getting data from PHP file with json is failing - php

Question

I can't get each data from PHP file, i'm always having AJAX failure, i tried many things, and looked for some pages for this problem but i can't find any solution, this is where i came last.
This is my jQuery function:
$(document).ready(function () {
$(function () {
$('a[class="someclass"]').click(function(){
var somedata = $(this).attr("id");
$.ajax({
type: "POST",
url: "foo.php",
data: {"id": somedata},
dataType:"json",
success: function(data){
e.preventDefault();
$("#data1").html(data[0]);
$("#data2").html(data[1]);
$("#data3").html(data[2]);
$("#data4").html(data[3]);
},
error:function(){
alert("AJAX request was a failure");
}
});
});
});
});
This is my PHP file:
$data = $_POST['id'];
$con = mysqli_connect('localhost','root','','database');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"database");
$sql="SELECT * FROM table WHERE id = '".$data."'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result)) {
$jsondata1 = array($row['data1'], $row['data2'], $row['data3'], $row['data4']);
echo json_encode($jsondata1);
}
mysqli_close($con);
I think there is no need to share HTML file, but if you want i can share with you.
Thank You!

Answer 1

You are overriding the array value in the loop and echo $jsondata1, so you are sending many different arrays with the echo inside the loop, try this code:
$jsondata1 = array();
while($row = mysqli_fetch_array($result)) {
$jsondata1[] = array($row['data1'], $row['data2'], $row['data3'], $row['data4']);
}
echo json_encode($jsondata1);

Answer 2

e.preventDefault(); is a problem. e is not defined and if it was the default action would have already occurred by the time the success function was called. Try
$('a[class="someclass"]').click(function(e){
e.preventDefault();
var somedata = $(this).attr("id");
$.ajax({
type: "POST",
url: "foo.php",
data: {"id": somedata},
dataType:"json",
success: function(data){
$("#data1").html(data[0]);
$("#data2").html(data[1]);
$("#data3").html(data[2]);
$("#data4").html(data[3]);
},
error:function(){
alert("AJAX request was a failure");
}
});
});
});
});

Answer 3

You need to encode the complete response.
when you echo each encoded it will not work.
Do something like
$jsonArray = array();
while(){
array_push($jsonArray, array($row['data1'], $row['data2']);
}
echo json_encode($jsonArray);

Answer 4

You're calling json_encode multiple times when what you need to be doing is assigning those intermediate variables to an array and encode that array using json encode for the return.
PHP:
$data = $_POST['id'];
$con = mysqli_connect('localhost','root','','database');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"database");
$sql="SELECT * FROM table WHERE id = '".$data."'";
$result = mysqli_query($con,$sql);
$jsondata1 = [];
while($row = mysqli_fetch_array($result)) {
$jsondata1[] = array($row['data1'], $row['data2'], $row['data3'], $row['data4']);
}
mysqli_close($con);
die(json_encode($jsondata1));
JavaScript:
$(document).ready(function () {
var doc = $(document);
doc.on('click', 'a.someclass', function(e){
e.preventDefault();
var somedata = $(this).attr("id");
$.ajax({
type: "POST",
url: "foo.php",
data: {"id": somedata},
dataType:"json",
success: function(data){
$("#data1").html(data[0]);
$("#data2").html(data[1]);
$("#data3").html(data[2]);
$("#data4").html(data[3]);
},
error:function(){
alert("AJAX request was a failure");
}
});
});
});