ajax calls return php array, json doesnt seem correct

Question

I have an ajax call that fires a php script that returns an array.

the php:

 $errors[1] = "you didn't enter name"; 
 $errors[2] = "your email is incorrect";
 $errors[3] = "You didnt enter password";
 echo json_encode($errors);

the javascript:

 .....
 datatype:'json',
 success: function(result)
 {
 alert(result);   
 }

I would expect to see:

 {"1":"you didn't enter name","2":"your email is incorrect","3":"You didnt enter password"}

instead i see: ["you didn't enter name","your email is incorrect","You didnt enter password]

it seems like the json_encode did something funky. what am i missing

Because what you wanted was $errors['1'] and what you have is $errors[1]. — aziz punjani
– aziz punjani, Commented May 10, 2013 at 18:09

wroniasty · Accepted Answer · 2013-05-10 19:13:30Z

1

Try this:

echo json_encode($errors, JSON_FORCE_OBJECT);

http://www.php.net/manual/en/json.constants.php

Also, yout PHP script should contain:

header("Content-Type: application/json");

before any echo statements.

edited May 10, 2013 at 19:13

answered May 10, 2013 at 18:01

wroniasty

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7 Comments

bart2puck Over a year ago

thanks @wroniasty, this seemed to help. how i have: {"0":"you didn't enter name","1":"your email is incorrect","2":"You didnt enter password"} if i try to ititerate through it: for (i=0;i<array.length;i++) when i print array[i] (array[0] = { array[1] = " array[2] = 0 i get each individual character, not what is in "0"....make sense?

wroniasty Over a year ago

To iterate over an Object use: for (var key in obj) { var element = obj[key]; console.log(element); }

bart2puck Over a year ago

I know i am noobing this all up. i get same result with the for loop. the console log shows { then " then 0 then " and so on...its almost like the javascript doesnt see it as an array but a long string.

wroniasty Over a year ago

Perhaps it IS a string, what does console.log(typeof result) output?

bart2puck Over a year ago

string....if i alert(result) i get this {"0":"firstName must contain a value.", "1":"lastName must contain a value.", "2":"callIdfirstName must contain a value.", "3":"callIdlastName must contain a value.", "4":"street must contain a value.", "5":"city must contain a value."} my code on the php page. echo json_encode($errors, JSON_FORCE_OBJECT); I even tried php::json_last_error() and got a 0.

|

Chris · Accepted Answer · 2013-05-10 18:00:50Z

1

Since you just used numbers as keys, it assumed you wanted an Array.

answered May 10, 2013 at 18:00

Chris

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Comments

Neftalí Yagua · Accepted Answer · 2016-07-17 04:24:56Z

1

Then setting this wrong, because that way will fail, by undefined vars.

$errors[1] = "you didn't enter name"; 
$errors[2] = "your email is incorrect";
$errors[3] = "You didnt enter password";
echo json_encode($errors);

Don't use this array configuration, instead use:

$errors=array(NULL,
  "you didn't enter name",
  "your email is incorrect",
  "You didnt enter password"
);
echo json_encode($errors, JSON_FORCE_OBJECT);

edited Jul 17, 2016 at 4:24

answered May 10, 2013 at 18:20

Neftalí Yagua

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