Develop Reference

How to upload image while using serialize Array in JavaScript?

I am working on a form which need to insert some data with an image without reloading the page. In my side everything is working fine when there is no image filed in the form, but if there is an <input type="file"> in my form, then my code is not passing the file/image information. Can you guys please teach me what to add in my code to upload or pass image please?
My Code
HTML Form
<form action="action.php" method="POST" enctype="multipart/form-data" id="myform">
<input type="hidden" value="access" name="label">
<input type="text" name="one">
<input type="text" name="two">
<input type="file" name="image">
<button type="submit" id="mybtn">Add Data</button>
</form>
<div id="myresult"></div>
My JavaScript
$('#mybtn').click( function(){
$.post(
$('#myform').attr('action'),
$('#myform:input').serializeArray(),
function(result) {
// Some Stuff...
}
);
});
My PHP
include 'database.php';
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
$label = ["access"];
if (in_array($_POST['label'], $label)) {
switch ($_POST['label']) {
case 'access':
$one = $_POST['one'];
$two = $_POST['two'];
$file_name = $_FILES['image']['name'];
$file_size = $_FILES['image']['size'];
$file_temp = $_FILES['image']['tmp_name'];
$error = [];
$valid = [];
$flag1 = $flag2 = $flag3 = false;
if (!empty($one)) {
// Some validation
$flag1 = true;
} else {
$flag1 = false;
}
if (!empty($two)) {
// Some validation
$flag2 = true;
} else {
$flag2 = false;
}
if (!empty($file_name)) {
// Some validation
$flag3 = true;
} else {
$flag3 = false;
}
if ($flag1 && $flag2 && $flag3) {
// move_uploaded_file() + Insert All data
if ($result) {
$valid[] = "Data added successfully!";
} else {
$error[] = "Please try again later!";
}
} else {
$error[] = "Something went wrong!";
}
// ALERT MESSAGE [error] and [valid]
if (!empty($error)) {
foreach ($error as $value) {
echo json_encode($value);
}
}
if (!empty($valid)) {
foreach ($valid as $value) {
echo json_encode($value);
}
}
break;
default:
# code...
break;
}
}
}
This code works perfectly without reloading page while there are no input type file. I want to know what code I have to add in my JavaScript section to execute the code successfully with an input type file.

$('#dataBtnIMG').click( function(){
var form = $('#dataFormIMG');
var formData = new FormData($('#dataFormIMG')[0]);
$.ajax({
url: form.attr('action'),
type: form.attr('method'),
data: formData,
cache: false,
contentType: false,
processData: false,
success:function(result) {
// Some Stuff
}
});
});

onsubmit return false is not working

The following script shows the error message correctly, but the form always submits whether confirm_shop_code() returns true or false. I tried in many ways to solve the bug but it still persists. I have to stop the form from submitting when it returns false, but allow it to submit when it returns true. Please can any one help me to solve this?
<h2 id="shop_data"></h2>
<!-- form -->
<form action="" class="form-horizontal form-label-left input_mask" method="post" onsubmit="return confirm_shop_code();">
<div class="col-md-4 col-sm-4 col-xs-8 form-group">
<input type="text" class="form-control" id="shop" name="code" value="<?php echo $account->code; ?>" placeholder="Enter Shop Code">
</div>
</form>
<!-- validation script -->
<script>
function confirm_shop_code(){
var code=document.getElementById( "shop" ).value;
if(code) {
$.ajax({
type: 'post',
url: 'validations.php',
data: {
shop_code:code,
},
success: function (response) {
$( '#shop_data' ).html(response);
if(response=="OK") {
return true;
} else {
return false;
}
}
});
} else {
$( '#shop_data' ).html("");
return false;
}
}
</script>
<!-- php code -->
<?php
include "system_load.php";
$code = $_POST['shop_code'];
global $db;
$query = "SELECT code from accounts WHERE code='".$code."'";
$result = $db->query($query) or die($db->error);
$count = $result->num_rows;
if($count > 0) {
echo "SHOP CODE already Exists";
} else {
echo "OK";
}
exit;
?>

The reason it is submitting is because AJAX calls are asynchronous by default. I wouldn't suggest making it synchronous because this will block the rest of the javascript execution. Also, you are returning false from the success method of $.ajax. This is not in the same scope as the parent function and therefore does not also cause the parent function to return false. So in fact, your confirm_shop_code() function is not returning anything unless code is false and that's why your form is always being submitted, no matter what happens with the AJAX call.
I would recommend using jQuery to bind to the form's submit event and just disable form submitting with preventDefault(). First, just add an id attribute to the form (e.g. "yourform") and do something like:
$("form#yourform").submit(function(e) {
e.preventDefault();
var form = $(this);
var code=document.getElementById( "shop" ).value;
if(code) {
$.ajax({
type: 'post',
url: 'validations.php',
data: {
shop_code:code,
},
success: function (response) {
$( '#shop_data' ).html(response);
if(response=="OK") {
form.unbind('submit').submit()
}
}
});
} else {
$( '#shop_data' ).html("");
}
});

