$.ajax({
type:"GET",
url: myurl, dataType:'json', data: 'username',
You should provide the accompanying HTML. It's possible your form is submitting POST variables (recommended) in which case $_GET['username'] being empty. Does it work with $_POST['username']?
I would suggest you to do some modifications, if it can work:
I. Take both $_GET and $_POST variables, if your form is using method="POST".
II. Using case insensitive match for "All" string as for all users.
<?php
include("dbconnect.php");
$username = isset($_GET['username']) ? $_GET['username']
: (isset($_POST['username']) ? $_POST['username'] : '');
if(preg_match("/^all$/i", $username))
{
$query = "select * from broons order by fname asc";
}
else
{
$query = "select * from broons where username = '$username' order by fname asc";
}
$link = mysql_query($query);
if (!$link) {
die($query);
}
$rows = array();
while($r = mysql_fetch_assoc($link)) {
$rows[] = $r;
}
$json = json_encode($rows);
echo $json;
echo $query;
?>
Good luck.
ADDED:
I might give a wrong way above.
At the calling side jQuery.ajax(), your username field seems not passed to the server, please try this:
$.ajax({
type:"GET",
url: myurl, dataType:'json', data: 'username=JohnDoe2'});
Related
I want to check if a user has favourited an item but I'm unsure how to return the result of a database query to ajax.
I will show different html depending on the result.
Php
$query = "SELECT itemID from favourites WHERE userid = '" . $user. "'";
$result = mysql_query($query);
echo json_encode($result);
Jquery
$.ajax({
url: "inc/functions.php",
type: "POST",
data: {--result--},
success: function () {
// if result found in database
$('favourite').hide();
// if result not found
$('favourite').show();
}
});
I can't figure out how to display $result in the jquery code.
Any help much appreciated.
$result in this case is a PHP object representing a result.
You will have to use a fetch() method in order to extract the result before sending it back to your JS.
See this link. There's a list of all fetch-family method right above the comments.
Also, you will need to make a connection with you database beforehand using mysqli_connect (or mysql_connect in your case).
As stated in the comments, you should however use mysqli* functions family instead of mysql*.
Thanks to the comments for info regarding mysqli. I updated the code and solved the ajax part.
For anyone else stuck, I got it working like this:
PHP
require ("../../connection.php");
$sql = "SELECT * FROM favourites WHERE userID = ? AND itemID = ?";
$user = $_POST['userID'];
$item = $_POST['itemID'];
$statement = $db->prepare($sql);
if($statement === false) {
trigger_error('Wrong SQL: ' . $sql . ' Error: ' . $db->error, E_USER_ERROR);
}
$statement->bind_param('ii',$user,$item);
$statement->execute();
$statement->bind_result($user,$item);
while($statement->fetch()){
echo 1;
}
$statement->close();
Jquery
$.ajax({
url: "inc/userList.php",
data: userList,
type: "POST",
success: function (result) {
if (result == 1){
$('#addItem').css('display', 'none');
$('#removeItem').css('display', 'inline-block');
} else {
$('#addItem').css('display', 'inline-block');
$('#removeItem').css('display', 'none');
}
}
});
I want to get a parameter from an url. The url looks like this:
www.example.com/?v=12345
I want to get the parameter and query my mysql database to get the right data with ajax.
So i have my ajax call here:
$.ajax({
type:"POST",
url:"ajax2.php",
dataType:"json",
success:function(response){
var id = response['id'];
var url = response['url'];
var name = response['name'];
var image = response['image'];
},
error:function(response){
alert("error occurred");
}
});
As you can see, the data which i want to get are in a json array and will be saved in javascript variables.
This is my php file:
<?php
// Connection stuff right here
$myquery = "SELECT * FROM mytable **WHERE id= **$myurlvariable**;
$result = mysql_query($myquery);
while($row = mysql_fetch_object($result))
{
$currentid = "$row->id";
$currentname = "$row->name";
$currenturl = "$row->url";
$currentimage = "$row->image";
$array = array('id'=>$currentid,'url'=>$currenturl, 'name'=>$currentname,'image'=>$currentimage);
echo json_encode($array);
}
?>
The part where i want to query the right variable is bolded. I don't know how to query that. And Furthermore how to even get the url parameter in the proper form.
