i have the following form:
<form name="register" id="register" action="include/process_registration.php" method="post">
<div class="form_error">Ooops! There is some missing or incorrect information. Please look back over this section. </div>
<div class="left_form2" >
<div class="inner_form1">
<p>Your First Name:*</p>
<p>Your Last Name:*</p>
<p>Date of Birth:*</p>
</div>
<div class="inner_form2">
<input type="text" name="firstname" id="firstname" class="login_form2" autocomplete="off"><br/>
<input type="text" name="lastname" id="lastname" class="login_form2" autocomplete="off"><br/>
<input type="text" name="dob" id="dob" class="login_form2"><br/>
</div>
</div>
<div class="left_form2" style="text-align:right;">
<div class="inner_form1">
<p>Email Address:*</p>
<p>Confirm Email:*</p>
</div>
<div class="inner_form2">
<input type="text" name="email" id="email" class="login_form2" autocomplete="off"><br/>
<input type="text" name="email2" id="email2" class="login_form2" autocomplete="off"><br/>
</div>
</div>
<input type="submit" id="register1" name="register" value="Register" class="buttons_register">
</form>
i am then using ajax to post my form data to my mysql query process_registration.php:
Ajax:
<script type="text/javascript">
$(function() {
//alert('Document is ready');
$('#register1').click(function() {
var a = $('#firstname').html();
var b = $('#lastname').html();
var c = $('#dob').html();
var d = $('#email').html();
var f = $('#email2').html();
//alert('You picked: ' + sel_stud);
$.ajax({
type: "POST",
url: "include/process_registration.php",
data: {theOption: a, theOption2: b, theOption3: c, theOption4: d, theOption5: f},
success: function(whatigot) {
//alert('Server-side response: ' + whatigot);
$('#LaDIV').html(whatigot);
$('#theButton').click(function() {
alert('You clicked the button');
});
} //END success fn
}); //END $.ajax
}); //END dropdown change event
}); //END document.ready
</script>
my php file process_registration which contains my mysql query looks like so:
<?php
session_start();
include("config.php");
include("verify.php");
//retrieve our data from POST
$firstname = $_POST['theOption'];
$lastname = $_POST['theOption2'];
$dob = $_POST['theOption3'];
$email = $_POST['theOption4'];
$email2 = $_POST['theOption5'];
$firstname = stripslashes($firstname);
$firstname = mysql_real_escape_string($firstname);
$lastname = stripslashes($lastname);
$lastname = mysql_real_escape_string($lastname);
$email = stripslashes($email);
$email = mysql_real_escape_string($email);
$email2 = stripslashes($email2);
$email2 = mysql_real_escape_string($email2);
$dob = stripslashes($dob);
$dob = mysql_real_escape_string($dob);
include '../dependables/secure.php';
$sql = "INSERT INTO supplier_registration (id, first_name, last_name, supplier_email, supplier_password, salt, dob, date) VALUES ('', '$firstname','$lastname','$email2', '$hash', '$salt', '$dob', now())";
$result2 = mysql_query($sql);
?>
for some reason i am getting taken to the process_registation.php page on the form submit and getting undefined index errors for all my form values. Can someone please show me where i am going wrong? Thanks
You are going to that page because the action of the form says it.
You can prevent that with javascript/jquery when submiting the form and then do the ajax code.
$("#register1").on("click", function(e) {
e.preventDefault();
//rest of your code
});
Also you won't get the inputs values with .html(), should use .val()
PS: you can get less code with serialize function for your forms. Take a look on jQuery API.
Remove this from your form...
action="include/process_registration.php"
Just make it action="" if you want it to reload the current page.
You might want to actually echo something from your process_registration.php too, otherwise the "whatigot" from the success function will always be empty.
You'll want to use e.preventDefault() to stop the submit click working as it is usually expected to.
You also need to use .val() instead of .html() as you should be getting the value of the input and not the html within the input.
$('#register1').click(function(e) {
e.preventDefault();
var a = $('#firstname').val();
var b = $('#lastname').val();
var c = $('#dob').val();
var d = $('#email').val();
var f = $('#email2').val();
// AJAX Call
});
Related
Having issues pulling the values put in a textbox with ajax and posting them to another .php file. Ive done this once before with a checkbox but wasn't able to duplicate my results. Here is the code for the text boxes in questions.
