using json_encode to return multiple data from the server

Question

I have a problem on my codes that uses jquery ajax to execute the php script and returns the data back to the browser. I want to return more one value from the 'echo' output of my php script. So I use json_encode to return the data inform of arrays. But my codes here does not return any value and no responses happening in my browser. Here is my codes.

// AJAX Code To Submit Form

// Get the value from selectbox
var so = $(this).find('option:selected').text();

var dataString = 'so1='+ so;

$.ajax({
    type: "POST",
    url: "functions/ajaxsubmit.php",
    data: dataString,
    cache: false,
    success: function(response){
        var pos1_text = $("#pos1").text(response.a);
    }
});

and heres my php code ajaxsubmit.php

<?php
include("connection.php");

//Fetching Values from URL
$sign1 = $_POST['so1'];

//Display Signing Position
$sql = "SELECT * FROM signing_officer WHERE signing_incharge = '$sign1'";
$result = mysql_query($sql);

while($row = mysql_fetch_array($result)) {
    $dbSigning_id = $row['signing_id']; 
    $dbSigning_position = $row['signing_position'];                                                 
}

$output = array('a' => $dbSigning_id, 'b' => $dbSigning_position);
echo json_encode($output, JSON_FORCE_OBJECT);

?>

Can anyone help me find the bug with this code? Really needed this.. Thankyou

Have you tried inspecting the request itself using your browser's developer tools/a third party extension like Firebug? Also, I'll have to check out the jQuery docs, but it's probable that jQuery's AJAX function doesn't automatically parse stringified JSON data, in which case, you'll have to use JSON.parse() — Brandon Anzaldi
– Brandon Anzaldi, Commented Jan 2, 2015 at 23:36

cram2208 · Accepted Answer · 2015-01-02 23:54:10Z

1

While I agree with @Aram Tchekrekjian comment, I think the mistake is on jQuery because your AJAX request is not made accordingly to receive JSON data

Two ways of achieving so are (using ajax function):

$.ajax({
    type: "POST",
    url: "functions/ajaxsubmit.php",
    data: dataString,
    cache: false,
    success: function(response){
        var pos1_text = $("#pos1").text(response.a);
    },
    dataType: 'json'
});

Or (by using post function):

$.post("functions/ajaxsubmit.php", dataString, function(response){
    var pos1_text = $("#pos1").text(response.a);
}, 'json');

Voila !

Additionally, I am targeting @Aram Tchekrekjian answer because I feel it completes mine:

$dbSigning_ids = array();
$dbSigning_positions = array();
while($row = mysql_fetch_array($result)) {
    $dbSigning_ids[] = $row['signing_id']; 
    $dbSigning_positions[] = $row['signing_position'];                                                 
}

$dbSigning_array = array($dbSigning_ids,$dbSigning_positions);
echo json_encode($dbSigning_array);

Latter seems to solve your issue, but I also see mistakes in the jQuery implementation.

Cheers,

edited Jan 2, 2015 at 23:54

answered Jan 2, 2015 at 23:48

cram2208

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1 Comment

Larcy Over a year ago

Thanks .. I see I did not specify what type of data the ajax function is trying to receive.. It's in the dataType:'json' . Works like charm !

Aram Tchekrekjian · Accepted Answer · 2015-01-02 23:44:05Z

0

You are only returning the last elements in your fetched array, you need to define arrays for your fields, and then put them into another container array then echo it through json_encode:

$dbSigning_ids = array();
$dbSigning_positions = array();
while($row = mysql_fetch_array($result)) {
    $dbSigning_ids[] = $row['signing_id']; 
    $dbSigning_positions[] = $row['signing_position'];                                                 
}

$dbSigning_array = array($dbSigning_ids,$dbSigning_positions);
echo json_encode($dbSigning_array);

Also, Brandon's comment is quite important, that you need to inspect the network from browser dev tools and see if you are getting the json response.

answered Jan 2, 2015 at 23:44

Aram Tchekrekjian

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