Can anybody help? Thank you!
You can get the query string using JavaScript and send it in the AJAX request.
Getting the query string(JavaScript) -
function query_string(variable)
{
var query = window.location.search.substring(1);
var vars = query.split("&");
for (var i=0;i<vars.length;i++) {
var pair = vars[i].split("=");
if(pair[0] == variable){return pair[1];}
}
return(false);
}
//Getting the parameter-
v = query_string('v'); // Will return '12345' if url is www.example.com/?v=12345
This needs to be passed as data in the AJAX call.
$.ajax(
{
type: "POST",
dataType: "json",
url: "ajax2.php",
data: "v="+v,
success: function(response){
var id = response['id'];
var url = response['url'];
var name = response['name'];
var image = response['image'];
},
error: function(jqXHR,textStatus,errorThrown){
//alert(JSON.stringify(jqXHR));
//alert(textStatus);
//alert(errorThrown);
alert(JSON.stringify(jqXHR)+" "+textStatus+" "+errorThrown);
//alert("error occurred");
}
}
);
This can be accessed as $_POST['v'] in the php form.
if(isset($_POST['v'])){
$myurlvariable = $_POST['v'];
$myquery = "SELECT * FROM mytable WHERE id= $myurlvariable";
...
And in php form, before you echo out the json response, change the content type. Something like this-
header("Content-Type: application/json");
echo json_encode($array);
If there is a database error, then it has to be handled.
So do this -
<?php
// Connection stuff right here
header("Content-Type: application/json");
if(isset($_POST['v'])){
$myurlvariable = $_POST['v'];
$myquery = "SELECT * FROM mytable WHERE id= $myurlvariable";
$result = mysql_query($myquery) or die(json_encode(Array("error": mysql_error()));
while($row = mysql_fetch_object($result))
{
$currentid = "$row->id";
$currentname = "$row->name";
$currenturl = "$row->url";
$currentimage = "$row->image";
$array[]= array('id'=>$currentid,'url'=>$currenturl, 'name'=>$currentname,'image'=>$currentimage);
}
echo json_encode($array);
}else{
echo json_encode(Array("error": "No POST values"));
}
?>
So this way, if the query has not executed properly, then you will know what exactly the error is.
Without any error checking, just the important part:
$myquery = "SELECT * FROM mytable WHERE id=" . $_POST['v'];
How to POST values from submit and check if they exist in mysql?
And what do I have to type in my .php file?
document.addEventListener("deviceready", onDeviceReady, false);
function onDeviceReady() {
$('#login').submit(function(){
var username = document.getElementById("username").value;
var password = document.getElementById("password").value;
});
}
function getData(sendData) {
$.ajax({
type: 'POST',
url: 'http://www.url.php',
data: { 'username': username, 'password': password },
success: afhandeling,
});
}
Call ajax like this:
jQuery.ajax({
type: "POST",
url: "http://www.url.php",
data: { username:username,password:password },
success: function( data )
{
}
});
and in ajax file:
if (isset($_POST['username']) && isset($_Post['password']))
{
$query = "SELECT * FROM users WHERE username='".$_POST['username']."' AND password=".$_POST['password'];
$result = mysql_query($query);
$row = mysql_fetch_assoc($result);
if($row)
{
echo 'login';
}
else
{
echo "error";
}
}
I think the URL has to be a local one, i.e. "/projects/blindchat/login.php".
On that page you can write something like this:
if (isset($_POST['username']) && isset($_POST['password'])) {
// MYSQL query:
SELECT 1 FROM users WHERE username = ? AND password = ?
}
Remember you have to escape the variables first to prevent SQL injection.
In login.php page you need to do something like this:
if(isset($_POST['username']) && isset($_Post['password'])) {
$q = "SELECT * FROM users WHERE username=$_POST['username'] AND password=$_POST['password']"
$r = mysql_query($q);
if(mysql_num_rows($r)==1) //Do Login
else echo "ERROR";
}
You submit the form which launches your ajax script that sends the data over to your PHP file that handles the input and gives you an answer.
Use PDO or MySqLi. Mysql is depreceated and no longer supported. My example below uses the PDO method.