<div align = "right">
<div class = ='text'>
<span style="float:right;"><strong> Status Report Date</strong>
<label for="from">From</label>
<input type="text" id="from" name="from">
<label for="to">to</label>
<input type="text" id="to" name="to">
<div id="dates"></div>
These are my date picker boxes since I attached a datepicker script to them but figured they act as normal text-boxes.
Here is the script for grabbing values from the textboxes
$(document).ready(function(){
$('input[type="text"]').click(function(){
var from = $(this).val();
$.ajax({
url:"sortByDates.php",
method:"POST",
data:{text:text},
success:function(data){
$('#dates').html(data);
}
});
});
});
</script>
Here is the .php file I am trying to send the values to.
<?php
if (isset($_GET['pageSubmit'])) {
$firstDate= $_POST['from'];
$lastDate= $_POST['to'];
echo $firstDate;
echo $lastDate;
}
I think you lost your focus... Check this code out
<div alight = "right">
<div class='text'>
<span ><strong> Status Report Date</strong>
<label for="from">From</label>
<input type="text" id="from" name="from">
<label for="to">to</label>
<input type="text" id="to" name="to">
<div id="dates"></div>
<button id="submit">click</button>
Jquery
<script>
$(document).ready(function(){
$('#submit').click(function(){
var from = $('#from').val();
var to = $('#to').val();
ps = "submit";
$.ajax({
url:"sortByDates.php",
method:"POST",
data:{pageSubmit: ps,from:from, to: to},
success:function(data){
$('#dates').html(data);
}
});
});
});
php script
<?php
if (isset($_POST['pageSubmit'])) {
$firstDate= $_POST['from'];
$lastDate= $_POST['to'];
echo $firstDate;
echo $lastDate;
}
?>
Firstly add value="" in the inputs
Then in your js code, you are sending "text" variable which is undefined as your variable is "from".
So try adding : data:{text:from}
You have not passed the values from,to and pagesubmit in ajax.Try this code hope it helps
In ajax
$(document).ready(function(){
$('input[type="text"]').click(function(){
var from = $("#from").val();
var to = $("#to").val();
$.ajax({
url:"sortByDates.php",
method:"POST",
data:{from:from,to:to,pageSubmit:1},
success:function(data){
$('#dates').html(data);
}
});
});
});
In PHP
<?php
if (isset($_GET['pageSubmit'])) {
$firstDate= $_POST['from'];
$lastDate= $_POST['to'];
echo $firstDate;
echo $lastDate;
}
?>
The variable "text" you use in the section
{text:text}
is undefined. So therefore no data will be sent.
Other than that there are other problems here, do you want both values to be sent? Your code doesn't look like it supports that. Are you sure you want this to occur when the user clicks the textbox?.
How to do coding of this Code With Using Ajax.Please Help.
I am Bignner here and i have written this code it's working but i want to use with ajax this because don't want to reload the page...?
PHP File
//Code For Making Form And getting Data…..
<html>
<body>
Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,
<form action="Form_Data.php" method="post">
ID: <input type="text" name="ID"><br><br>
NAME: <input type="text" name="NAME"><br><br>
PASSWORD: <input type="text" name="PASSWORD"><br><br>
CREDITS: <input type="text" name="CREDITS"><br><br>
E_mail: <input type="text" name="EMAIL_ID"><br><br>
CREATED_ON:<input type="text" name="CREATED_ON"><br><br>
MODIFIED_ON:<input type="text" name="MODIFIED_ON"><br><br>
<input type="submit">
</form>
</body>
</html>
//code for taking data from form data.