You need to add async:false to your ajax code
function confirm_shop_code(){
var code=document.getElementById( "shop" ).value;
var stopSubmit = false;
if(code) {
$.ajax({
type: 'post',
url: 'validations.php',
async:false,
data: {
shop_code:code,
},
success: function (response) {
$( '#shop_data' ).html(response);
if(response=="OK") {
stopSubmit = false;
} else {
stopSubmit = true;
}
}
});
} else {
$( '#shop_data' ).html("");
stopSubmit = true;
}
if(stopSubmit){
return;
}
}

You should call return false; function on the click event of the submit button.
<button type="submit" id="submit" onclick="return false;" class="btn btn-primary col-4">Proceed</button>
or you can use:
document.getElementById("submit").addEventListener("click", function (e) {
//your logic here
//this return false will not work here
return false;
//this will work
e.preventDefault();
});

JQuery Display Alert Message on Form Submit

I am trying to display alert messages on jquery from the client side. The jquery will be called once the submit button is clicked. The form then will call the server side php. Here is my code:
FORM
<form action="branch_add_verifier.php" method="POST" id="formAdd">
<input type="text" name="id" id="id">
<input type="submit" value="submit">
</form>
JQUERY
$(document).ready(function(){
var $form = $('#formAdd');
$form.submit(function(){
var id= $("#id").val();
if (id.length < 12) {
alert("INPUT ERROR");
return false;
}
$.post($form.attr('action'), $(this).serialize(), function(response){
alert("DATA SUCCESSFULLY ADDED");
},'json');
return false;
});
});
But the alert message does not pop up inside the $.postmethod
And I also want to know how I can pop up the alert message from the server side. Here is my sample code:
SERVER SIDE
<?php $query = mysqli_query($conn, "SELECT * FROM table1
INNER JOIN table2
ON table1.col1= table2.col1
WHERE table2.col3= '".$_REQUEST['id']."'");
if (mysqli_num_rows($query) != 0) {
echo "<script>alert('ERROR')</script>";
return false;
} ?>
In summary, the code above works but I need to display messages that would tell me if the query is successful or not. Thanks
My new problem is that the code below bring me to another page:
FORM
<form action="branch_add_verifier.php" method="POST" id="formAdd">
<input type="text" name="id" id="id">
<input type="submit" value="submit">
</form>
JQUERY
$(document).ready(function(){
$('#formAdd').on('submit', function (e) {
e.preventDefault();
var id= $("#id").val();
if (id.length < 12) {
alert("INPUT ERROR");
return false;
}
$.ajax({
context: this,
url: $(this).attr('action'),
type: 'POST',
data: new FormData(this),
dataType: 'json'
}).done(function (data) {
if(data == 'ok') {
alert("DATA SUCCESSFULLY ADDED");
}
if(data == 'no') {
alert("ERROR");
}
}).fail(function (data) {
console.log('failed');
});
});
});
SERVER
$query = mysqli_query($conn, "SELECT * FROM table1
INNER JOIN table2
ON table1.col1= table2.col1
WHERE table2.col3= '".$_REQUEST['id']."'");
if (mysqli_num_rows($query) != 0) {
mysqli_close($conn);
echo json_encode('no');
return false;
}
I need to return after the json_encode because there are still methods below that.

HTML Form
<form action="branch_add_verifier.php" method="POST" id="formAdd">
<input type="text" name="id" id="id">
<input type="submit" value="submit">
</form>
Script :
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script>
$(function() {
$(document).on('submit', "#formAdd", function(e) {
e.preventDefault();
$.ajax({
url: $(this).attr('action'),
type: "post",
data: $(this).serialize(),
error:function(){
alert("ERROR : CANNOT CONNECT TO SERVER");
},
success: function(data) {
alert(data);
}
});
return false;
});
});
</script>
PHP server side like this:
<?php
$insert = mysqli_query($conn, "insert query here");
if($insert) {
echo json_encode('ok');
} else {
echo json_encode('no');
}
?>

Just you need to put id="id" in input type text.
<input type="text" name="id">
Working Code
$(document).ready(function(){
var $form = $('#formAdd');
$form.submit(function(){
var id= $("#id").val();
if (id.length < 12) {
alert("INPUT ERROR");
return false;
}
$.post($form.attr('action'), $(this).serialize(), function(response){
alert("DATA SUCCESSFULLY ADDED");
},'json');
return false;
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form action="branch_add_verifier.php" method="POST" id="formAdd">
<input type="text" id="id" name="id">
<input type="submit" value="submit">
</form>

