Develop Reference

Prevent echo line from AJAX response

I want to prevent an echo from an AJAX response. I have 2 buttons and I need to enable and disable them using JS by AJAX responses. The JS code to enable/disable the HTML elements has been already written inside the PHP if condition of the AJAX URL Page. From AJAX I can display the results on <span id="dupmsg"></span>.
The result will be "Already Exists" and "Not Exist". I only want to display the message and enable/disable the buttons based on the condition. Here it's not working:
Index Page in php:
<h2>Enabling and Disabling text field using JavaScript</h2>
<form id="registration-form">
Enter your name: <input type="text" id="name">
</form>
<button onclick="disable()">Disable the text field</button>
<button onclick="enable()">Enable the text field</button>
<p>Ajax Response is: <span id="dupmsg"></span></p>
<script>
function check_dup()
{
var barcode=$("#memb_barcode").val();
$.ajax({
type: 'POST',
url: "ajax_attendance.php",
data: {
barcode: barcode
},
success: function(msg)
{
//alert(msg); // your message will come here.
$('#dupmsg')
.css('color', 'red')
.html(msg)
},
error: function(jqxhr, status, exception) {
alert('Exception:', exception);
}
})
}
</script>
Ajax URL Page:
<?php
$reg_no = mysqli_real_escape_string($con, $_POST['reg_no']);
$barcode = mysqli_real_escape_string($con, $_POST['barcode']);
$sql = "SELECT id from tblstudent where reg_no = '$reg_no' && barcode like '$barcode' ";
$query = mysqli_query($con, $sql);
$ecount = mysqli_num_rows($query);
if($ecount!=0)
{
printf("Already Exists");
echo'
<script>
function disable() {
document.getElementById("name").disabled = true;
}
</script> ';
}
else
{
printf("Not Exists");
echo'
<script>
function enable() {
document.getElementById("name").disabled = false;
}
</script> ';
}
?>
The problem is that the JS written inside the PHP echo is reflecting back to span id="dupmsg" with AJAX response. I don't want to bring it in AJAX response. Please help.

<h2>Enabling and Disabling text field using JavaScript</h2>
<form id="registration-form">
Enter your name: <input type="text" id="name">
</form>
<button id="disable" onclick="disable()">Disable the text field</button>
<button id="enable" onclick="enable()">Enable the text field</button>
<p>Ajax Response is: <span id="dupmsg"></span></p>
<script>
function check_dup()
{
var barcode=$("#memb_barcode").val();
$.ajax({
type: 'POST',
url: "ajax_attendance.php",
data: {
barcode: barcode
},
success: function(msg)
{
if(msg=='true'){
document.getElementById("name").disabled = true;
$("#disable").attr("disabled", true); // write the id of the button u want to hide
$('#dupmsg')
.css('color', 'red')
.html("Already Exists")
}
else if(msg=='false')
{
document.getElementById("name").disabled = false;
$("#enable").attr("disabled", false); // write the id of the button u want to hide
$('#dupmsg')
.css('color', 'red')
.html("Not Exists")
}
},
error: function(jqxhr, status, exception) {
alert('Exception:', exception);
}
})
}
function enable() {
document.getElementById("name").disabled = false;
}
function disable() {
document.getElementById("name").disabled = true;
}
</script>
ajax_attendance.php
<?php
$reg_no = mysqli_real_escape_string($con, $_POST['reg_no']);
$barcode = mysqli_real_escape_string($con, $_POST['barcode']);
$sql = "SELECT id from tblstudent where reg_no = '$reg_no' && barcode like '$barcode' ";
$query = mysqli_query($con, $sql);
$ecount = mysqli_num_rows($query);
if($ecount!=0)
{
return true;
}
else
{
return false;
}
?>

Inactive and active PHP MySQL

Having problem to change the state if its active or inactive everytime I clicked one of them nothings happened.
This is my code on ajax.
<script src="//code.jquery.com/jquery-1.10.2.min.js"></script>
<script type="text/javascript">
$(document).on('click','.status_checks',function(){
var status = ($(this).hasClass("btn-success")) ? '0' : '1';
var msg = (status=='0')? 'Deactivate' : 'Activate';
if(confirm("Are you sure to "+ msg)){
var current_element = $(this);
url = "ajax.php";
$.ajax({
type:"POST",
url: url,
data: {id:$(current_element).attr('data'),status:status},
success: function(data)
{
location.reload();
}
});
}
});
</script>
my php code ajax.php
<?php $db= new mysqli('localhost','root','password','dbname');
extract($_POST);
$user_id=$db->real_escape_string($id);
$status=$db->real_escape_string($status);
$sql=$db->query("UPDATE user SET status='$status' WHERE id='$id'");
echo 1;
?>
this is my code for displaying it, on this part everything works fine, when the value are 0 it will display inactive and 1 for active, however when clicking the status theres nothing happen only the notification and reload the page.
<td><i data="<?php echo $user['id'];?>" class="status_checks btn
<?php echo ($user['status'])?
'btn-success': 'btn-danger'?>"><?php echo ($user['status'])? 'Active' : 'Inactive'?>
</i></td>