<html>
<?php
include 'connnect.php';
mysql_set_charset('utf8');
//query for insert data into tables
$ID = $_POST['ID'];
$NAME =$_POST['NAME'];
$EMAIL_ID =$_POST['EMAIL_ID'];
$PASSWORD =$_POST['PASSWORD'];
$CREDITS =$_POST['CREDITS'];
$CREATED_ON=$_POST['CREATED_ON'];
$MODIFIED_ON=$_POST['MODIFIED_ON'];
$query = "INSERT INTO `user_table`
(`ID`,`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
VALUES
('$ID','$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
$query_run= mysql_query($query);
$retval=mysql_query($query,$conn);
if ($query_run)
{ echo 'It is working';
}
mysql_close($conn);
?>
</html>
I have Tried Yet ... Is Blewo...
file for html and ajax
<html>
<HEAD>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
</HEAD>
<body>
<div id="status_text">
Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,
<form onsubmit="return false" method="post">
ID: <input type="text" id="ID" name="ID"><br><br>
NAME: <input type="text" id="NMAE" name="NAME"><br><br>
PASSWORD: <input type="text" id= "PASSWORD"name="PASSWORD"><br><br>
CREDITS: <input type="text" Id= "CREDITS"name="CREDITS"><br><br>
Email_ID: <input type="text" id="Email_ID"name="EMAIL_ID"><br><br>
CREATED_ON:<input type="text" id="CREATED_ON" name="CREATED_ON"><br><br>
MODIFIED_ON:<input type="text" id="MODIFIED_ON" name="MODIFIED_ON"><br><br>
<input type="submit" id="btn_submit" name="submit" value="Send">
</div>
<script>
//on the click of the submit button
$("#btn_submit").click(function(){
//get the form values
var ID = $('#ID').val();
var NAME = $('#NAME').val();
var PASSWORD = $('#PASSWORD').val();
var CREDITS = $('#CREDITS').val();
var EMAIL_ID = $('#EMAIL_ID').val();
var CREATED_ON = $('#CREATED_ON').val();
var MODIFIED_ON = $('#MODIFIED_ON').val();
//make the postdata
var postData = '&ID='+ID+'&NAME='+NAME+'&PASSWORD='+PASSWORD+'&REDITS'+CREDITS+'&EMAIL_ID'+EMAIL_ID+'&CREATED_ON'+CREATED_ON+'&MODIFIED_ON'+MODIFIED_ON;
//call your .php script in the background,
//when it returns it will call the success function if the request was successful or
//the error one if there was an issue (like a 404, 500 or any other error status)
});
$.ajax({
url : "Form_Data.php",
type: "POST",
data : postData,
success: function(data,status, xhr)
{
//if success then just output the text to the status div then clear the form inputs to prepare for new data
$("#status_text").html(data);
$('#ID').val();
$('#NAME').val('');
$('#PASSWORD').val('');
$('#EMAIL_ID').val('');
$('#CREATED_ON').val('');
$('#MODIFIED_ON').val('');
}
});
</script>
</form>
</body>
</div>
</html>
code for query...
<html>
<?php
include 'connnect.php';
mysql_set_charset('utf8');
//query for insert data into tables
$ID = $_POST['ID'];
$NAME =$_POST['NAME'];
$EMAIL_ID =$_POST['EMAIL_ID'];
$PASSWORD =$_POST['PASSWORD'];
$CREDITS =$_POST['CREDITS'];
$CREATED_ON=$_POST['CREATED_ON'];
$MODIFIED_ON=$_POST['MODIFIED_ON'];
$query = "INSERT INTO `user_table`
(`ID`,`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
VALUES
('$ID','$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
$query_run= mysql_query($query);
$retval=mysql_query($query,$conn);
if ($query_run)
{ echo 'It is working';
}
mysql_close($conn);
?>
</html>
I have Solved It ... How To Use Ajax and MYSQL...
PHP CODE
<?php
include 'connnect.php';
mysql_set_charset('utf8');
//query for insert data into tables
$ID = $_POST['ID'];
$NAME =$_POST['NAME'];
$EMAIL_ID =$_POST['EMAIL_ID'];
$PASSWORD =$_POST['PASSWORD'];
$CREDITS =$_POST['CREDITS'];
$CREATED_ON=$_POST['CREATED_ON'];
$MODIFIED_ON=$_POST['MODIFIED_ON'];
$query = "INSERT INTO `user_table`
(`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
VALUES
('$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
$query_run= mysql_query($query);
// $retval=mysql_query($query,$conn);
if ($query_run)
{
echo 'It is working';
}
mysql_close($conn);
?>
HTML FILE....
