Hello i have an recaptcha working with ajax but it's not inserting my form infos into database, it inserts empty rows and i dont have any clue why
Here is my code:
require_once('inc/recaptchalib.php');
$publickey = "6LdW1N0SAAAAADMIPPkOGd939meXV9a9qDwwcxbu"; // you got this from the signup page
$privatekey = "";
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js" type="text/javascript"></script>
<script type="text/javascript">
var RecaptchaOptions = {
theme : 'clean'
};
</script>
<script type="text/javascript">
function validateCaptcha()
{
challengeField = $("input#recaptcha_challenge_field").val();
responseField = $("input#recaptcha_response_field").val();
//alert(challengeField);
//alert(responseField);
//return false;
var html = $.ajax({
type: "POST",
url: "ajax.recaptcha.php",
data: "recaptcha_challenge_field=" + challengeField + "&recaptcha_response_field=" + responseField,
async: false
}).responseText;
if(html == "success")
{
$("#captchaStatus").html("Success. Submitting form.");
return false;
// Uncomment the following line in your application
//return true;
}
else
{
$("#captchaStatus").html("Your captcha is incorrect. Please try again");
Recaptcha.reload();
return false;
}
}
</script>
Here is the form that i want to insert into database:
<form id="loginform" name="loginform" action="" method="post" enctype="multipart/form-data" onSubmit="return validateCaptcha()" >
<fieldset class="step">
<p>
<label for="nick">Nickname:</label>
<input type="text" name="nick" value="" id="nicknamee" value="" >
</p>
<p>
<label for="email">Email:</label>
<input type="text" name="email" value="" id="email_comm" value="" >
</p>
<p>
<textarea class="text-input textarea wysiwyg" name="add_comment" id="text_commentt"></textarea>
</p>';?>
<div id="login">
<?php echo recaptcha_get_html($publickey);?>
<p style="color: red;" id="captchaStatus"> </p>
</div>
<?php
echo'
<p style="margin-top:200px;"><button id="vote_petiton_submit" type="submit" ></button>
</p>
</fieldset>
</form>
Now the verification of the reCaptcha and database insert file. Validation is ok just inserting datas in database not working proper, it doesnt post the values.
<?php
require_once('inc/recaptchalib.php');
require_once('engine/db.php');
$publickey = "6LdW1N0SAAAAADMIPPkOGd939meXV9a9qDwwcxbu"; // you got this from the signup page
$privatekey = "";
mysql_query("SET NAMES utf8");
$resp = recaptcha_check_answer ($privatekey,
$_SERVER["REMOTE_ADDR"],
$_POST["recaptcha_challenge_field"],
$_POST["recaptcha_response_field"]);
if ($resp->is_valid) {
?>success<?
$nick=$_REQUEST['nick'];
$type=0;
$email=$_REQUEST['email'];
$descr=$_REQUEST['add_comment'];
$descriere=strip_tags($descr,'<br><p><a><i><u><b>');
$id=150;
$insert = "INSERT INTO comments (`nickname`,`desc`,`email`,`type`,`id_pet_dem_des`)
VALUES ('$nick','$descriere','$email','$type',$id)";
$result = mysql_query($insert)
or die("query failed: " . mysql_error());
$queryz = "update petition set comments=comments+1 where id_petition='$id'";
$resultz = mysql_query($queryz)
or die("query failed: " . mysql_error());
}
else
{
die ("The reCAPTCHA wasn't entered correctly. Go back and try it again." .
"(reCAPTCHA said: " . $resp->error . ")");
}
?>
Thnx in advance.
