Warning: json_decode() expects parameter 1 to be string, array given

Question

Here is the code:

<? $fields = json_decode($this->item->extra_fields); 
$allowed_fields= array('3','5');  
if ( $fields != NULL ) { 
foreach ( $fields as $field ) { 
if ( in_array($field->id, $allowed_fields ) && $field->value != NULL) {  
?>
<span class = "list_for<? echo $field->id; ?>">
<? echo $field->value;?></span><? }   } }?>

When I change json_decode to json_encode I get the following error:

Warning: Invalid argument supplied for foreach()...
----------

Everything worked before the upgrade of PHP to 5.4 from 5.3.

I am not a developer so please try to not be too technical. Thanks in advance!

it already says in the error message, feed it with a json string, but instead you fed it with an array — Kevin
– Kevin, Commented Sep 9, 2014 at 12:08

Wezy · Accepted Answer · 2014-09-09 12:13:19Z

2

as stated in the manual, do it like this

$fields = json_decode($this->item->extra_fields, true);

When TRUE, returned objects will be converted into associative arrays.

$fields will be an array and the foreach() will work

answered Sep 9, 2014 at 12:13

Wezy

6671 gold badge5 silver badges14 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Dale Over a year ago

Was just about to answer this myself, good answer and should keep the rest of the code working

JimL Over a year ago

inside the foreach he uses field as an object, he will have to rewrite it if changing to an assoc array.

Manos Krokos Over a year ago

I added the true bit, but I still get the original error message.

Wezy Over a year ago

exactly what is in '$this->item->extra_fields', what do you get when you execute var_dump($fields); ?

Paul Roub · Accepted Answer · 2017-06-11 03:02:57Z

1

Test this. before sending to PHP change your jsonObj to string.

jsonObj{..}

jsonObj2=JSON.stringify(jsonObj);
$.ajax({
   type: "POST",
   url:"Methods.php",
   data : { 'Json_data':jsonObj2},
   success: function(response) {
    console.log("Result:" + response);
   }
});

edited Jun 11, 2017 at 3:02

Paul Roub

36.5k27 gold badges88 silver badges95 bronze badges

answered Jun 11, 2017 at 2:55

SW_IR

7095 silver badges5 bronze badges

Comments

Ahmed Saeed · Accepted Answer · 2014-09-09 12:47:37Z

0

Note that json_decode accepts string and according to the above error you are trying to pass an array to it so if you need to iterate on the fields array you may want to ignore json_decode as it seems that you already have an array.

json_encode accepts array but it gives a string so it give you a warning when you try to iterate on sting

so try the below code.

<? $fields = $this->item->extra_fields; 
$allowed_fields= array('3','5');  
if ( $fields ) { 
foreach ( $fields as $field ) { 
if ( in_array($field->id, $allowed_fields ) && $field->value != NULL) {  
?>
<span class = "list_for<? echo $field->id; ?>">
<? echo $field->value;?></span><? }   } }?>

answered Sep 9, 2014 at 12:47

Ahmed Saeed

566 bronze badges

Comments

René Höhle · Accepted Answer · 2014-09-09 12:14:40Z

0

You can't check if its NULL.

<? $fields = json_decode($this->item->extra_fields); 
$allowed_fields= array('3','5');  
if($fields) {
foreach ( $fields as $field ) { 
if ( in_array($field->id, $allowed_fields ) && $field->value != NULL) {  
?>
<span class = "list_for<? echo $field->id; ?>">
<? echo $field->value;?></span><? }   } }?>

Try if($fields) {

You get true if he can decode the json string.

answered Sep 9, 2014 at 12:14

René Höhle

27.4k22 gold badges78 silver badges91 bronze badges

Comments

WordpressCoder · Accepted Answer · 2014-09-09 12:41:38Z

0

<?php 
$fields = json_decode($this->item->extra_fields); 
$allowed_fields= array('3','5');  
if ( !empty($fields) ) { 
  foreach ( $fields as $key => $field ) { 
    if ( in_array($field->id, $allowed_fields ) && !empty($field->value)) {  
   ?>
      <span class = "list_for<?php echo $field->id; ?>">
      <?php echo $field->value;?></span>
  <?php 
  }   
 } 
} 
?>

Please copy paste this code and check that error remove or not. You can't check the $fields with NULL.So you have to check it only by using !empty($fields)

answered Sep 9, 2014 at 12:41

WordpressCoder

16

Collectives™ on Stack Overflow

Warning: json_decode() expects parameter 1 to be string, array given

5 Answers 5

4 Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

4 Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related