<html>
<HEAD>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
</HEAD>
<body>
<div id="status_text">
Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,
ID: <input type="text" id="ID" name="ID"><br><br>
NAME: <input type="text" id="NAME" name="NAME"><br><br>
PASSWORD: <input type="text" id= "PASSWORD"name="PASSWORD"><br><br>
CREDITS: <input type="text" Id= "CREDITS"name="CREDITS"><br><br>
Email_ID: <input type="text" id="EMAIL_ID"name="EMAIL_ID"><br><br>
CREATED_ON:<input type="text" id="CREATED_ON" name="CREATED_ON"><br><br>
MODIFIED_ON:<input type="text" id="MODIFIED_ON" name="MODIFIED_ON"><br><br>
<input type="submit" id="btn_submit" name="submit" value="Send"/>
</div>
<script>
<!--
//Browser Support Code
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
} }
//on the click of the submit button
$("#btn_submit").click(function(){
//get the form values
var ID = $('#ID').val();
var NAME = $('#NAME').val();
var PASSWORD = $('#PASSWORD').val();
var CREDITS = $('#CREDITS').val();
var EMAIL_ID = $('#EMAIL_ID').val();
var CREATED_ON = $('#CREATED_ON').val();
var MODIFIED_ON = $('#MODIFIED_ON').val();
// make the postdata
// var postData = '&ID='+ID+'&NAME='+NAME+'&PASSWORD='+PASSWORD+'&CREDITS'+CREDITS+'&EMAIL_ID'+EMAIL_ID+'&CREATED_ON'+CREATED_ON+'&MODIFIED_ON'+MODIFIED_ON;
// alert(postData);
var myData={"ID":ID,"NAME":NAME,"PASSWORD":PASSWORD,"CREDITS":CREDITS,"EMAIL_ID":EMAIL_ID,"CREATED_ON":CREATED_ON,"MODIFIED_ON":MODIFIED_ON};
//call your .php script in the background,
//when it returns it will call the success function if the request was successful or
//the error one if there was an issue (like a 404, 500 or any other error status)
$.ajax({
url : "Form_Data.php",
type: "POST",
data : myData,
success: function(data,status,xhr)
{
//if success then just output the text to the status div then clear the form inputs to prepare for new data
$("#status_text").html(data);
$('#ID').val();
$('#NAME').val('');
$('#PASSWORD').val('');
$('#EMAIL_ID').val('');
$('#CREATED_ON').val('');
$('#MODIFIED_ON').val('');
}
});
});
</script>
</body>
</div>
</html>
change your script because your ajax was outside the click function
//on the click of the submit button
$("#btn_submit").click(function(){
//get the form values
var ID = $('#ID').val();
var NAME = $('#NAME').val();
var PASSWORD = $('#PASSWORD').val();
var CREDITS = $('#CREDITS').val();
var EMAIL_ID = $('#EMAIL_ID').val();
var CREATED_ON = $('#CREATED_ON').val();
var MODIFIED_ON = $('#MODIFIED_ON').val();
//make the postdata
var postData = '&ID='+ID+'&NAME='+NAME+'&PASSWORD='+PASSWORD+'&REDITS'+CREDITS+'&EMAIL_ID'+EMAIL_ID+'&CREATED_ON'+CREATED_ON+'&MODIFIED_ON'+MODIFIED_ON;
//call your .php script in the background,
//when it returns it will call the success function if the request was successful or
//the error one if there was an issue (like a 404, 500 or any other error status)
$.ajax({
url : "Form_Data.php",
type: "POST",
data : postData,
success: function(data,status, xhr)
{
//if success then just output the text to the status div then clear the form inputs to prepare for new data
$("#status_text").html(data);
$('#ID').val();
$('#NAME').val('');
$('#PASSWORD').val('');
$('#EMAIL_ID').val('');
$('#CREATED_ON').val('');
$('#MODIFIED_ON').val('');
}
});
});
</script>
and change your php code to this
<?php
include 'connnect.php';
mysql_set_charset('utf8');
//query for insert data into tables
if(isset($_POST['NAME'])){
$ID = $_POST['ID'];
$NAME =$_POST['NAME'];
$EMAIL_ID =$_POST['EMAIL_ID'];
$PASSWORD =$_POST['PASSWORD'];