Can not get ajax callback to a function

I have a form for user to register new account. I use jquery + ajax to check availability of email address on form submission. In Jquery code I used e.preventDefault(); to prevent form submission if there is any error occurs. I tried the existed email address in the email input and click submit the form. It allows form to submit. It should not do this because ajax reponseText return true means that the email address is already existed in database.
Could anyone please tell me how to fix my code so that if ajax response returns true, it will prevent form submission and shows up errors.
I tried to read and follow this article but fails after so many attempts.
Here is my form:
<form role="form" method="post" id="signupForm" action="index.php?view=signup-gv">
<div class="col-xs-6 border-right">
<div class="form-group">
<label for="exampleInputEmail1">Full Name</label>
<input type="text" class="form-control" id="regname" name="regname" placeholder="Full Name">
</div>
<div class="form-group">
<label for="exampleInputEmail1">Email Address</label><span id="emailcheck"></span>
<input type="email" class="form-control" id="regemail" name="regemail" placeholder="Enter email">
</div>
</div>
<div class="form-group col-xs-6">
<label for="exampleInputPassword1">Password</label>
<input type="password" class="form-control" id="regpass" name="regpass" placeholder="Password">
</div>
<button style="position:relative; left: 15px; top: 10px;" class="btn btn-default" name="register" id="register">Register</button>
</form>
Here my jquery code:
$(document).ready(function(){
$('#regname').focus();
$('#signupForm').submit(function(e) {
var regname = $('#regname');
var regemail = $('#regemail');
var regpass = $('#regpass');
var register_result = $('#register_result');
register_result.html('Loading..');
if(regname.val() == ''){
regname.focus();
register_result.html('<span class="errorss"> * Full name can not be blank</span>');
e.preventDefault();
}
else if ($.trim(regemail.val()).length == 0) {
regemail.focus();
register_result.html('<span class="errorss">* Email address can not be blank</span>');
e.preventDefault();
}
else if(regpass.val() == ''){
regpass.focus();
register_result.html('<span class="errorss">* Password can not be blank</span>');
e.preventDefault();
}
emailCheck().done(function(r){
if(r){
$('#regemail').focus();
$('#register_result').html('<span class="errorss"> This email address is already existed. Please choose another one </span>');
e.preventDefault();
}
});
});
});
function emailCheck() {
var regemail = $('#regemail');
var emailcheck = $('#emailcheck');
emailcheck.html('');
var UrlToPass = {regemail:regemail.val()} ;
$.ajax({
type : 'POST',
cache: false,
data : UrlToPass,
url : 'emailcheck.php',
success: function(responseText){
if(responseText == 0){
return false; // good to go
}
else{
emailcheck.html('<span class="errorss"> This email is existed.</span>');
return true; // This email is registered. Please try different one
}
}
});
}

First you are not returning anything from the emailCheck() function, but you are using it as if it is returning a promise object.
So
$(document).ready(function () {
$('#regname').focus();
$('#signupForm').submit(function (e) {
var regname = $('#regname');
var regemail = $('#regemail');
var regpass = $('#regpass');
var register_result = $('#register_result');
register_result.html('Loading..');
//prevent the form submit
e.preventDefault();
if (regname.val() == '') {
regname.focus();
register_result.html('<span class="errorss"> * Full name can not be blank</span>');
} else if ($.trim(regemail.val()).length == 0) {
regemail.focus();
register_result.html('<span class="errorss">* Email address can not be blank</span>');
} else if (regpass.val() == '') {
regpass.focus();
register_result.html('<span class="errorss">* Password can not be blank</span>');
} else {
emailCheck().done(function (r) {
if (r) {
$('#regemail').focus();
$('#register_result').html('<span class="errorss"> This email address is already existed. Please choose another one </span>');
} else {
$('#signupForm')[0].submit();
}
});
}
});
});
function emailCheck() {
var regemail = $('#regemail');
var emailcheck = $('#emailcheck');
emailcheck.html('');
var UrlToPass = {
regemail: regemail.val()
};
var deferred = jQuery.Deferred();
$.ajax({
type: 'POST',
cache: false,
data: UrlToPass,
url: 'emailcheck.php',
success: function (responseText) {
if (responseText == 0) {
deferred.resolve(false);
} else {
emailcheck.html('<span class="errorss"> This email is existed.</span>');
deferred.resolve(true);
}
},
error: function () {
deferred.reject();
}
});
return deferred.promise();
}