i fixed it with changeing data from ajax post with
data: $('#loginform').serialize() + "recaptcha_challenge_field=" + challengeField + "&recaptcha_response_field=" + responseField,
Related
I have this form
<form id="home" class="validate-form" method="post" enctype="multipart/form-data">
<!-- Form Item -->
<div class="form-group">
<label>How much money do you need? (Kenya Shillings)</label>
<div class="input-group">
<div class="input-group-addon">Ksh</div>
<input id="moneyAmount" type="number" id="amount" name="amount" class="form-control slider-control input-lg" value="100000" min="10000" max="1000000" data-slider="#moneySlider" required>
</div>
<div id="moneySlider" class="form-slider" data-input="#moneyAmount" data-min="10000" data-max="1000000" data-value="100000"></div>
</div>
<!-- Form Item -->
<div class="form-group">
<label>How long? (months)</label>
<div class="input-group">
<input id="monthNumber" type="number" id="duration" name="duration" class="form-control slider-control input-lg" value="10" min="6" max="12" data-slider="#monthSlider" required>
<div class="input-group-addon">months</div>
</div>
<div id="monthSlider" class="form-slider" data-input="#monthNumber" data-min="6" data-max="12" data-value="10"></div>
</div>
<div class="form-group">
<label>Telephone Number</label>
<!-- Radio -->
<input type="number" id="telephone" name="telephone" class="form-control" required/>
</div>
<!-- Form Item -->
<div class="form-group">
<label>3 Months Bank or Paypal or Mpesa Statements</label>
<!-- Radio -->
<input type="file" name="image" class="ml btn btn-primary btn-lg" /><span>Upload</span>
</div>
<!-- Form Item -->
<div class="form-group">
<label>Monthly repayment</label>
<span id="formResult" class="form-total">Ksh<span>262.99</span></span>
</div>
<div class="form-group form-submit">
<button type="submit" class="btn-submit btn-lg"><span>Send a request!
</span></button>
</div>
</form>
This is the Jquery Script.
$( "#home" ).on( "submit", function( event ) {
event.preventDefault();
alert('subsequent clicks');
function chek(fData) {
var reg = new RegExp("^[-]?[0-9]+[\.]?[0-9]+$");
return reg.test(fData)
}
var phone = $('#telephone').val();
var amount = $('#amount').val();
var duration = $('#duration').val();
var ch = chek(phone);
if(phone == ""){
alert('phone cannot be empty');
return;
}
if(amount == ""){
alert('amount cannot be empty');
return;
}
if(duration == ""){
alert('duration cannot be empty');
return;
}
if(ch == false){
alert("Phone number must be a number");
return;
}
if(phone.length < 10 || phone.length > 12 ){
alert("Phone number must have 10 digits");
return;
}
if(ch == true && phone !== "" && amount !== "" && duration !== "" && phone.length == 10){
var s = phone;
s = s.replace(/^0+/, '');
var cc = 254;
var p = cc+s;
var pn = p.toString();
$('#telephone').val(p.toString());
var formData = new FormData($(this)[0]);
$.ajax({
url: 'http://example.com/home.php', //<== just add it to the end of url ***
type: 'POST',
data: formData,
async: true,
success: function (data) {
console.log(data)
},
cache: false,
contentType: false,
processData: false
});
return false;
}
});
This is my PHP code:
<?php
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
header('Access-Control-Allow-Origin: *');
header('Access-Control-Allow-Methods: GET, POST');
header("Access-Control-Allow-Headers: X-Requested-With");
function random_str($length, $keyspace = '0123456789abcdefghijklmnopqrstuvwxyz')
{
$str = '';
$max = mb_strlen($keyspace, '8bit') - 1;
for ($i = 0; $i < $length; ++$i) {
$str .= $keyspace[random_int(0, $max)];
}
return $str;
}
$pass = random_str(4);
/**
Generic Customer Shown Interest
*/
$servername = "localhost";
$username = "root";
$password = "123456";
$dbname = "algo";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//Posted Variables
$amount = $_POST['amount'];
$duration = $_POST['duration'];
$telephone = $_POST['telephone'];
$date = date('Y-m-d H:i:s');
//Check If User Exists
$result = $conn->query("select id from users where telephone=$telephone");
if($result->num_rows == 0) {
//Insert New User
$sql = "INSERT INTO users (telephone, password, service_name,date_submitted) VALUES ('$telephone', '$pass', 'loans','$date')";
if ($conn->query($sql) === TRUE) {
echo "User Is Inserted";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
} else {
//Insert New User
$sql2 = "INSERT INTO loans (amount, duration, telephone,documents,status,date)
VALUES ('$amount', '$duration','$telephone','logan2','on-hold','$date')";
if ($conn->query($sql2) === TRUE) {
echo "Loan Is Inserted";
} else {
echo "Error: " . $sql2 . "<br>" . $conn->error;
}
$conn->close();
}
?>
As you can tell the form is pretty basic and its only posting data to the server. When I load the page, I am able to insert data into the database but when I click the link again, nothing is inserted.
