Ajax not being called by select - php

Hi I have a select box that when it is changed I want the value in a database to be updated via Ajax. Using the console I can see that my saveedit2.php file is not being called.
Select Box
<form><select id="workingpattern">
<?php
if(isset($workingpatterns) && !empty($workingpatterns)){
foreach($workingpatterns as $k4=>$v4) {
?>
<option value="<?php echo $workingpatterns[$k4]["workingpatternid"]; ?>">
<?php echo $workingpatterns[$k4]["text"]; ?></option>
<?php }}?>
</select></form>
Ajax:
<script>
$(document).ready(function(){
$('#workingpattern').change(function(){
var e = document.getElementById("workingpattern");
var value = e.options[e.selectedIndex].value;
$.ajax({
url: "saveedit2.php",
type: "post",
data: value,
success: function(data) {
console.log(data);
}});
});
</script>
SaveEdit2.php
<?php
require_once("connect_db.php");
$value=$_POST['value'];
$sql = "UPDATE employmenthistory SET workingpatternid = '$value' WHERE employmenthistoryid=1";
$result = mysqli_query ($dbc, $sql) or die(mysqli_error ($dbc));
?>

There are a few issues that I see. First, I would use 'this' to get the element and use jQuery to get the value since you are using it already. Secondly, you need a name for the value in the data set:
$('#workingpattern').change(function(){
var value = $(this).val();
$.ajax({
url: "saveedit2.php",
type: "post",
data: 'value='+value,
success: function(data) {
console.log(data);
}
});
});

Try
Ajax
$('#workingpattern').change(function(){
var value = $("#workingpattern").val();
$.ajax({
dataType: "json",
url: "./saveedit2.php",
data: {'value':value},
success: function(data){
if(data['result']=="ok")
alert("Done");
else
alert("Error");
}
});
SaveEdit2.php
<?php
require_once("connect_db.php");
$ajax_result = "error";
$value=$_POST['value'];
$sql = "UPDATE employmenthistory SET workingpatternid = '$value' WHERE employmenthistoryid=1";
$result = mysqli_query ($dbc, $sql) or die(mysqli_error ($dbc));
if($result)
$ajax_result = "ok";
echo json_encode(array('result'=>$ajax_result));
?>

Related

Passing dataType as json failing to execute

I am trying to perform the below SQL query on page load via JQuery/AJAX. It will not post when I write dataType: 'json', and does not provide an error in the console log.
I have tested the SQL is executing by placing a echo statement. Without the line it echos, with the line it does not.
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
$(document).ready(function() {
//Load Questions
var getQuestions= true;
$.ajax({
type: "POST",
url: "comment.php",
context: document.body,
data:{getQuestions:getQuestions},
dataType: 'json', //<--This line here
success: function(data){
$("#updateDisplay").html(data);
}
});
});
SQL (comment.php)
<?php
//connect to db
include 'connect.php';
$getQuestions = $_POST['getQuestions'];;
if($getQuestions==TRUE){
$sqlQuery = "SELECT *
FROM Questions
ORDER BY QuestionID DESC";
$runQuery = mysql_query($sqlQuery) or die(mysql_error());
echo 'Test echo';
echo json_encode($runQuery);
}
After the I have a DIV to place the results:
<div id="updateDisplay">
//PHP loop will go here
I wasn't correctly handling the SQL output to match JSON.
Fixed SQL
$runQueryArray=array();
$sqlQuery = "SELECT *
FROM Questions
ORDER BY QuestionID DESC";
$runQuery = mysql_query($sqlQuery) or die(mysql_error());
while (($row = mysql_fetch_array($runQuery, MYSQL_ASSOC)) !== false){
$runQueryArray[] = $row;
}
echo json_encode($runQueryArray);
}
Also needed to fix my JSON call:
$(document).ready(function() {
var getQuestions= true;
$.ajax({
type: "POST",
url: "comment.php",
context: document.body,
data:{getQuestions:getQuestions},
dataType: 'JSON',
success: function(JSON){
$.each(JSON,function(i,val){
$('#updateDisplay').append('<p>QuestionID: '+ val.QuestionID + ' Title: '+ val.Title+ ' Description: '+ val.Description+' Date Created: '+val.DateCreated+'</p><p><button type="submit" class="postReply" value='+val.QuestionID+'>Reply</button></p>');
});
}
});
});