You are confusing yourself with sync and async functions. An ajax function makes an Async call and returns output in its callback. You are trying to wrap an Async function inside a normal function and expecting it to behave synchronously.
Your function returns before the Ajax call receives its output. Use
async: false
$.ajax({
type : 'POST',
cache: false,
async: false,
data : UrlToPass,
Refer to following for dettails:
How to make JQuery-AJAX request synchronous

PHP/JavaScript login script refreshing when form is submitted

I've created a login page using the following JavaScript:
function handleLogin() {
var form = $("#loginForm");
//disable the button so we can't resubmit while we wait
//$("#submitButton",form).attr("disabled","disabled");
var e = $("#username", form).val();
var p = $("#password", form).val();
console.log("click");
if(e != "" && p != "") {
//var str = form.serialize();
//alert(str);
$.ajax({
type: 'POST',
url: 'http://localhost/php/log.php',
crossDomain: true,
data: {username: e, password :p},
dataType: 'json',
async: false,
success: function (response){
alert ("response");
if (response.success) {
alert("you're logged in");
window.localStorage["username"] = e;
window.localStorage["password"] = md5(p);
//window.localStorage["UID"] = data.uid;
window.location.replace("http://www.google.co.uk");
}
else {
alert("Your login failed");
//window.location("main.html");
}
},
error: function(error){
//alert(response.success);
alert('Could not connect to the database' + error);
window.location = "index.html";
}
});
}
else {
//if the username and password is empty
alert("You must enter username and password");
}
return false;
}
and the PHP in log.php is:
$link = mysql_connect("$host", "$username", "$password") or die("Could not connect to host.");
mysql_select_db("$db_name", $link) or die("Could not find database.");
$uname = isset($_POST['username']) ? $_POST['username'] : '';
$password = isset($_POST['password']) ? $_POST['password'] : '';
$sql = "SELECT * FROM user WHERE username = '$uname' AND password = '$password'";
$result=mysql_query($sql);
$num_row=mysql_num_rows($result);
$row=mysql_fetch_array($result);
if (is_object($result) && $result->num_rows == 1) {
$response['success'] = true;
}
else
{
$response['success'] = false;
}
echo json_encode($response);
//echo 'OK';
I would expect the page to show some error if the combination is wrong, or redirect to a different page if it's correct. However it just refreshes with the username/password in the url.
Log in form
<form id="loginForm" method="post">
<div data-role="fieldcontain" class="ui-hide-label">
<label for="username">Username:</label>
<input type="text" name="username" id="username" value="" placeholder="Username" />
</div>
<div data-role="fieldcontain" class="ui-hide-label">
<label for="password">Password:</label>
<input type="password" name="password" id="password" value="" placeholder="Password" />
</div>
<input type="submit" value="Login" id="submitButton">
</form>

Since you're using jQuery (version 1.10.1), you can bind the click event directly to your forms submit button
$(document).on('click', '#submitButton', function(e) {
e.preventDefault();
... See detailed jsFiddle example ...
return false;
});
With preventDefault(), you remove the default submit behavior of the Submit button, so the page won't be reloaded after form submit.
Also use form attributes like action to provide ajax URL, or method, for setting ajax request type.
P.S This type of binding won't work for jQuery 1.6
jsFiddle example

Please change your html container as follows
<div data-role="fieldcontain" class="ui-hide-label">
<label for="username">Username:</label>
<input type="text" name="username" id="username" value="" placeholder="Username" />
</div>
<div data-role="fieldcontain" class="ui-hide-label">
<label for="password">Password:</label>
<input type="password" name="password" id="password" value="" placeholder="Password" />
</div>
<input type="button" value="Login" id="submitButton" onclick="handleLogin()">
Since, you are submitting data using ajax, then you dont need to writt "form" tag in your html
Total page would be refresh when click "Submit" button. so I have changed as "button" and provide the ajax method in "onclick" attribute
Sample PHP file (index.php) that I tried without connecting DB for testing yours scenario
<?php
if($_POST){
if($_POST['username'] == "admin" && $_POST['password'] == "admin"){
$response['success'] = true;
}else{
$response['success'] = false;
}
echo json_encode($response);
exit;
}
?>
modified your js code
function handleLogin() {
var e = $("#username").val();
var p = $("#password").val();
if(e != "" && p != "") {
$.ajax({
type: 'POST',
url: 'index.php',
crossDomain: true,
data: {username: e, password :p},
dataType: 'json',
async: false,
success: function (response){
alert ("response");
if (response.success) {
alert("you're logged in");
}
else {
alert("Your login failed");
}
},
error: function(error){
alert('Could not connect to the database' + error);
}
});
}
else {
alert("You must enter username and password");
}
return false;
}

Try this: if(response.responseText)