Is form data blocking me from posting duplicate data to the server?
change ajax part of your code and replace to this code shown below:
<script type="text/javascript">
$.ajax({
type:'POST',
url:'testing2.php',
data:new FormData($('#svf-form-4')[0]),
cache: false,
contentType: false,
processData: false,
success:function(msg){
$('#message').html(msg);
}
});
return false;
</script>
Hope it will work .
I cant explain what really worked but it seems clearing the form did allow for more post submission although i relied on this comment What does "async: false" do in jQuery.ajax()?
and this page What does "async: false" do in jQuery.ajax()? for inspiration.
This is the success callback
success: function (data) {
$("#home").trigger('reset');
console.log(data);
},
I'm trying to save some data into db using AJAX.It says it's successfully done but nothing happens.No data is being saved.Just keeps saying "Success".
index.php // form
<div id="cont">
<div id="kayit" style="display:none;">
Success
</div>
<div id="basarisiz" style="display:none;">
Empty or Fail
</div>
<form method="post" id="yaz" onsubmit="return false">
<input type="text" name="title" id="baslik" placeholder="Başlık" required><br/>
<textarea name="content" ></textarea><br/>
<input type="submit" name="gonder" value="Gönder" id="gonder" onclick="kayit()">
</form>
yaz.js
function kayit()
{
var baslik = $("input[name=title]").val();
var icerik = $("input[name=content]").val();
if (baslik =="" || icerik == "")
{
$('#basarisiz').show(1);
$('#kayit').hide(1);
}else
{
$.ajax ({
type: "POST",
url: "yazikaydet.php",
data: $("#yaz").serialize(),
success: function (sonuc) {
if (sonuc == "hata") {
alert ("unable to connect to db");
}else {
$('#kayit').show(1);
$('#basarisiz').hide(1);
$("input[name=title]").val("");
$("input[name=content]").val("");
}
}
}) }}
yazikaydet.php //the file which will save the data,not processing
<?php
include 'giris.php'; //sessions and connection strings
if(isset($_POST["gonder"]))
{
$baslik = $_POST["title"];
$icerik = $_POST["content"];
$tarih = date("d/m/20y");
$kull = $_SESSION["username"];
$kaydet = mysqli_query($connect,"INSERT INTO gonderiler (yazar,tarih,baslik,icerik) VALUES ('$kull','$tarih','$baslik','$icerik')");
if($kaydet)
{
echo "Yes";
}
else
{
echo "No";
}
}
?>
when I echo the php variable it work properly , but when I try to insert the data into database it doesn't work , what is the solution please I get stuck
I got this error on console
POST http://localhost/validate.php 500 (Internal Server Error)
send # jquery-3.1.1.min.js:4
ajax # jquery-3.1.1.min.js:4
(anonymous) # jquery.PHP:26
dispatch # jquery-3.1.1.min.js:3
q.handle # jquery-3.1.1.min.js:3
HTML/JQUERY
<form action="" id="myForm">
<input type="text" id="name" ><br/>
<input type="text" id="age" ><br/>
<input type="submit" value="Submit">
</form>
<div id="result"></div>
<script>
$(function() {
$("#myForm").submit(function(e) {
e.preventDefault();
var name = $('#name').val();
var age = $('#age').val();
$.ajax({
url: 'validate.php',
method: 'POST',
data: {postname:name, postage:age},
success: function(res) {
$("#result").append(res);
}
});
});
});
</script>
PHP
<?php
include 'mysqldb.php';
$name = $_POST['postname'];
$age = $_POST['postage'];
$sql = "insert into uss (first, last) values('".$name."','".$age."')";
$result = $conn->query($sql);
echo $result ;
?>
mysqldb.php
<?php
$conn = mysql_connect('localhost', 'root', 'password' , 'datab');
if (!$conn) {
die("Connection failed: ".mysqli_connect_error());
}
?>
Please add the details of the error message you get.
