What I'm trying to do is calling some database data via ajax and php. But the ajax call doesn't work, and I can't find out a solution on the web.
So here is my code:
test.php
<?php
include_once 'db_class.php';
$cat = $_GET['cat'];
$dbconn = new dbconn('localhost', 'root', 'somepsw', 'blog');
$dbconn->set_query("select * from posts where category = '".$cat."'");
echo '<br/>'.$dbconn->query.'<br/>';
$result = $dbconn->result;
$num = $dbconn->num_results;
$array = mysqli_fetch_assoc($result);
echo json_encode($array);
?>
If i type that url on browser: http://127.0.0.1:82/blog/ws/test.php?cat=css
The data returned via jsonEncode is correct, but when i'm loading it on a html page with jquery he can't read the data.
test.html
<html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<script>
function ajaxCall() {
var css;
$.ajax({
url: 'test.php',
type: "GET",
data: {cat: css},
dataType: 'json',
success: function(rows)
{
alert(rows);
},
error: function() { alert("An error occurred."); }
});
}
ajaxCall();
</script>
</head>
<body></body>
</html>
Thanks in advance.
I just rewrote the php code using PDO, should be more safe now.
db.php
<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpsw = "somepsw";
$dbname= "blog";
try {
#$dbh = new PDO("mysql:host=$dbhost;dbname=$dbname", $dbuser, $dbpsw);
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$dbh->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
}
catch(PDOException $e) {
echo "Connection failed, an error occured! Please contact server administrator."; //user friendly message
getErrorsLog($e->getMessage());
}
function closeDbConn () {
$dbh = null;
}
function getErrorsLog($message) {
$file = 'dberrors.log';
$date = date("d/m : H:i :");
// Open the file to get existing content
$current = file_get_contents($file);
// Append a new error message to the file
$current .= $date.$message;
$current .= "\r\n";
// Write the contents back to the file
file_put_contents($file, $current);
exit();
}
?>
blogdata.php
<?php
include_once "db.php";
$tableName = "posts";
$data = array();
#$view = $_GET["view"];
if (isset($_GET["view"])) {
$stmt = $dbh->prepare("SELECT * FROM $tableName WHERE category =? ORDER BY created DESC");
}
else {
try {
$stmt = $dbh->prepare("SELECT * FROM $tableName ORDER BY created DESC");
}
catch (PDOException $e) {
getErrorsLog($e->getMessage());
}
}
$stmt->bindValue(1, $view, PDO::PARAM_STR);
$stmt->execute();
$affected_rows = $stmt->rowCount(); //Rows count
if ($affected_rows == 0) {
echo "The data you looking for no longer exist, please contact the administrator.";
exit();
}
foreach ($stmt->fetchAll(PDO::FETCH_ASSOC) as $row) {
$data[] = $row;
}
echo json_encode($data);
closeDbConn();
?>
Your variable css has no value. You wanted to use the string 'css'. Maybe you want to be able to load other categories, too. So change your ajaxCall function to
function ajaxCall(category)
{
$.ajax({
url: 'test.php',
type: "GET",
data: {cat: category},
dataType: 'json',
success: function(rows) {
alert(rows);
},
error: function() {
alert("An error occurred.");
}
});
}
and call it using
ajaxCall('css');
Related
I am passing the $place variable to a query in listplace.php using a ajax call. The ajax call works perfectly in php1.php code, but the $place value is not passed over the query. Please help!
listplace.php too works perfectly but when i try to pass $place in where condition it fails.
php1.php code
<select id="name">
<option selected disabled>Please select</option>
</select>
<?php if (isset($_GET['place']) && $_GET['place'] != '') { ?>
<script src="https://code.jquery.com/jquery-3.2.1.min.js"></script>
<script>
$.ajax({
type: "POST",
data: {place: '<?= $_GET['place'] ?>'},
url: 'listplace.php',
dataType: 'json',
success: function (json) {
if (json.option.length) {
var $el = $("#name");
$el.empty(); // remove old options
for (var i = 0; i < json.option.length; i++) {
$el.append($('<option>',
{
value: json.option[i],
text: json.option[i]
}));
}
}else {
alert('No data found!');
}
}
});
</script>
<?php } ?>
listplace.php
<?php
//connect to the mysql
$db = #mysql_connect('localhost', 'root', 'password') or die("Could not connect database");
#mysql_select_db('test', $db) or die("Could not select database");
$place = $_POST['place'];
$sql = #mysql_query("select product_name from products_list where product_name = '$place'");
$rows = array();
while($r = mysql_fetch_assoc($sql)) {
$rows[] = $r['product_name'];
}
if (count($rows)) {
echo json_encode(['option'=> $rows]);
}else {
echo json_encode(['option'=> false]);
}
?>
An improvement will be to start using prepared statements. This is just an addition to Exprator's answer
This will prevent SQL injection attacks.
