Why when i typed "+-1-23$%^&sdfsdf/><" in the textarea but it save only "-1-23$%^" into database?
Code :
function postingMsg (){
$('.error').hide();
var messageposting2= $("textarea#messageposting").val();
var dataString = 'messageposting2='+ messageposting2;
$.ajax({
type: "POST",
url: "note-send.php",
data: dataString,
success: function(msg) {
msg = parseFloat(msg)
}
});
return false;
}
if ((isset($_POST['messageposting2'])) && (strlen($_POST['messageposting2']) > 0)) {
$messageposting3 = $_POST['messageposting2'];
$sql = "UPDATE users
SET my_note=?
WHERE user_id=?";
$q = $conn->prepare($sql);
$q->execute(array($messageposting3, $_SESSION['user_id']));
echo "1";
} else {echo "0";}
It has nothing to do with PDO or your database. You must URL-encode your string before sending it through Ajax.
var dataString = 'messageposting2='+ encodeURIComponent(messageposting2);
Related
My jQuery Ajax code:
$(".comment").click(function(){
var id = $(this).data("id");
var name = $("#username_"+id).val();
if(name==''){
alert("Please Fill All Fields");
}else{
$.ajax({
type: "POST",
url: "comments",
data:{username:name},
dataType: "JSON",
success: function(jsonStr){
$("#cmt_output").html(JSON.stringify(jsonStr));
}
});
}
});
My Php Code:
............
$con = mysqli_connect("localhost","root","","quotes") or die("Error".mysqli_error($con));
$name = $_POST['username'];
$user = get_userdetails();
$vw_id = $user->id;
$query = mysqli_query($con,"INSERT INTO tlb_comments(user_id,comments) values ('$vw_id','$name')");
$comments = mysqli_query($con,"SELECT * FROM tlb_comments");
$avatar = mysqli_query($con,"SELECT * FROM tlb_avatar WHERE `vw_id`='".$vw_id."'");
$avatar_select = mysqli_fetch_all($avatar);
$comment_select = mysqli_fetch_all($comments);
array_push($avatar_select,$comment_select);
echo json_encode($avatar_select);
I am trying to insert data using AJAX JSON but it's not working. I tried without JSON and it works, but an alert box shows with some HTML code.
HTML:
Short Break
AJAX:
$(document).ready(function() {
$('#sbreak').on('click', function() {
var name = $("SBreak").val();
$.ajax({
type: "POST",
dataType: 'json',
url: "brkrequest.php",
data: {
sname: name
}
cache: false,
success: function(server_response) {
if (server_response.status == '1') //if ajax_check_username.php return value "0"
{
alert("Inserted ");
} else if (server_response == '0') //if it returns "1"
{
alert("Already Inserted");
}
},
});
return false;
});
});
PHP: :
session_start();
date_default_timezone_set('Asia/Kolkata');
$sname=$_POST['sname'];
$sname= $_SESSION['myusername'];
$reqdate = date("Y-m-d H:i:s");
include("connection.php");
//Insert query
$query = sprintf("SELECT * FROM `breakqueue` WHERE (`sname` ='$sname')");
$result = mysql_query($query);
if(mysql_num_rows($result) > 0){
$data['status']= '1';//If there is a record match Already Inserted
}
else { // if there is no matching rows do following
$query = mysql_query("INSERT INTO `breakqueue`(`id`, `sname`, `btype`, `reqdate`, `apdate`, `status`) VALUES ('','$sname','Sbreak','$reqdate','','Pending')");
$data['status']= '0';//Record Insered
}
echo json_encode($data);
}
use it in php
header('Content-Type:application/json');
and write
success: function(server_response){
console.log(typeof server_response);
...
for finding response type,
if type of server_response isn't object
use it for convert it to object :
server_response = JSON.parse(server_response);
php Code:
session_start();
//Here added...