Ajax get value from php

Ajax
$(document).ready(function(){
$("#diklat").change(function(){
var diklat = $("#diklat").val();
$.ajax({
url: "function.php",
data: {'action': 'diklat'},
cache: false,
success: function(msg){
$("#angkatan").html(msg);
}
});
});
PHP
$get_action = $_GET['action'];
if($get_action=='diklat'){
$diklat = $_GET['diklat'];
$angkatan = mysql_query("SELECT id,name FROM batches WHERE IdMasterDiklat='$diklat' order by id");
echo "<option>-- Pilih Angkatan --</option>";
while($p = mysql_fetch_array($angkatan)){
echo "<option value=\"".$p['id']."\">".$p['name']."</option>\n";
}
}
The value didnt include in my ajax, ajax only read echo. how to get that value
You have written data: {'action': 'diklat'} but it should dilkat without quotes as its variable so in php you will get value in $_GET['action'].
Ajax
$(document).ready(function(){
$("#diklat").change(function(){
var diklat = $("#diklat").val();
$.ajax({
url: "function.php",
data: {'action': diklat},
cache: false,
success: function(msg){
$("#angkatan").html(msg);
}
});
});
PHP
if($_GET['action'] == 'diklat'){
$diklat = $_GET['action'];
$angkatan = mysql_query("SELECT id,name FROM batches WHERE IdMasterDiklat='$diklat' order by id");
echo "<option>-- Pilih Angkatan --</option>";
while($p = mysql_fetch_array($angkatan)){
echo "<option value=\"".$p['id']."\">".$p['name']."</option>\n";
}
}
You should pass 2 variables, one action and another id (diklat)
Ajax:
$(document).ready(function(){
$("#diklat").on('change', function(){
var diklat = $("#diklat").val();
$.ajax({
type: "POST",
url: "function.php",
data: {'action': 'diklat', 'diklat':diklat},
cache: false,
success: function(msg){
$("#angkatan").html(msg);
}
});
});
PHP
$action = isset($_POST['action']) ? $_POST['action'] : '';
if ($action == 'diklat')
{
$diklat = isset($_POST['diklat']) ? $_POST['diklat'] : '';
$angkatan = mysql_query("SELECT id, name FROM batches WHERE IdMasterDiklat='$diklat' order by id");
echo "<option>-- Pilih Angkatan --</option>";
while($p = mysql_fetch_array($angkatan))
{
echo "<option value=\"".$p['id']."\">".$p['name']."</option>\n";
}
}
Two things:
For a GET request, to ask a PHP server, data should be URL-encoded (as you see here, query string is just appended to the URL...) So tell 'data' : '?action=diklat' ...
EDIT : Jquery automatically converts object to Url-encoded query string... So It work with array ! I'm confusing with angular's $http...
And also, the parameter will be in $_GET['action'] (because action is parameter name, and diklat the value... So PHP convert query string to associative array, with parameters names as keysn and values as values...

Return JSON from PHP to ajax on button click

I have a page with list of buttons, when each button is clicked, it's value is captured and ajax call in made. PHP does DB updates on ajax call. I want to return data to ajax call. The data is obtained from DB. But I'm unable to point out what's the error in below code.
Here is PHP code:
if (isset($_GET['val']))
{
$chapter_id=$_GET['val'];
$sql= "SELECT file_name,edit_link,name,email FROM `chapter_list` where chapter_id='$chapter_id'";
$result = mysql_query($sql,$rst);
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
$vol_name = $row["name"];
$vol_email= $row["email"];
$vol_link= $row["edit_link"];
}
$update=mysql_query("UPDATE `chapter_list` SET `status` = '$update_status' WHERE `chapter_list`.`chapter_id` = '$chapter_id';");
header('Content-Type: application/json');
echo json_encode(array("name"=>$vol_name,"email"=>$vol_email,"link"=>$vol_link));
}
Here is the AJAX request
$(document).ready(function(){
$('.btn').click(function(){
var clickBtnValue = $(this).val();
$.ajax ({
url: '',
data: { val : clickBtnValue},
dataType:'JSON',
success: function(res) {
alert(res.name);
}
});
});
});
I'm not getting the alert!
Try like this.
Maybe response data is null.check your php code(query lines).
Here My php code is :
if (isset($_GET['val'])) {
$vol_name = 'dummy_name';
$vol_email = 'dummy_email';
$vol_link = 'dummy link';
header('Content-Type: application/json');
echo json_encode(array("name"=>$vol_name,"email"=>$vol_email,"link"=>$vol_link));
exit;
}
My javascriptcode is :
<input type="text" class="btn" value="test" />
<script type="text/javascript">
if('addEventListener' in document){
document.addEventListener("DOMContentLoaded", function(e){
//dom loaded
$(document).on("click",".btn",function(e){
e.preventDefault()
var e_val = $(this).val();
console.log('my value is :' + e_val);
if(e_val){
$.ajax({
type: "get",
dataType: 'json',
url: 'here your url or slash',
data: { // request e_val
val : e_val,
}
}).done(function(xhr) {
// console.log(xhr);
if(xhr.name){
alert('response data is '+ xhr.name);
}
})
}
})
},false)
}
</script>
try this..
while($row = mysql_fetch_assoc($result))
{
$vol_name = $row["name"];
$vol_email= $row["email"];
$vol_link= $row["edit_link"];
$ret[$vol_name]= array(
'email'=>$vol_email,
'link'=>$vol_link
);
}
then use in the return statement..
echo json_encode($ret);
You can send parameters in HTML
<button class="btn" atribute_id="21543">Button</button>
$(document).ready(function() {
$('.btn').click(function() {
var Value_of_Btn= $(this).attr("atribute_id"); <-------
$.ajax({
url: '',
data: {
val: clickBtnValue
},
dataType: 'JSON',
success: function(res) {
alert(res.name);
}
});
});
});

