How to get String value from json_decode in php

Question

My json-data is as follows -

[{"name": "ram"}]

What I want is the value of name in a variable e.g., $fname. I have tried -

<?php
$jsondata = '[{"name": "ram"}]';
//$obj = json_decode($jsondata);
$obj = json_decode($jsondata,true);
print_r($obj); // This line outputs as :- Array ( [0] => Array ( [name] => ram ) ) 
//What I need is value of key
print_r($obj['name']);
foreach ($obj as $k=>$v){
echo $v;
}
?>

But I am unable to get desired output.

print_r($obj); And you will see the structure of your decoded json string. (Wait a minute and you will have tones of vampires here) — Rizier123
– Rizier123, Commented Apr 4, 2015 at 17:10
@Rizier123 it shows Array ( [0] => Array ( [name] => ram ) ) using print_r($obj), but I don't know how to use it, to get value of name — Dr. Atul Tiwari
– Dr. Atul Tiwari, Commented Apr 4, 2015 at 17:12
@Rizier123, I just read the manual. I think in layman language, it can be said that - output of my decoded json is Array inside an array. thanks for the link — Dr. Atul Tiwari
– Dr. Atul Tiwari, Commented Apr 4, 2015 at 17:44
^ You got it! That's it! This is also why I let you look at the structure of the decoded json: Array ( [0] => Array ( [name] => ram ) ) <-- See 2x array! So it's multidimensional. And the first dimension is 0 then name — Rizier123
– Rizier123, Commented Apr 4, 2015 at 17:48

Caal Saal VI · Accepted Answer · 2015-04-04 17:44:59Z

1

Here how to get that value

<?php
$jsondata = '[{"name": "ram"}]';
$obj = json_decode($jsondata,true);
//In case have multiple array
foreach ($obj as $k){
echo $k['name'];
}
//else
$obj[0]['name'];

//if 
$jsondata = '{"name": "ram"}';
$obj = json_decode($jsondata,true);
//use 
echo $obj['name'];
?>

edited Apr 4, 2015 at 17:44

answered Apr 4, 2015 at 17:18

Caal Saal VI

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3 Comments

Dr. Atul Tiwari Over a year ago

it works, but one question though, since my json-data contains only one name, wouldn't it be un-necessary to use foreach? if yes, then, what's the alternative for it? thanks.

Rizier123 Over a year ago

@Dr.AtulTiwari ^^ Look at the output again what your decoded json string prints! And read my manual link! (And that's why an answer should include an explanation)

Caal Saal VI Over a year ago

@Dr.AtulTiwari if $jsondata = '{"name": "ram"}'; you can use $obj['name'] or you can use $obj[0]['name'] for your current jsondata

Tom Fenech · Accepted Answer · 2015-04-04 17:42:36Z

1

As your output indicates, your JSON string represents an array containing one object. As you know that you want a value contained within the first element, you can get it directly:

$jsondata = '[{"name": "ram"}]';
$obj = json_decode($jsondata, true);
$name = $obj[0]['name'];
echo $name;

answered Apr 4, 2015 at 17:42

Tom Fenech

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1 Comment

Dr. Atul Tiwari Over a year ago

thanks for the answer. But @Call Sall VI has pointed out the same in his comment and modified the answer too accordingly (can't select 2 best answers), though I can upvote it :)

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