How to assign ajax response to Variables - php

Question

I have drop down Select box as follows
<?php
$sql = "SELECT scheduleName FROM schedule";
$result = mysqli_query($link,$sql);
echo "<select name='schedule' id='schedule'>";
echo "<option value=''>-- Select Schedule --</option>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['scheduleName'] . "'>" . $row['scheduleName'] . "</option>";
}
echo "</select>";
?>
And I have a file called processClg.php as follows
<?php
include "config.php";
if ($_POST['type']=='POST')
{
$qry = "SELECT * FROM schedule WHERE scheduleName LIKE 'Row id of drop down selection'";
$res = mysqli_query($link,$qry);
}
?>
How can I call processClg.php file on $("#schedule").change(function ());by assigning Row id of drop down selection as where condition.
Update
Am getting Response from processClg.Php as follows
[{"id":"2","scheduleName":"shanth","subject":"Patho","university":"Dali","facultyName":"Dr","scheduleStartDate":"2015-06-05","scheduleEndDate":"2015-06-09"}]
How to assign response values from ajax call to the following Php variables
<?php
$scheduleStartDate = '';
$scheduleEndDate = '';
?>
Any help my greatly appreciated.

Answer 1

$("#schedule").change(function() {
var value = $('#schedule option:selected').text();
var ajaxCheck = $.ajax({
url: 'processClg.php',
type: 'POST', // had mention post bcoz u mention in processClg.php
dataType: 'json', // processClg.php will return string means change to text
data: { id: value },
success: function(data){
console.log('success');
itrToRead(data);
}
});
});
function itrToRead(data) {
$(data).each(function(key, value){
console.log('key is: '+key+' and value is: '+value);
});
}
processClg.php
<?php
include "config.php";
if ($_POST['type']=='POST') {
$qry = "SELECT * FROM schedule WHERE scheduleName LIKE '".$_POST['id']."'";
$res = mysqli_query($link,$qry);
echo $res;
}
?>

Answer 2

You can call ajax as follows:
var RowId;
$.ajax({
type: "POST",
async: false,
url: url,
data: postdata,
//dataType: "json",
success: function (data) {
RowId = data;
}
});
The fact is that to assign response to variable is pass the parameter async: false,
Then its work.

Answer 3

I guess you have your dropdown in your rendered page and you want to send the selected value to the php page:
$("#schedule").change(function(){
var val2pass = $(this).find(':selected').val(); // get the value
$.ajax({
url: 'processClg.php',
type:'post', // <-----you need to use post as you are using $_POST[]
data: { rowid : val2pass }, //<---pass the value
success: function(data){
itrToRead(data);
}
});
});
So now on the php side you need to do this:
<?php
include "config.php";
if ($_POST['type']=='POST')
{
$qry = "SELECT * FROM schedule WHERE scheduleName LIKE '".$_POST['rowid']."'";
$res = mysqli_query($link,$qry);
}
?>

Answer 4

Your procellClg.php will be
<?php
include "config.php";
if (isset($_REQUEST['qid']))
{
$qry = "SELECT * FROM schedule WHERE scheduleName LIKE '".$_REQUEST['qid']."'";
$result = mysqli_query($link,$qry);
echo "<select name='schedule' id='schedule'>";
echo "<option value=''>-- Select Schedule --</option>";
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['scheduleName'] . "'>" . $row['scheduleName'] . "</option>";
}
echo "</select>";
}
?>
and then make Ajax function call
$("#schedule").change(function() {
val = $(this).val();
$.ajax({
type: "POST",
async: false,
url: 'processClg.php',
data: {qid:val},
success: function(data){
$(this).html(data);
}
});
});