Problem parsing through Ajax a JSON data fetched by PHP file

Question

I have difficulty parsing JSON date from my PHP file

{"date":"20\/12\/2022","result":"£13000.00","medias":"BBC","country":"UK"}

but when I try to parse it and to see the data in the console.log - it's empty

Please help

My Ajax Function

function ajax_call(){
    const style2 = $("#style1").val();
    const radio_btn = $('input[name="drone"]:checked').val();
    if(style2==""){
        document.getElementById("error").innerHTML = "Error: Please enter style code !"; 
        return false;
    } 
    {
        $.ajax({ 
            type: 'post',
            url: "t.php",  
            data: { styles: style2 , country: radio_btn},
            dataType: "json",
            success: function(data){  
                var jsondata = $.parseJSON(data);
                console.log(jsondata);
            }
        })
    }
}

My PHP

<php

header('Content-type: application/json');
$date = "20/12/2020";
$end_result = "£13000.00";
$medias = "BBC";
$country = "UK";

$sortjson = array('date' => $date,  
                    'result' =>iconv('Windows-1252', 'UTF-8', $end_result), 
                    'medias' => $medias, 
                    'country' => $country
            );

echo json_encode($sortjson, JSON_UNESCAPED_UNICODE);
?>

Most likely you are not waiting for an answer to the promise. — Gabriel Lupu, Nov 12 '21 at 12:05
What is the output of the console.log(data) you may see it as a string — Akshay Venugopal, Nov 12 '21 at 12:06
Also try adding this to Ajax call dataType: "text" or dataType: "json" — Akshay Venugopal, Nov 12 '21 at 12:08
I dont have any output in the console.log i have the response in devtool for the ajax call which return the thing i want . — Alev Mustafa, Nov 12 '21 at 12:08
What have you tried to resolve the problem? If dumping `data` yields nothing, did you check that the request is sent properly? — Nico Haase, Nov 12 '21 at 12:48
Good code indentation would help us read the code and more importantly it will help **you debug your code** [Take a quick look at a coding standard](https://www.php-fig.org/psr/psr-12/) for your own benefit. You may be asked to amend this code in a few weeks/months and you will thank me in the end. — RiggsFolly, Nov 12 '21 at 12:49
May be worth a read https://stackoverflow.com/questions/1580647/json-why-are-forward-slashes-escaped — RiggsFolly, Nov 12 '21 at 13:07

Terry Lennox · Accepted Answer · 2021-11-12T12:49:23.673

I believe there is a typo in the PHP code, you need to change the first line from:

<php

to

<?php

If the first line is <php, the output won't be valid JSON.

Complete PHP code:

<?php

header('Content-type: application/json');
$date = "20/12/2020";
$end_result = "£13000.00";
$medias = "BBC";
$country = "UK";

$sortjson = array('date' => $date,  'result' =>iconv('Windows-1252', 'UTF-8', $end_result), 'medias' => $medias, 'country' => $country);


echo json_encode($sortjson, JSON_UNESCAPED_UNICODE);

?>

Also, I'm not 100% sure you need the iconv call, you could try the below code:

<?php

header('Content-type: application/json');
$date = "20/12/2020";
$end_result = "£13000.00";
$medias = "BBC";
$country = "UK";

$sortjson = array('date' => $date,  'result' => $end_result, 'medias' => $medias, 'country' => $country);

echo json_encode($sortjson, JSON_UNESCAPED_UNICODE);

?>

score -1 · Answer 2 · answered Nov 12 '21 at 12:12

-1

You have a error in your Success method:

use JSON.parse() method instead of $.parseJSON()

success: function(data){  

   var jsondata = JSON.parse(data);
   console.log(jsondata);
}

answered Nov 12 '21 at 12:12

Shahid Rasheed

371
4
10

Same , empty console.log – Alev Mustafa Nov 12 '21 at 12:20
$.parseJSON(data) === JSON.parse(data); – mplungjan Nov 12 '21 at 12:27
Please add some explanation to your answer such that others can learn from it. What did you change, and why? – Nico Haase Nov 12 '21 at 12:49

score -1 · Answer 3 · answered Nov 12 '21 at 12:15

-1

You have to use JSON.parse()

function ajax_call(){
const style2 = $("#style1").val();
const radio_btn = $('input[name="drone"]:checked').val();
if(style2==""){document.getElementById("error").innerHTML = "Error: Please enter style code !"; return false;} {
$.ajax({ 
    type: 'post',
    url: "t.php",  
    data: { styles: style2 , country: radio_btn},
                  dataType: "json",
                 success: function(data){  
   var jsondata = JSON.parse(data);
        console.log(jsondata);
}
})
}
}

answered Nov 12 '21 at 12:15

Sachin Sanchaniya

996
1
8
16

same thing empty console.log – Alev Mustafa Nov 12 '21 at 12:20
$.parseJSON(data) === JSON.parse(data); – mplungjan Nov 12 '21 at 12:27
Please add some explanation to your answer such that others can learn from it. What did you change, and why? – Nico Haase Nov 12 '21 at 12:48

Problem parsing through Ajax a JSON data fetched by PHP file

3 Answers3