Try to configure your code to:
$(document).on('click','.status_checks',function(){
var status = '1';
var msg = 'Activate';
if($(this).hasClass("btn-success")){
status = '0';
msg = 'Deactivate';
}
if(confirm("Are you sure to "+ msg)){
var id= $(this).data('id');
url = "/ajax.php";
$.ajax({
type:"POST",
url: url,
data: {id:id,status:status},
dataType: "json",
success: function(data)
{
console.log(data);
location.reload();
}
});
}
});
<?php $db= new mysqli('localhost','root','password','dbname');
$user_id=$_POST['id'];
$newStatus=$_POST['$status'];
$sql = "UPDATE user SET status=".$newStatus." WHERE id=".$user_id."
";
if($db->query($sql) === TRUE){
echo json_encode(1);
}else{
echo json_encode(0);
}
?>
<td><i data-id="<?php echo $user['id'];?>" class="status_checks btn
<?php echo ($user['status'])?
'btn-success': 'btn-danger'?>"><?php echo ($user['status'])? 'Active' :
'Inactive'?>
</i></td>

Generic function to update DIV

In my display page I have a script that updates a div from another php page.
How do I re-write the below script to:
Be a function I can call with updateadiv(divid,content) in my display page (in php) where content is a php variable. (the script will then not call the php page but take the input variable).
<!DOCTYPE html>
<html>
<body>
function updatediv(divId, content)
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script>
(function($)
{
$(document).ready(function()
{
$.ajaxSetup(
{
cache: false,
beforeSend: function() {
$('#content').hide();
$('#loading').show();
},
complete: function() {
$('#loading').hide();
$('#content').show();
},
success: function() {
$('#loading').hide();
$('#content').show();
}
});
var $container = $("#content");
$container.load("http://192.168.1.90/json_output.php");
var refreshId = setInterval(function()
{
$container.load('http://192.168.1.90/json_output.php');
}, 9000);
});
})(jQuery);
</script>
<?php
function createarray() {
global $reindexed_devices_a ;
$feed_url = "http://192.168.1.90/JSON?request=getstatus&ref=all&location1=all&location2=all";
//$feed_url = "demoxml.xml";
$json = file_get_contents($feed_url);
$devices_a = json_decode($json, true);
//echo $devices_a[Devices][2][name] ;
//echo "<BR> ReIndexed:";
$reindexed_devices_a = array(Devices => array());
foreach ($devices_a[Devices] as $value) {
$reindexed_devices_a[Devices][$value[ref]] = $value;
}
//echo $reindexed_devices_a[Devices][59][name] ;
//need to do some conditional formatting before updating to DIV's
if ($reindexed_devices_a[Devices][59][name] == 'Coffee maker') {
$reindexed_devices_a[Devices][59][name] = "overwritten";
}
echo time();
//echo $reindexed_devices_a[Devices][59][name] ;
}
createarray();
$name59=$reindexed_devices_a[Devices][59][name];
//check if $name59 has changed - if yes update div
updatediv('59');
echo "This is a test <div id = 'name59'>.....</div>";
?>
</body>
</html>

function updateadiv(DivID , Url){
$('#loading').hide();
$('#' + DivID ).html('');
$.ajax({
url:url,
success:function(rData){
$('#' + DivID ).html(rData);
}
});
}

How to update database thruogh dropdown without page reload

Please look at this code first:
$(document).ready(function() {
$("#alternatecolor [type=button]").each(function() {
$(this).on('click', function() {
btnObj = $(this);
rowId = $(this).attr("rowId");
changeStatus = $(this).attr("changeStatus");
$.get("changeStatus.php?rowId="+rowId+"&changeStatus="+changeStatus,function(data,status){
if(changeStatus == 0){
str ="Unverify";
btnText = "Verify";
newStatus = 1;
}
else{
str ="Verify";
btnText = "Unverify";
newStatus = 0;
}
if(data == 'success'){
alert("Status updated successfully to "+str+".");
btnObj.val(btnText);
btnObj.attr("changeStatus",newStatus);
}
else{
alert("some error");
}
});
});
});
});
</script>
Here is my change status page:
$dbhost = 'xxxxxx';
$dbuser = 'xxxxxxx';
$dbpass = 'xxx';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db('xxxxxxxx');
$sql ="update experiment set verification=".$_GET['changeStatus']." where row=".$_GET['rowId'];
$retval = mysql_query( $sql, $conn );
if(!($retval))
{
die('error');
}
else{
echo "success";
}
mysql_close($conn);
I was using a button in this code to query my database with values 0,1. If pressed once, database queried with 1, if pressed again, database queried with 0.
Now, I have to put a dropdown in place of button with 3 values to query database with: 0,1,2. If selected first value, database row to be updated with value 0 and so on.
How would I do that?

