Null";
 }
?>

Here key a and d is not null so i want take this key value like Kani and Raja

Your example array doesn't contain any null values but some empty strings. — insertusernamehere
– insertusernamehere, Commented Feb 15, 2017 at 17:19

Niklesh Raut · Accepted Answer · 2017-02-15 17:31:19Z

0

If you want not empty not null array try like this

  <?php 
     print_r(array_filter(array('a' => '','b' => 'Kani' , 'c' => '', 'd' => 'Raja')));
  ?>

Check here : https://eval.in/737817

answered Feb 15, 2017 at 17:31

Niklesh Raut

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Comments

akond · Accepted Answer · 2017-02-15 17:25:49Z

0

I've forgotten to specify the third parameter to in_array.

 if (in_array(null, $array, true)) {
  echo "There are null values.";
 }else{
  echo "Not Null";
 }

This way it will tell if there is an actual null value in the array.

answered Feb 15, 2017 at 17:25

akond

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Comments

Roman · Accepted Answer · 2017-02-15 17:27:28Z

0

hey there you can use this :

$notNulvals = array();
$index =0;
foreach ($array as $key => $value) {
    if ($value) { 
        array_push($notNulvals, $value);
        $index=1;
    }
}

if ($index!=0) {
    echo "all values are null";
} else {
    echo $notNulvals; //you can display it the way you want 
}

edited Feb 15, 2017 at 17:27

Roman

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answered Feb 15, 2017 at 17:23

ouqas mohamed

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