$CREDITS =$_POST['CREDITS'];
$CREATED_ON=$_POST['CREATED_ON'];
$MODIFIED_ON=$_POST['MODIFIED_ON'];
$query = "INSERT INTO `user_table`
(`ID`,`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
VALUES
('$ID','$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
$query_run= mysql_query($query);
$retval=mysql_query($query,$conn);
if ($query_run)
{ echo 'It is working';
}
}
mysql_close($conn);
?>
In this code I'm just submitting your two input fields, the rest you can add by yourself. Try this:
<html>
<body>
Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,
<form action="Form_Data.php" method="post">
NAME: <input id="name" type="text" name="NAME"><br><br>
PASSWORD: <input id="password" type="text" name="PASSWORD"><br><br>
<input type="submit" id="submit">
</form>
</body>
</html>
$("#submit").click(function() {
var name= $("#name").val();
var password= $("#password").val();
$.ajax({
type: "POST",
url: "your_php_path.php",
data: 'name=' + name+ '&password=' + password,
success: function(result) {
alert(result);
}
});
});
Im sending form data from employee.php to a new_employee.php, new_employee.php inserts the data to my database.
After submitting the form it redirects the user to new_employee.php and let them know if it where successful or not.
What I want is that it clears the form and let the user know if it was successful or not on employee.php rather than redirecting the user.
NOTE! I know that this isnt sql injection protected or anything but i simplified it when i posted it here.
And im also aware that i should use mysqli.
employee.php
<body>
<form id="add-emp" method="POST" action="add_employee.php">
<input type="text" id="name" name="name" placeholder="Full Name" />
<input type="text" id="email" name="email" placeholder="Email" />
<input type="submit" value="Register Employee" class="button" onClick="fun();"/>
</form>
<!--<div id="result"></div>-->
</body>
<script>
$('#add-emp').submit(function(event) {
event.preventDefault();
var form = $(this),
value1 = $form.find( 'input[name="name"]' ).val(),
value2 = $form.find( 'input[name="email"]' ).val(),
url = $form.attr( 'action' );
var posting = $.post( url, { name: value1, email: value2 } );
});
});
</script>
And this is my new_employee.php
<?php
$name = $_POST['name'];
$email = $_POST['email'];
mysql_connect("localhost", "****", "****");
mysql_select_db("****");
$stmt = "INSERT INTO employee (name, email) VALUES ('$name', '$email')";
$result = mysql_query($stmt);
if($result){
echo("success!");
}else{
echo("failed!");
}
?>
There are a few things going wrong here. You have a }) to many in your code and defined var form but using $form:
$('#add-emp').submit(function(event) {
event.preventDefault();
var form = $(this),
value1 = form.find( 'input[name="name"]' ).val(),
value2 = form.find( 'input[name="email"]' ).val(),
url = form.attr( 'action' );
var posting = $.post( url, { name: value1, email: value2 }, function(data) { alert(data)} );
});
And with the callback function in $.post you can see what happens and get your message (failed or success) back.
https://api.jquery.com/jQuery.post/
You almost there, all that is left is to grab the string of you echo in php file, and add it to the dom.
success: function(data){
var div = document.getElementById('#messageDiv');
div.innerHTML(data);
}
I am trying to insert value in database from jquery ajax and i want whenever data insertion is successfull, a result output comes true other wise "error:failed". My entry in database successfully updated, but when i alert(msg), its doesnt give me message.
