not able to carry the form values through post method - php

Question

There is a form where the user enters a number and according to the condition applied on the number, a list of addresses are displayed. I would like to store the data that is returned through AJAX. The code on the page that has a form:
index.php
<script>
$(document).ready(function() {
$("#phone").keyup(function() {
var number = $("#phone").val();
$.ajax({
url: "t_fetchaddr.php",
type: 'POST',
data: 'number='+number,
cache: false,
}).done(function(html) {
$('#results').html(html);
});
});
});
<script>
<form action="insert_temp.php" method="POST">
<input type="text" name="phoneno" id="phone" value="" />
<div id="results"></div>
<button class="button btn btn-primary btn-large" type="submit" name="submit" value="submit">Submit</button>
</form>
code on t_fetchaddr.php page
$val = $_REQUEST['number'];
$sql2 = "SELECT * FROM user_address where number='".$val."' ";
$result2 = mysqli_query($con, $sql2);
if (mysqli_num_rows($result2) > 0)
{ ?>
<div class="span6" >
<div class="span3">
<? while($row2 = mysqli_fetch_assoc($result2))
{ ?>
<input type="radio" name="old_address" value="<? echo $row2['address']; ?>" ><? echo $row2['address']; ?><br>
<? } ?>
</div>
</div>
<? } ?>
code on insert_temp.php page
$old_address = mysqli_real_escape_string($con, $_POST['old_address']);
echo $old_address;
Everything is working fine until displaying of the address through number, but when I submit the form it is not going to the back end. I tried to echo $old_address but got nothing.
other input values in index page inside the form are going to backend but value that is being fetched from t_fetchaddr.php page is not getting carried, Can anyone please tell where I went wrong

Answer 1

Try this and watch your console :
$(document).ready(function() {
$("#phone").keyup(function() {
var number = $(this).val();
$.ajax({
url: "t_fetchaddr.php",
type: 'POST',
data: {number:number},
cache: false,
success : function(html) {
$('#results').html(html);
},
error : function(err){
console.log(err);
}
});
});
});

Answer 2

<script>
$(document).ready(function()
{
$("#phone").keyup(function()
{
var number = $("#phone").val();
$.ajax({
url: "t_fetchaddr.php",
type: 'POST',
data: {number :number}, //modified
cache: false,
success:function(html)
{
$('#results').html(html);
}
});
});
});
</script>//Missing closing
<form action="insert_temp.php" method="POST">
<input type="text" name="phoneno" id="phone" value="" />
<div id="results"></div>
<button class="button btn btn-primary btn-large" type="submit" name="submit" value="submit" >Submit</button>
</form>
and in php
$val = $_POST['phoneno'];
$sql2 = "SELECT * FROM user_address where number='".$val."' ";
$result2 = mysqli_query($con, $sql2);
if (mysqli_num_rows($result2) > 0)
{ ?>
<div class="span6" >
<div class="span3">
<? while($row2 = mysqli_fetch_assoc($result2))
{ ?>
<input type="radio" name="old_address" value="<? echo $row2['address']; ?>" ><? echo $row2['address']; ?><br>
<? } ?>
</div>
</div>
<? } ?>
Note:
Missing closing tag </script>
And this line changed data: 'number='+number,

Answer 3

just try this code on fetchaddr.php
i have just removed the in between php tags.
<?
$val = $_REQUEST['number'];
$sql2 = "SELECT * FROM user_address where number='".$val."' ";
$result2 = mysqli_query($con, $sql2);
if (mysqli_num_rows($result2) > 0) {
echo '<div class="span6" >
<div class="span3">';
while($row2 = mysqli_fetch_assoc($result2))
{
echo '<input type="radio" name="old_address" value="'.$row2['address'].'" >'.$row2['address'].'<br>';
}
echo '</div>
</div>';
} ?>
hopefully this will solve your problem.