Make little changes to your code so that it can show the query error if any
<?php
include 'mysqldb.php';
$name = $_POST['postname'];
$age = $_POST['postage'];
$sql = "INSERT INTO `uss` (`first`, `last`) VALUES('{$name}','{$age}')";
if($conn->query($sql))
{
echo "Record inserted";
}
else
{
echo $conn->error;
}
?>
Sugesstions: Your query have the chances of the SQL Injection. Make it secure.
if you are using ajax , try the following,
<form >
<input type="text" id="name" ><br/>
<input type="text" id="age" ><br/>
<input type="submit" value="Submit" id="submit">
</form>
<div id="result"></div>
$("#submit").click(function(){
var name = $('#name').val(); // getting name
var age = $('#age').val();
$.ajax({
url : "validate.php",
type: "POST",
data: {name:name, age:age},
success: function(data)
{
$("#result").html(data);
}
});
});
in your controller function,echo the result
<?php
include 'mysqldb.php';
$name = $_POST['postname'];
$age = $_POST['postage'];
$sql = "insert into uss (first, last) values('$name','$age')";
$result = $conn->query($sql);
echo $result;
?>
jQuery Ajax
Form with id myFrom
<form action="" id="myForm">
<input type="text" id="name" ><br/>
<input type="text" id="age" ><br/>
<input type="submit" value="Submit">
</form>
<div id="result"></div>
jQuery Ajax section
$(function() {
$("#myForm").submit(function(e) {
e.preventDefault();
var name = $('#name').val(); // getting name
var age = $('#age').val(); // getting age
/* Ajax section */
$.ajax({
url: 'validate.php',
method: 'POST',
data: {postname:name, postage:age},
success: function(res) {
$("#result").append(res);
}
});
});
});
validate.php
<?php
include 'mysqldb.php';
$name = $_POST['postname'];
$age = $_POST['postage'];
//check ajax response by `echo $name` and `$age`
$sql = "insert into uss (first, last) values('".$name."','".$age."')";
$result = $conn->query($sql);
echo $result ;
?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
<form >
<input type="text" id="name" ><br/>
<input type="text" id="age" ><br/>
<input type="button" value="Submit" onclick="postdata();">
</form>
<div id="result"></div>
<script type="text/javascript">
function postdata() {
alert("ashad");
var name = $('#name').val();
var age = $('#age').val();
$.post('validate.php',{postname:name,postage:age},
function(data){
$('#result').html(data);
});
}
</script>
<?php
include 'mysqldb.php';
$name = $_POST['postname'];
$age = $_POST['postage'];
//check ajax response by `echo $name` and `$age`
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} else
{
$sql = "insert into uss(first, last) values('$name','$age')";
$result = $conn->query($sql);
}
echo $result ;
?>
I have created a simple Login Register program using PHP.
Now I am trying to validate if username already exists or not using jquery ajax. The jquery code runs but keeps on showing 'Checking Availability'.
Here is the code I have used. Please ignore the vulnerability and other errors in my PHP code ( which may not affect jquery ajax process ) as I am new to this. I'm working for improving those things.
Register.php
<?php
include('config.php');
if(isset($login_session))
{
header("Location: login.php");
}
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
$username = mysqli_real_escape_string($obj->conn,$_POST['username']);
$password = mysqli_real_escape_string($obj->conn,$_POST['password']);
$name = mysqli_real_escape_string($obj->conn,$_POST['name']);
$email = mysqli_real_escape_string($obj->conn,$_POST['email']);
$password = md5($password);
$sql ="SELECT uid from users WHERE username = '$username' or email = '$email'";
$register_user = mysqli_query($obj->conn,$sql) or die(mysqli_error($sql));
$no_rows = mysqli_num_rows($register_user);
if($no_rows == 0)
{
$sql2 = "INSERT INTO users(username, password, name, email) values ('$username', '$password', '$name', '$email')";
$result = mysqli_query($obj->conn, $sql2) or die(mysqli_error($sql2));
echo "Registration Successfull!";
}
else{
echo "Registration Failed.";
}
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Register</title>
<script type="text/javascript" src="js/jquery.js"></script>
<script type="text/javascript" src="js/username.js"></script>
</head>
<body>
<form action="register.php" method="post">
<label>UserName:</label>
<input type="text" id="username" name="username" required/>
<span id="status"></span><br />
<label>Password :</label>
<input type="password" name="password" required/><br/>
<label>Full Name :</label>
<input type="text" name="name" required/><br/>
<label>Email :</label>