retrieve mysql result to jquery

My problem is that I can't retrieve the result of the mysql result via ajax, please help
ajax code:
$.ajax({
type: "POST",
url: "do_find_courses.php",
//data:{question_id:question_id,answer:answer},
data:{user_id:user_id}, dataType:'json',
success:function(msg) {
alert ('asdasd')
// $("#quiz_form,#demo1").addClass("hide");
// $('#result').show();
$('p').html(msg);
}
});
PHP code:
$final=array();
$sql_courses=mysql_query("SELECT course_id, course_name FROM course") or die (mysql_error());
$row_courses = mysql_fetch_array($sql_courses);
$result=$row_courses['course_name'];
//array_push($final,$result);
//print_r($result);
echo json_encode($result);
Change PHP Code as below
$final = array();
$sql_courses = mysql_query("SELECT course_id, course_name FROM course") or die(mysql_error());
$row_courses = mysql_fetch_array($sql_courses);
echo json_encode($row_courses);
change php code as below:
$.ajax({
type: "POST",
url: "do_find_courses.php",
//data:{question_id:question_id,answer:answer},
data: {
user_id: user_id
},
dataType: 'json',
success: function (msg) {
$('p').html(msg.course_name);
}
});
Its better to send it with a key value like this:
And its better to use console.log(variable); to check variable's content
ajax code:
$.ajax({
type: "POST",
url: "do_find_courses.php",
//data:{question_id:question_id,answer:answer},
data:{user_id:user_id}, dataType:'json',
success:function(msg) {
alert ('asdasd');
console.log(msg);//You should check output of this in browser console
// $("#quiz_form,#demo1").addClass("hide");
// $('#result').show();
$('p').html(msg.cname);
}
});
PHP code:
$final=array();
$sql_courses=mysql_query("SELECT course_id, course_name FROM course") or die (mysql_error());
$row_courses = mysql_fetch_array($sql_courses);
$result=$row_courses['course_name']; // this will have first_coures_name (an string)
$final['cname']=$result;
//print_r($result);
echo json_encode($final); //the output should be this {'cname':'first_course_name'}

how can i create a success back function?

$(function() {
$(".follow").click(function(){
var element = $(this);
var I = element.attr("id");
var info = 'id=' + I;
$.ajax({
type: "POST",
url: "listen.php",
data: info,
success: function(){}
});
$("#follow"+I).hide(); ///showing the remove button after the data has been entered
$("#remove"+I).show();
return false;
});
});
The PHP file listen.php
<?php session_start();
include_once ('includes/connect.php');
$id = $_POST['id'];
$follower = $_SESSION['user_id'];
$registerlistener = mysql_query("INSERT INTO relationships (leader, listener) VALUES('".$id."', '".$follower."')");
?>
what I want to do is when I click the follow button, I want to check if the data has been entered into the database, before showing the remove button, basically checking on the background.
mysql_query will return TRUE or FALSE. You can echo that from the PHP script, and have the ajax call read it.
listen.php:
<?php session_start();
include_once ('includes/connect.php');
$id = $_POST['id'];
$follower = $_SESSION['user_id'];
$registerlistener = mysql_query("INSERT INTO relationships (leader, listener) VALUES('".$id."', '".$follower."')");
echo json_encode(array('response'=>$registerlistener));
?>
In your JavaScript:
$.ajax({
type: "POST",
url: "listen.php",
data: info,
dataType: 'json',
success: function(data){
if(data.response){
// mysql_query returned TRUE
$("#follow"+I).hide();
$("#remove"+I).show();
}
else{
// FALSE
}
}
});
If you want, you can use the $.post shorthand:
$.post('listen.php', info, function(data){
if(data.response){
// mysql_query returned TRUE
$("#follow"+I).hide();
$("#remove"+I).show();
}
else{
// FALSE
}
}, 'json');
Put the code you want to execute inside your 'success' callback function.
$.ajax({
type: "POST",
url: "listen.php",
data: info,
success: function(){
$("#follow"+I).hide();
$("#remove"+I).show();
}
});
do it like this:
listen.php
<?php session_start();
include_once ('includes/connect.php');
$id = $_POST['id'];
$follower = $_SESSION['user_id'];
if($registerlistener = mysql_query("INSERT INTO relationships (leader, listener) VALUES('".$id."', '".$follower."')")):
echo "true";
else:
echo "false";
endif;
?>
pass parameter in success function, example "msg". whatever was echo'ed in listen.php will be in the msg variable now
success: function(msg){}
if(msg == "true")
{
//do something
}
else
{
//show error message
}
});

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