remove button and add select like this:
<select class="my-select">
<option value="0">update</option>
<option value="1">create</option>
<option value="2">delete</option>
</select>
you also need to add data-rowId attribute.
jquery:
$(document).ready(function() {
$(".my-select").change(function() {
btnObj = $(this);
rowId = $(this).attr("data-rowId");
changeStatus = $(this).val();
$.get("changeStatus.php?rowId="+rowId+"&changeStatus="+changeStatus,function(data,status){
if(data == 'success'){
alert("Status updated successfully to "+str+".");
}
else{
alert("some error");
}
});
});
});

update
create
delete
You can do using class name or ID. If it's id then do $("#id1").click(function () {... in below code.
Awlad has done using GET method, below is using POST method.
$(function () {
$(".my-select").click(function () {
var category = $(this).text();
//$('label').css('color', selText);
$.ajax({
url: "invite_db.php",
type: "POST",
data: {"category": category},
success: function(data) {
$(".articleContent").html(data);
//setInterval("messageDisable()", 5000);
}
});
});
});
I have used parameter name according to my code. Please you change it

Here is your HTML and Javascript code
<select onchange="getval(this);">
<option value="1">One</option>
<option value="2">Two</option>
</select>
<script type="text/javascript">
function getval(sel) {
$.ajax({
type: "POST",
url: url, // url is page where you want to process this post request
data: 'val=' + sel.value,
});
}
</script>
On PHP side
<?php
// Your database connection code
$val = $_POST['val'];
mysql_query("UPDATE 'tablename' SET 'column name' = $val 'Your WHere caluse'")
?>

Error in Ajax submit

I am new to Yii and ajax I want to read some inputs for a model from view and save it by using ajax. I used the following code inside form
<input type="button" name="save" value="save" onclick="saveFile()" id="profile-update" class="btn button-main" live="false">
and saveFile() is
function saveFile()
{
var data;
data = new FormData();
data.append('file', $('#UserProfile_profile_picture')[0].files[0]);
data.append('UserProfile', $('#profile-update-form').serialize());
$.ajax({
url: '<?php echo CHtml::normalizeUrl(array('user/profileupdate?rand=' . time())); ?>',
type:"POST",
data: data,
processData: false,
contentType: false,
success: function (data) {
$("#AjaxLoader1").hide();
if(data.status=="success"){
$("#formResult1").html("Profile settings changed successfully.");
$("#profile-update-form")[0].reset();
}
else{
$.each(data, function(key, val) {
$("#profile-update-form #"+key+"_em_").text(val);
$("#profile-update-form #"+key+"_em_").show();
});
}
},
beforeSend: function(){
$("#AjaxLoader1").show();
}
}
)
return false;
}
and the code in controller is
$profile = UserProfile::model()->findByAttributes(array('user_id' => Yii::app()->user->id));
if (!$profile) {
$profile = new UserProfile;
$profile->create_time = time();
$profile->update_time = time();
}
if (isset($_POST['UserProfile'])) {
$profile->attributes = $_POST['UserProfile'];
$profile->profile_picture=$_FILES['file']['name'];
$images = CUploadedFile::getInstance($profile,'profile_picture');
// print_r($_POST);
// print_r($profile->phone);
// print_r($images);
// exit();
if (isset($images))
{
if(!is_dir(Yii::getPathOfAlias('webroot').'/images/profilepic/'. 'quy'))
{
mkdir(Yii::getPathOfAlias('webroot').'/images/profilepic/'. $profile->profile_picture);
// the default implementation makes it under 777 permission, which you could possibly change recursively before deployment, but here�s less of a headache in case you don�t
}
foreach ($images as $image => $pic)
{
echo $pic->name;if ($pic->saveAs(Yii::getPathOfAlias('webroot').'/images/profilepic/'.$pic->name))
{
$profile->profile_picture = $pic->name;
}
}
}
$profile->user_id = Yii::app()->user->id;
$profile->update_time = time();
$valid = $profile->validate();
$error = CActiveForm::validate(array($profile));
if ($error == '[]') {
$profile->save(false);
echo CJSON::encode(array('status' => 'success'));
Yii::app()->end();
} else {
$error = CActiveForm::validate(array($profile));
if ($error != '[]')
echo $error;
Yii::app()->end();
exit();
}
}
But here only the profile_picture is saving to the database all other fields are not changing. and the profile picture is not copying into the folder($images is blank) Please somebody help me to solve this problem. Thanks in advance

The code $profile->attributes = $_POST['UserProfile']; doesnt worked so i send it seperately by data.append('UserProfile[about_me]', $('#UserProfile_about_me').val());
data.append('UserProfile[city]', $('#inputCity').val());
data.append('UserProfile[phone]', $('#inputPhone').val()); and in controller i used $profile->profile_picture=$_FILES['file']['name'];
$profile->about_me = $_POST['UserProfile']['about_me'];
$profile->city = $_POST['UserProfile']['city'];
$profile->phone = $_POST['UserProfile']['phone']; I know this is not the correct way but may be helpful for someone who is hurry.