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"> </script>
<body>
<div class="wrapper">
<div id="main" style="padding:50px 0 0 0;">
<!-- Form -->
<form id="contact-form" method="post">
<h3>Paypal Payment Details</h3>
<div class="controls">
<label>
<span>TagId</span>
<input placeholder="Please enter TagId" id="tagid" type="text" tabindex="1" >
</label>
</div>
<div class="controls">
<label>
<span>Paypal Email: (required)</span>
<input placeholder="All Payment will be collected in this email address" id="email" type="email" tabindex="2">
</label>
</div>
<div class="controls">
<label>
<span>Amount</span>
<input placeholder="Amount you would like to charged in GBP" id="amount" type="tel" tabindex="3">
</label>
</div>
<div class="controls">
<div id="error_div"></div>
</div>
<div>
<button name="submit" type="submit" id="form-submit">Submit Detail</button>
</div>
</form>
<!-- /Form -->
</div>
</div>
<script type="text/javascript">
$(document).ready(function(){
$('#form-submit').click(function()
{
var tagid = $("#tagid").val();
var email = $("#email").val();
var amount = $("#amount").val();
var param = 'tagid='+ tagid + '&email=' + email + '&amount=' + amount;
param = param + '&type=assign_amount';
locurl = 'dbentry.php';
$.ajax({
url: locurl,
type:'post',
data:param,
success:function(msg)
{
alert(msg);
}
});
});
});
dbentry.php
<?php
$vals = $_POST;
include 'dbconfig.php';
if($vals['type'] == "assign_amount")
{
$values = assign_amount();
echo json_encode(array('status' =>$values));
}
function assign_amount()
{
global $con;
global $vals;
$sql = "INSERT INTO `dynamic_url`(`tagid`,`email`,`amount`) VALUES('".$vals['tagid']."','".$vals['email']."','".$vals['amount']."')";
$result = mysql_query($sql,$con);
if($result){
if( mysql_affected_rows() > 0 ){
$status="success";
}
}else{
$status="failed";
}
return $status;
}
?>
Try to echo it like
if($result){
if( mysql_affected_rows() > 0 ){
$status="success";
}
} else {
$status="failed";
}
return $status;
And in your if statement code like
if($vals['type'] == "assign_amount")
{
$values = assign_amount();
echo $values;
}
For the ajax return purpose you better to echo or print rather than return it.
In order to see alert() message, you have to prevent default behaviour of clicked submit button:
$('#form-submit').click(function(e)
{
e.preventDefault();
//....
}
Otherwise, the FORM is submited and page is reloaded.
Display $status at last in php file instead of return statement
You will get it in alert
echo $status;
Can you try this,
var locurl = 'dbentry.php';
$.ajax({
url: locurl,
type:'post',
data:param,
dataType:'json',
success:function(msg)
{
alert(msg.status.sql);
}
});
Your code has a lot of flaws in it. For instance you are contatenating the string to create a data object. But if somebody would enter a & or = or any other special charactor in it, your form would fail.
Also you are binding on the click function on a button. While this works, it would be useless for people without javascript. This might not be an issue, but its easily prevented with some minor changes.
I would change the <button name="submit" to <input type="submit" and then bind jQuery to the form it self. Also add the action attribute to the form to include 'dbentry.php'
$(function(){
$('#contact-form').submit(function(){
var $form = $(this);
var data = $form.serialize();
var locurl = 'dbentry.php';
$.post(locurl,data, function(msg) {
alert(msg.status)
}, 'json');
return false; //prevent regular submit
});
});
Now to make it work PHP has to return JSON data.
<?php
header('Content-type: application/json');
//your code that includes
echo json_encode(array('status' =>$sql));
//also notice that your code only returns data on success. Nothing on false.
