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

Question

What causes this error in my code?

$query = $this->db->query("SELECT * FROM tour_foreign ORDER BY id desc");
        $data = array();
        foreach ($query->result() as $row)
            $data[] = array('guide' => $row->guide);

            echo json_decode($data); //Line 167

error:

erro: json_decode() expects parameter 1 to be string, array given: Line Number: 167

UPDATE:

If I use json_encode instead of json_decode, my output is this:

[{"guide":["\u0633\u06cc\u062f \u0633\u0639\u06cc\u062f \u062f\u0627\u062f\u0627\u0634\u0632\u0627\u062f\u0647"]},{"guide":["\u0633\u06c‌c\u062f \u0633\u0639\u06cc\u062f \u062f\u0627\u062f\u0627\u0634\u0632\u0627\u062f\u0647"]},{"guide":null}]

They are persian words.

You are passing an array where a string is expected. You are creating the array in the line before - is that necessary? Why not just use $row->guide directly? — Pekka
– Pekka, Commented Sep 15, 2011 at 20:15

gen_Eric · Accepted Answer · 2012-04-15 14:46:46Z

55

I think you want json_encode, not json_decode.

edited Apr 15, 2012 at 14:46

user142162

answered Sep 15, 2011 at 20:15

gen_Eric

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7 Comments

Jennifer Anthony Over a year ago

if use of json_encode my output is this:

[{"guide":["\u0633\u06cc\u062f \u0633\u0639\u06cc\u062f \u062f\u0627\u062f\u0627\u0634\u0632\u0627\u062f\u0647"]},{"guide":["\u0633\u06cc\u062f \u0633\u0639\u06cc\u062f \u062f\u0627\u062f\u0627\u0634\u0632\u0627\u062f\u0647"]},{"guide":null}]

, they are persian word

gen_Eric Over a year ago

@Selena: The \u0633 characters are just Unicode characters that are encoded. The first guide is سید سعید داداشزاده. Is that right?

gen_Eric Over a year ago

@Selena: Show it where? If you echo the string (in JavaScript) they will print correctly. You don't need to do anything special, just echo them normally.

Jennifer Anthony Over a year ago

i not want echo they in JavaScript, What do you mean of just echo them normally!?

gen_Eric Over a year ago

@Selena: What are you trying to do here? What do you want to do with the words?

|

Kareem · Accepted Answer · 2019-04-25 09:51:27Z

34

Set decoding to true

Your decoding is not set to true. If you don't have access to set the source to true. The code below will fix it for you.

$WorkingArray = json_decode(json_encode($data),true);

edited Apr 25, 2019 at 9:51

answered Oct 19, 2014 at 11:46

Kareem

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Comments

macintosh264 · Accepted Answer · 2011-09-15 20:42:53Z

3

json_decode() is used to decode a json string to an array/data object. json_encode() creates a json string from an array or data. You are using the wrong function my friend, try json_encode();

answered Sep 15, 2011 at 20:42

macintosh264

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Ayman Elshehawy · Accepted Answer · 2019-12-10 11:04:45Z

3

Make an object

$obj = json_decode(json_encode($need_to_json));
Show data from this $obj

$obj->{'needed'};

Reference

answered Dec 10, 2019 at 11:04

Ayman Elshehawy

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Akhil Jain · Accepted Answer · 2015-06-17 04:58:59Z

1

here is the solution for similar problem which i was facing while extracting name from user profile facebook json object

$uname=json_encode($userprof);
$uname=json_decode($uname);
echo "Welcome " . $uname -> name  ;

edited Jun 17, 2015 at 4:58

answered Feb 28, 2012 at 7:17

Akhil Jain

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ZaMy · Accepted Answer · 2020-06-11 15:15:30Z

0

Ok I was running into the same issue. What I failed to notice is that I was using json_decode() instead of using json_encode() so for those who are going to come here please make sure you are using the right function, which is json_encode()

Note: Depends on what you are working on but make sure you are using the right function.

edited Jun 11, 2020 at 15:15

answered Jun 11, 2020 at 14:41

ZaMy

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