$sql_con = new mysqli('localhost', 'root', 'password', 'test');//get connection
$place = $_POST['place'];//posted variable
if($stmt = $sql_con->prepare("select product_name from products_list where product_name =?")) {//prepare returns true or false
$stmt->bind_param("s", $place); //bind the posted variable
$stmt->execute(); //execute query
$stmt->bind_result($product_name);//bind the result from query securely
$rows = array();//create result array
while ($stmt->fetch()) {//start loop
$rows[] = $product_name;//grab everything in array
}
if (count($rows)) {//check for number
echo json_encode(['option'=> $rows]);
} else {
echo json_encode(['option'=> false]);
}
change this line
data: {place: '<?= $_GET['place'] ?>'},
to
data: {place: '<?= $_GET["place"] ?>'},
I'm trying to convert a PHP variable to a JS variable so I can use it in a game I'm making. When I check the map code it is just undefined. Thanks in advance. FYI the PHP works.
<script>
var mapCode;
var used;
var active;
function downloadCode() {
$.ajax({
type: 'GET',
url: 'getMapCode.php',
data: {
mapCode: $mapCode,
used: $used,
active: $active,
},
dataType: "text",
});
}
</script>
<?php
ini_set('display_errors', 1);
error_reporting(E_ALL);
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "database";
// Create connection
$conn = mysqli_connect($servername, $username, $password);
mysqli_select_db($conn, $dbname);
// Check connection
if (!$conn)
{
die("Connection failed: " . mysqli_connect_error());
}
// echo "Connected successfully";
$query = "SELECT mapCode FROM mapCodes";
$result = mysqli_query($conn, $query);
$mapCode = mysqli_fetch_row($result);
$query1 = "SELECT used FROM mapCodes";
$result1 = mysqli_query($conn, $query1);
$used = mysqli_fetch_row($result1);
$query2 = "SELECT active FROM mapCodes";
$result2 = mysqli_query($conn, $query2);
$active = mysqli_fetch_row($result2);
mysqli_close($conn);
?>
I understand that the PHP Code is hideous but it works and I'm going to 'pretty it up' later when the whole thing is working
If the file extension is .php and not .js then this should work
<script>
function downloadCode() {
$.ajax({
type: 'GET',
url: 'getMapCode.php',
data: {
mapCode: "<?php echo $mapCode; ?>",
used: "<?php echo $used; ?>",
active: "<?php echo $active; ?>",
},
dataType: "text",
});
}
</script>
If you have .js file then declare javascript variable before including your js in .php file
<script>
var mapCode = "<?php echo $mapCode; ?>";
var used = "<?php echo $used; ?>";
var active = "<?php echo $active; ?>";
</script>
then in .js file you will get easily
<script>
function downloadCode() {
$.ajax({
type: 'GET',
url: 'getMapCode.php',
data: {
mapCode: mapCode,
used: used,
active: active,
},
dataType: "text",
});
}
</script>
You only need to use <?php echo $mapCode;?> instead $mapCode. .... php variables can't be reed whithout open Php tag
My current project is actually dealing with lots of ajax calls,
here is the simplified version of what I use to communicate with server:
// php
// needed functions
function JSONE(array $array)
{
$json_str = json_encode( $array, JSON_NUMERIC_CHECK );
if (json_last_error() == JSON_ERROR_NONE)
{
return $json_str;
}
throw new Exception(__FUNCTION__.': bad $array.');
}
function output_array_as_json(array $array)
{
if (headers_sent()) throw new Exception(__FUNCTION__.': headers already sent.');
header('Content-Type: application/json');
print JSONE($array);
exit();
}
// pack all data
$json_output = array(
'mapCode' => $mapCode,
'used' => $used,
'active' => $active
);
// output/exit
output_array_as_json( $json_output );
// javascript
function _fetch()
{
return $.ajax({
url: 'getMapCode.php', // url copied from yours
type: 'POST',
dataType: 'json',
success: function(data, textStatus, req){
console.log('server respond:', data);
window.mydata = data;
},
error: function(req , textStatus, errorThrown){
console.log("jqXHR["+textStatus+"]: "+errorThrown);
console.log('jqXHR.data', req.responseText);
}
});
}
window.mydata = null;
_fetch();
I have not tested this, but let me know I'll fix it for you.
