How to use json_decode in php

Question

I am trying to use this code:

 <?php  echo json_decode('"\uD83D\uDE00"'); ?>

When I am using this code I am getting this value :

But when I am trying to use this code using some variable like this:

 <?php
 $var = "\uD83D\uDE00";
 echo json_decode('"{$var}"');
 echo "{$var}";
 ?>

I am getting this as output:

 {$var}\uD83D\uDE00

I have tried many things but nothing is working what can be the best way to do this.

how this is showing output: here I am asking about some basic php concatination. — , Feb 03 '17 at 10:57
`'$var'` print `$var` while `"$var"` print value of `$var` Note the outer single and double quote — M A SIDDIQUI, Feb 03 '17 at 10:59
echo json_decode('"$var"'); this code is giving me this value $var — , Feb 03 '17 at 11:02
Possible duplicate of [json\_decode to array](http://stackoverflow.com/questions/5164404/json-decode-to-array) — Luzan Baral, Feb 03 '17 at 11:42

score 1 · Accepted Answer · answered Feb 03 '17 at 11:07

1

It is strange as demand but funny, try this :)

<?php
    $var = "\uD83D\uDE00"; 
    echo json_decode('"'.$var.'"');
?>

answered Feb 03 '17 at 11:07

score 1 · Answer 2 · answered Feb 03 '17 at 11:07

1

Try this:

 <?php
    $var  ='"\uD83D\uDE00"';
      echo json_decode($var); 
?>

answered Feb 03 '17 at 11:07

B. Desai

score 0 · Answer 3 · answered Feb 03 '17 at 11:07

0

Please try this

  <?php 
     $var = "\uD83D\uDE00";
     echo json_decode($var);
     echo $var;
    ?>

answered Feb 03 '17 at 11:07

3 Answers3