header('Content-Type:application/json');
date_default_timezone_set('Asia/Kolkata');
$sname=$_POST['sname'];
$sname= $_SESSION['myusername'];
$reqdate = date("Y-m-d H:i:s");
include("connection.php");
//Insert query
$query = sprintf("SELECT * FROM `breakqueue` WHERE (`sname` ='$sname')");
$result = mysql_query($query);
if(mysql_num_rows($result) > 0){
$data['status']= '1';//If there is a record match Already Inserted
}
else{ // if there is no matching rows do following
$query = mysql_query("INSERT INTO `breakqueue`(`id`, `sname`, `btype`, `reqdate`, `apdate`, `status`) VALUES ('','$sname','Sbreak','$reqdate','','Pending')");
$data['status']= '0';//Record Insered
}
echo json_encode($data);
}
Javascript Code:
$(document).ready(function()
{
$('#sbreak').on('click', function(){
var name = $("SBreak").val();
$.ajax({
type: "POST",
dataType:'json',
url: "brkrequest.php",
data: {sname: name}
cache: false,
success: function(server_response){
//TODO:REMOVE IT After seeing. alert or console.log for seeing type
alert(typeof server_response);
if(typeof server_response){
server_response = JSON.parse(server_response);
}
if(server_response.status == '1')//if ajax_check_username.php return value "0"
{
alert("Inserted ");
}
else if(server_response == '0')//if it returns "1"
{
alert("Already Inserted");
}
},
});
return false;
i want to add data to my database by passing as a json string then using php what i have done but adds an empty line in the database instead of adding the data that i sent and i get the alert("fail") message where is the mistake please
here is my save function
function save(){
var eml = document.getElementById("tbemail").value;
var mp = document.getElementById("tbmdp").value;
var data = {email: eml, mdp: mp};
$.ajax({
url: "http://localhost:800/test/insert.php",
type: 'POST',
dataType: 'json',
data: JSON.stringify(data),
contentType: "application/json; charset=utf-8",
success: function (data) {
alert('success');
},
error: function () {
alert("fail");
}
});
and here is my php file insert.php
<?php
$json = isset($_POST['data']) ? $_POST['data'] : "";
$new=json_decode($json, true);
$conn= mysqli_connect("localhost","root","") or die ("could not connect to mysql");
mysqli_select_db($conn,"bd") or die ("no database");
$sql = "INSERT INTO user (email,mdp) VALUES ('".$new['email']."','".$new['mdp']."') ";
$insert=mysqli_query($conn, $sql);
if ($insert) {
echo "created ";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
?>
Can you try this:
JS:
function save(){
var eml = document.getElementById("tbemail").value;
var mp = document.getElementById("tbmdp").value;
var data = {email: eml, mdp: mp};
$.ajax({
url: "http://localhost:800/test/insert.php",
type: 'POST',
dataType: 'json',
data: data,
contentType: "application/json; charset=utf-8",
success: function (data) {
alert('success');
},
error: function () {
alert("fail");
}
});
}
PHP Code:
<?php
$email = isset($_POST['email']) ? $_POST['email'] : "";
$mdp = isset($_POST['mdp']) ? $_POST['mdp'] : "";
$new=json_decode($json, true);
$conn= mysqli_connect("localhost","root","") or die ("could not connect to mysql");
mysqli_select_db($conn,"bd") or die ("no database");
$sql = "INSERT INTO user (email,mdp) VALUES ('".$email."','".$mdp."') ";
$insert=mysqli_query($conn, $sql);
if ($insert) {
echo "created ";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
?>
Get data using this way in PHP
$data = json_decode(file_get_contents('php://input'));
$email = $data ->email;
$mdp = $data ->mdp;
In your PHP, each individual variable that you're passing through (i.e. email and mdp) is passed as individual $_POST data, not into a single $_POST variable called 'data'. Right after your opening PHP tag, check for the email and mdp:
$email = (isset($_POST['email']) ? $_POST['email'] : "");
$mdp = (isset($_POST['mdp']) ? $_POST['mdp'] : "");
$conn= ....
You can try this:
function save(){
var eml = document.getElementById("tbemail").value;
var mp = document.getElementById("tbmdp").value;
var data = {'email': eml,'mdp': mp}; //json
$.ajax({
url: "http://localhost:800/test/insert.php",
type: 'POST',
dataType: 'json',
data: data,//pass it here
contentType: "application/json; charset=utf-8",
success: function (data) {
alert('success');
},
error: function () {
alert("fail");
}
});
}
php:
<?php
if(isset($_POST['email'],$_POST['mdp']) {
$conn= mysqli_connect("localhost","root","") or die ("could not connect to mysql");
mysqli_select_db($conn,"bd") or die ("no database");
$sql = "INSERT INTO user (email,mdp) VALUES ('".$_POST['email']."','".$_POST['mdp']."') ";
$insert=mysqli_query($conn, $sql);
if ($insert) {
echo "created ";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
}
?>
try like this,
function save(){
var eml = document.getElementById("tbemail").value;
var mp = document.getElementById("tbmdp").value;
var data = {email: eml, mdp: mp};
$.ajax({
url: "http://localhost:800/test/insert.php",
type: 'POST',
// dataType: 'json',
data: {"data":JSON.stringify(data)},
// contentType: "application/json; charset=utf-8",
success: function (data) {
alert('success');
},
error: function () {
alert("fail");
}
});
}
Try this :
function save() {
var eml = document.getElementById("tbemail").value;
var mp = document.getElementById("tbmdp").value;
var data = {email: eml, mdp: mp};
$.ajax({
type: 'POST',
url: "http://localhost:800/test/insert.php",
//dataType: 'json',
data: {"data": JSON.stringify(data)},
//contentType: "application/json; charset=utf-8",
success: function (data) {
if (data == 'created')
alert('Success');
else
alert('Fail');
}
});
}
Apologies if this is a repeat question, but any answer I have found on here hasn't worked me. I am trying to create a simple login feature for a website which uses an AJAX call to PHP which should return JSON. I have the following PHP:
<?php
include("dbconnect.php");
header('Content-type: application/json');
$numrows=0;
$password=$_POST['password'];
$username=$_POST['username'];
$query="select fname, lname, memcat from members where (password='$password' && username='$username')";
$link = mysql_query($query);
if (!$link) {
echo 3;
die();
}
$numrows=mysql_num_rows($link);
if ($numrows>0){ // authentication is successfull
$rows = array();
while($r = mysql_fetch_assoc($link)) {
$json[] = $r;
}
echo json_encode($json);
} else {
echo 3; // authentication was unsuccessfull
}
?>
AJAX call:
$( ".LogIn" ).live("click", function(){
console.log("LogIn button clicked.")