?>
I have a small commenting system that I have modified and try implement into the site. It's in 'ajax'. When the jQuery with HTML is embedded into the page the commenting system works fine - i.e. when the user clicks on a submit button the code returns 'false', stops the page from refreshing and submits data. BUT when I implemented the code within my site and placed it in a seperate .js file the code for some reason doesn't work properly. I mean - the page after the onclick refreshes. Why is that so ? The code is not changed at all - when on its own, it works but not in the index.php site when implemented. I tried to change input type to 'button' and call a function from onclick - the page doesn't refresh but also doesn't insert the input..I'm running out of ideas as to why it is that so. Here's the code:
$(document).ready(function () {
$(".submit").click(function () {
var name = $("#name").val();
var email = $("#email").val();
var comment_area = $("#comment_area").val();
var dataString = 'name=' + name + '&email=' + email + '&comment_area=' + comment_area;
var emailReg = /^([\w-\.]+#([\w-]+\.)+[\w-]{2,4})?$/;
var emailaddressVal = $("#email").val();
if (name == '' || !emailReg.test(emailaddressVal) || comment == '') {
alert('Please enter valid data and type in message'); return false;
}
else {
$.ajax({
type: "POST",
url: "comments.php",
data: dataString,
cache: false,
success: function (html) {
$("#com_list").append(html);
$("#com_list").fadeIn("slow");
$("#flash").fadeOut('fast');
}
});
} return false;
});
});
//END
//COM LIST
//HTML / PHP
<div class="slider">
<form id="comment_form" name="comment_form" method="post" action="#"
enctype="multipart/form-data">
<input type="text" id="name" name="name" maxlength="16"/> Name<br /><br/>
<input type="text" id="email" name="email"/> Email (will not show)<br /><br/>
<textarea id="comment_area" name="comment_area" maxlength="1000"></textarea><br /><br/>
<input type="submit" class="submit" name="submit_comment" value="submit"/> &
nbsp;comment or <a href="index.php" id="cancel"
onmousedown="$('.slider').hide();$('#com_list').show();"><u>cancel</u></a>
</form>
</div>
//comments.php
if($_POST) {
$name=$_POST['name'];
$email=$_POST['email'];
$comment_area=$_POST['comment_area'];
//$lowercase = strtolower($email);
//$image = md5( $lowercase );
$insert = mysqli_query($connect,"INSERT INTO comments (name,email,comment,com_date)
VALUES ('$name','$email','$comment_area',curdate())");
}
////////////////
Thanks for any suggestions..
aha!
there is an error in your js:
in my console i'm getting "comment is not defined "
if(name=='' || !emailReg.test(emailaddressVal) || comment=='')
and earlier you have:
var comment_area = $("#comment_area").val(); //<--
change this to comment and it'll get past that at least.
EDIT: a little background. when firefox hits an error, sometimes it'll swallow it, and just stop running any javascript after that error, so your return false and or prevent default code isn't fire, so it's still going to post the form and refresh the page.
Change this line:
$(".submit").click(function () {
To this:
$("#comment_form").submit(function () {
The submit event gets triggered on the <form> element, not on the submit button.
Keep your damn code clean, so you can understand what you are cooking...
This will work for you:
$(document).ready(function(){
$("#comment_form").submit(function(e){
e.preventDefault(); // stop refresh
var name = $("#name").val();
var email = $("#email").val();
var comment_area = $("#comment_area").val();
var dataString = 'name='+ name + '&email=' + email + '&comment_area=' + comment_area+'&submit_comment=true';
var emailReg = /^([\w-\.]+#([\w-]+\.)+[\w-]{2,4})?$/;
var emailaddressVal = $("#email").val();
if(name=='' || !emailReg.test(emailaddressVal) || comment==''){
alert('Please enter valid data and type in message');
} else{
$.ajax({
type: "POST",
url: "comments.php",
data: dataString,
cache: false,
success: function(html){
$("#com_list").append(html);
$("#com_list").fadeIn("slow");
$("#flash").fadeOut('fast');
}
});
}
});
$('#cancel').click(function(e){
e.preventDefault();
$('.slider').hide();
$('#com_list').show();
});
});
Here is some more clean code...
<div class="slider">
<form id="comment_form" name="comment_form" method="post" action="#" enctype="multipart/form-data">
<input type="text" id="name" name="name" maxlength="16"/> Name<br /><br/>
<input type="text" id="email" name="email"/> Email (will not show)<br /><br/>
<textarea id="comment_area" name="comment_area" maxlength="1000"></textarea><br /><br/>
<input type="submit" class="submit" name="submit_comment" value="submit"/> comment or <u>cancel</u>
</form>
</div>
Here is some other clean and SECURE code
<?php
if(isset($_POST['submit_comment'])){
$name = mysql_real_escape_string($_POST['name']);
$email = mysql_real_escape_string($_POST['email']);
$comment_area = mysql_real_escape_string($_POST['comment_area']);
//$lowercase = strtolower($email);
//$image = md5( $lowercase );
$query = 'INSERT INTO comments (name,email,comment,com_date) '.
"VALUES ('$name','$email','$comment_area',CURDATE())";
$insert = mysqli_query($connect, $query);
}
?>