Your PHP should look something like this(this is untested code, so there might be typos):
<?php
$username = $_POST['username'];
$password = $_POST['password'];
if (!empty($username) && !empty($password)) {
// We create a PDO connection to our database
$con = new PDO("mysql:host=yourhost;dbname=yourdatabase", "username", "password");
// We prepare our query, this effectively prevents sql injection
$query = $con->prepare("SELECT * FROM table WHERE username=:username AND password=:password LIMIT 1");
// We bind our $_POST values to the placeholders in our query
$query->bindValue(":username", $username, PDO::PARAM_STR);
$query->bindValue(":password", $password, PDO::PARAM_STR);
// We execute our query
$query->execute();
$result = $query->fetch(); // Grab the matches our query produced
// Here we check if we found a match in our DB
if (!empty($result)) {
echo "Matches were found";
} else {
echo "No matches found";
}
} else {
echo "Please fill out all fields";
}
?>
As for getting a reply from your AJAX script you can simply alert the response or show it as you please.
success: function(data) {
alert(data);
}
I have jquery pop form . It takes one input from the user ,mapping_key , Once the user enters the mapping key ,i make an ajax call to check if there is a user in the database with such a key.
This is my call .
Javascript:
$.ajax({
url : base_url+'ns/config/functions.php',
type: 'POST',
data : {"mapping_key":mapping_key} ,
success: function(response) {
alert(response)
}
});
PHP:
$sql = "select first_name,last_name,user_email,company_name from registered_users where mapping_key = '$mapping_key'";
$res = mysql_query($sql);
$num_rows = mysql_num_rows($res);
if($num_rows == 0)
{
echo $num_rows;
}
else{
while($result = mysql_fetch_assoc($res))
{
print_r($result);
}
}
Now i want to loop through the returned array and add those returned values for displaying in another popup form.
Would appreciate any advice or help.
In your php, echo a json_encoded array:
$result = array();
while($row = mysql_fetch_assoc($res)) {
$result[] = $row;
}
echo json_encode($result);
In your javascript, set the $.ajax dataType property to 'json', then you will be able to loop the returned array:
$.ajax({
url : base_url+'ns/config/functions.php',
type: 'POST',
data : {"mapping_key":mapping_key} ,
dataType : 'json',
success: function(response) {
var i;
for (i in response) {
alert(response[i].yourcolumn);
}
}
});
change
data : {"mapping_key":mapping_key} ,
to
data: "mapping_key=" + mapping_key,
You have to take the posted mapping_key:
$mapping_key = $_POST['mapping_key'];
$sql = "select first_name,last_name,user_email,company_name from registered_users
where mapping_key = '$mapping_key'";
or this:
$sql = "select first_name,last_name,user_email,company_name from registered_users
where mapping_key = $_POST['mapping_key']";
How will i process this Ajax in php.
What i want to do is send the data to process.php and if mode=loadlinks it will do a mysql query
function PresentLinks(div_id){
$("#loading-status").fadeIn(900,0);
$("#loading-status").html("<img src='img/bigLoader.gif' />");
$.ajax({
type: "POST",
url: "process.php",
data: "mode=loadlinks",
success: function(msg){
$("#loading-status").fadeOut(900,0);
$("#"+div_id).html(msg);
}
});}
What i want to process is
if($_POST['mode'] == loadlinks){ // this is what i want to ask
$query = "SELECT * FROM site ORDER BY link_id DESC";
$result = MYSQL_QUERY($query) or die (mysql_error());
while($data = mysql_fetch_row($result)){
echo ("$data[1]");
}}
else {
}
You need to quote strings in PHP. Otherwise they will be assumed to be constants. You should also be using PDO.
if($_POST['mode'] == 'loadlinks'){
$pdo = new PDO('mysql:host=HOST;dbname=DATABASE'), 'username', 'password');
$stmt = $pdo->execute('SELECT * FROM site ORDER BY link_id DESC');
$sites = $stmt->fetchAll();
foreach($sites as $site) {
echo "<div>" . $site['name'] . "</div>"; // Or whatever info you want to output
}
}
For performance you should be specifying table column names to retrieve instead of using *.
you need to quote the string value
if($_POST['mode'] == 'loadlinks'){.....