Post php form jquery ajax - php

Question

i have form with input field generated from loop, ie
<form name="" method='post' action=''>
<?php
for($i=0;$i<10;$i++){
echo "<input type='text' name='data[]' class='data_cls' value='".$i."'>";
}
?>
<input type='submit' id='btn' value='save'>
i want to submit the form using jquery ajax.
$('.btn').click(function(){
var datString = "HOW TO GET THESE VALUES";
$.ajax({
url: "data_process.php",
type: "post",
data: dataString,
success: function(data) {
alert('OK');
}
});
});

Answer 1

You can use .serialize() jQuery method to get the form data. Like this,
$('.btn').click(function(){
var dataString = $('#FORM_ID').serialize();
//replace FORM_ID with the ID of the form.
$.ajax({
url: "data_process.php",
type: "post",
data: dataString,
success: function(data) {
alert('OK');
}
});
});

Answer 2

Try this:
var datString = "";
$(".data_cls")
.map(function () {
datString += $(this).val();
})
.get();

Answer 3

Try
var Data = {};
$('input[name=data]').each(function(i) {
Data[i] = $(this).val();
});

Answer 4

I think you have forgot to bind parameter in data ajax post
$('.btn').click(function(){
var datString = "'HOW TO GET THESE VALUES'";
$.ajax({
url: "data_process.php",
type: "post",
data: "'USERPARAMETR' : " dataString,
success: function(data) {
alert('OK');
}
});
});
If you not use parameter in data_process.php page. Please use it.
Please update and check

Answer 5

First give the form an ID or class i.e. form
<form name="" method='post' action='' id='form'>...
Then you may capture all inputs from the from using the general serialize method
<script type="text/javascript">
$(document).on("click", ".btn", function(e) {
e.preventDefault();
var datastring = $("#form").serialize();
{
$.ajax({
url: "data_process.php",
type: "post",
data: dataString,
...
And on the PHP page you will receive the data as posted
$values = $_POST['data'];
$N = count($values);
for($i=0; $i < $N; $i++)
{
$currentvalue = $values[$i];
//now this value will loop and you will do with it the way you want to
//For example
$save = mysqli_query($link, "UPDATE table SET `column` = '$currentvalue' WHERE ...") or die(mysqli_error($link));
}