var username=$("#username").val();
var password=$("#password").val();
var dataString = 'username='+username+'&password='+password;
$.ajax({
type: "POST",
url: "scripts/sendLogDetails.php",
data: dataString,
dataType: "JSON",
success: function(data){
if (data == '3') {
alert("Invalid log in details - please try again.");
}
else {
sessionStorage['username']=$('#username').val();
sessionStorage['user'] = data.fname + " " + data.lname;
sessionStorage['memcat'] = data.memcat;
storage=sessionStorage.user;
alert(data.fname);
window.location="/awt-cw1/index.html";
}
}
});
}
As I say, whenever I run this the values from "data" are undefined. Any idea where I have gone wrong?
Many thanks.
hi i am working on an authentification page , so my code is the following
$(document).ready(function(){
var form = $("#connexion");
var login =$("#logins");
var password=$("#passe");
$("#go").click(function(e){
e.preventDefault();
$.ajax({type: "POST",
url: "check_con.php",
data: { email:login.val() , password:password.val() },
success:function(result){
if(result == 'true')
{
alert(result);
}
}});
});
});
i get the form , the login and password and i pass them to my php script .
<?php
//data connection file
//require "config.php";
require "connexion.php";
extract($_REQUEST);
$pass=crypt($password);
$sql = "select * from Compte where email='$email'";
$rsd = mysql_query($sql);
$msg = mysql_num_rows($rsd); //returns 0 if not already exist
$row = mysql_fetch_row($rsd);
if($msg == 0)
{
echo"false1";
}
else if($row[1] == crypt($password,$row[1]))
{
echo"true";
}
else
{
echo"false2";
}
?>
everything is goood , when i give the good email and password i get true otherwise i get false, that's not the problem , the problem is i am trying to redirect the user to another page called espace.php if the result is true so i've tried this .
$(document).ready(function(){
var form = $("#connexion");
var login =$("#logins");
var password=$("#passe");
$("#go").click(function(e){
$.ajax({type: "POST",
url: "check_con.php",
data: { email:login.val() , password:password.val() },
success:function(result){
if(result == 'true')
{
form.submit(true);
}
else form.submit(false);
}});
});
});
now even if the login and password are not correct the form is submitted , how could i manage to do that i mean , if the informations are correct i go to another page , otherwise i stay in the same page.
use json to get result from authanication page
<?php
//data connection file
//require "config.php";
require "connexion.php";
extract($_REQUEST);
$pass=crypt($password);
$sql = "select * from Compte where email='$email'";
$rsd = mysql_query($sql);
$msg = mysql_num_rows($rsd); //returns 0 if not already exist
$row = mysql_fetch_row($rsd);
$result = array();
if($msg == 0)
{
$result['error'] = "Fail";
}
else if($row[1] == crypt($password,$row[1]))
{
$result['success'] = "success";
}
else
{
$result['error'] = "try again";
}
echo json_encode($result); die;
?>
And in the ajax,, check what is the response.
$(document).ready(function(){
var form = $("#connexion");
var login =$("#logins");
var password=$("#passe");
$("#go").click(function(e){
$.ajax({type: "POST",
url: "check_con.php",
data: { email:login.val() , password:password.val() },
success:function(result){
var response = JSON.parse(result);
if(response.error){
//here provide a error msg to user.
alert(response.error);
}
if(response.success){
form.submit();
}
}});
});
});