<input type="email" name="email" required/><br/>
<input type="submit" value=" Submit "/><br />
</form>
</body>
</html>
username.js
$(document).ready(function()
{
$("#username").change(function()
{
var username = $("#username").val();
var msgbox = $("#status");
if(username.length > 3)
{
$("#status").html('<img src="img/loader.gif" align="absmiddle"> Checking availability...');
$.ajax({
type: "POST",
url: "php/username-check.php",
data: "username="+ username,
success: function(msg){
$("#status").ajaxComplete(function(event, request){
if(msg == 'OK')
{
msgbox.html('<img src="img/yes.png" align="absmiddle"> <font color="Green"> Available </font> ');
}
else
{
$("#username").removeClass("green");
$("#username").addClass("red");
msgbox.html(msg);
}
});
}
});
}
else
{
$("#status").html('<font color="#cc0000">Enter valid User Name</font>');
}
return false;
});
});
username-check.php
<?php
include("config.php");
if(isSet($_POST['username']))
{
$username = $_POST['username'];
$username = mysqli_real_escape_string($obj->conn,$username);
$sql = "SELECT username FROM users WHERE username='$username'";
$sql_check = mysqli_query($obj->conn,$sql);
if (!$sql_check)))
{
echo 'could not complete query: ' . mysqli_error($obj->conn,$sql_check);
}else{
echo 'query successful!';
}
if(mysqli_num_rows($obj->conn,$sql_check))
{
echo '<font color="#cc0000"><b>'.$username.'</b> is already in use.</font>';
}
else
{
echo 'OK';
}
}
?>
and I want to know if there is a way to check if jQuery Ajax sent the POST request to that file or not?
You are confusing ajax functions...Syntax will be like this
$.ajax({
url: url,
data: data,
type: "POST",
beforeSend: function () {
},
success: function (returnData) {
},
error: function (xhr, ajaxOptions, thrownError) {
},
complete: function () {
}
});
Examine the request using a browser utility
- Launch the chrome browser
- Right click and select inspect element menu
- click on Network tab
- Load your URL
- Perform the Ajax request
- You can see the request here (new request will be last in the list).
- Click on it
- Right side window shows you request and response data
You did correct.Easy way to check them is use firebug tool on your browser...I recommend firefox with firebug.install it first and then open it before you post your form.then goto console log and send your form...Check it out,best software.
I'm having a problem with my ajax call. I'm submitting some info via php to mySQL, the submission part works perfectly, it's adding the data to the database, but the ajax function isn't loading that php in the background, it's loading it in the window and showing the php file results.
Here's the HTML code.
<form action="upload.php" class="addItem" method="post">
Firstname:<br><input type="text" class="firstname" name="firstname" /><br>
Lastname:<br><input type="text" class="lastname" name="lastname" /><br>
Age:<br><input type="text" class="age" name="age" /><br>
Photo:<br><input type="file" name="image" accept="image/jpeg" /><br><br>
<input type="submit" class="submitItem" value="Add Row" />
</form>
Logout
</div>
<script>
$(document).ready(function(){
$(".submitItem").click(function(){
// Start AJAX send
jQuery.ajax({
// Enter below the full path to your "send" php file
url: "upload.php",
type: "POST",
data: data,
cache: false,
success: function (html) {
// If form submission is successful
if ( html == 1 ) {
$('.successMessage').show(200);
$('.successMessage').delay(2000).hide();
}
// Double check if maths question is correct
else {
$('.errorMessage').show(200);
$('.errorMessage').delay(2000).hide();
}
}
});
});
});
</script>
Here's the PHP code
<?php
$name = $_POST['firstname'];
$surname = $_POST['lastname'];
$age = $_POST['age'];
if(($name === "") || ($surname === "") || ($age === "")){
echo "please fill in all fields";
} else {
$con = mysql_connect("localhost","user","password");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("my_db", $con);
$sql="INSERT INTO Persons (FirstName, LastName, Age)
VALUES
('$_POST[firstname]','$_POST[lastname]','$_POST[age]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
if ($sql) { echo "1"; }
else{echo "error";}
}
mysql_close($con);
?>
Your handler needs to return false; to instruct the browser not to do its regular submission action.
(Also, you should really consider using the submit event of the form, rather than the click event of the button.)
<script type="text/javascript">
$(function(){
$("form.addItem").submit(function(){
// Start AJAX send
jQuery.ajax({
// ... your parameters ...
});
return false;
});
});
</script>