How did i get you, you need to get the result from ajax request, to do it, you should first setup your php outputs your results, so the ajax can get outputed results from php like this:
<?php
ini_set('display_errors', 1);
error_reporting(E_ALL);
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "database";
// Create connection
$conn = mysqli_connect($servername, $username, $password);
mysqli_select_db($conn, $dbname);
// Check connection
if (!$conn)
{
die("Connection failed: " . mysqli_connect_error());
}
// echo "Connected successfully";
$query = "SELECT mapCode FROM mapCodes";
$result = mysqli_query($conn, $query);
$mapCode = mysqli_fetch_row($result);
$query1 = "SELECT used FROM mapCodes";
$result1 = mysqli_query($conn, $query1);
$used = mysqli_fetch_row($result1);
$query2 = "SELECT active FROM mapCodes";
$result2 = mysqli_query($conn, $query2);
$active = mysqli_fetch_row($result2);
mysqli_close($conn);
// Outputing results:
echo json_encode(array('mapCode'=>$mapCode[0], 'used'=>$used[0], 'active'=>$active[0]));
?>
Then in ajax, use success for listening return message after ajax finished:
<script>
var mapCode;
var used;
var active;
function downloadCode() {
$.ajax({
type: 'GET',
url: 'getMapCode.php',
data: {
/** Your data to send to server **/
},
dataType: "text",
success: function(data) { /** Here is data returned by php echo **/
var temp = $.parseJSON(data);
mapCode = temp['mapCode'];
used = temp['used'];
active = temp['active'];
}
});
}
</script>
I have two file, test.html & test.php. I would like to display the result of an SQL query via jQuery AJAX.
test.php outputs proper JSON, but I'm not able fetch the same on clicking upon the button "Fetch Data". Is it wrong way of using AJAX?
Once fetching the data in test.html, how do I access the contents?
test.html
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("button").click(function(){
$ajax({
url:'test.php',
type:'get',
dataType:'json',
success:function(data){
alert(data);
console.log(data['success']);
console.log(data.success);
}
});
});
});
</script>
</head>
<body>
<button>Fetch Data</button>
</body>
</html>
test.php
<?php
$dbuser="root";
$dbname="test";
$dbpass="root";
$dbserver="localhost";
// Make a MySQL Connection
$con = mysql_connect($dbserver, $dbuser, $dbpass) or die(mysql_error());
mysql_select_db($dbname) or die(mysql_error());
// Create a Query
$sql_query = "SELECT ID, UserName, Status FROM user_details1";
// Execute query
$result = mysql_query($sql_query) or die(mysql_error());
$jsonArray = array();
while ($row = mysql_fetch_array($result)){
$jsonArrayItem = array();
$jsonArrayItem["ID"] = $row["ID"];
$jsonArrayItem["UserName"] = $row["UserName"];
$jsonArrayItem["Status"] = $row["Status"];
array_push($jsonArray, $jsonArrayItem);
//echo '<option value='. $row['id'] . '>'. $row['login'] . '</option>';
}
mysql_close($con);
$tableData = array(
"data" => $jsonArray
);
header('Content-Type: application/json');
echo json_encode($tableData,JSON_UNESCAPED_SLASHES);
die();
?>
How do I display/access/print the fetched result contents (AJAX section)?
Make a function like this
var dataToSend = "My Data";
$(button).on("click",function(event){
$.ajax({
method: "POST",
url: "test.php",
data: { pDataToSend: dataToSend }
}).done(function( data ) {
$('.results').empty();
$('.results').append(data);
});
});
And make a div like this
<div class="results></div>
In your PHP file you can read the POST using this code.
$foo = $_POST['pDataToSend'];
Also, your test.php file is all over the place. Use a PDO like this
//connect and setup database example
try {
$db = new PDO("mysql:host=localhost;dbname=second;port=8889","root","root");
$db->setAttribute(PDO::ATTR_ERRMODE,PDO::ERRMODE_EXCEPTION);
$db->exec("SET NAMES 'utf8'");
} catch (Exception $e){
echo 'Could not connect to the database.105';
exit();
}
//select,insert,delete, update from database example
try{
$results = $db->prepare("SELECT * FROM articles WHERE article_id = ? AND user_id = ?");
$results->bindParam(1,$var);
$results->bindParam(2,$var2);
$results->execute();
$hold = $results->fetchAll(PDO::FETCH_ASSOC);
} catch (Exception $e) {
echo "Data could not be retrieved from the database.";
exit();
}
I'm new using ajax and I have a code to display from wordpress some information from database columns.
I have this PHP code to connect with the database and create the JSON file:
<?php
$username = $_REQUEST['username'];
$password = $_REQUEST['password'];
if (isset($username) && isset($password)) {
//CONEXION
$host="localhost";
$user="DB_Username";
$pass="DB_Password";
$dbname="DB_Name";
//Conexion
$conexion = mysqli_connect($host, $user, $pass,$dbname)
or die("unexpected error");
//gWe made the search
$sql = "SELECT * FROM Column WHERE A_Login='$username'";
mysqli_set_charset($conexion, "utf8");
if(!$result = mysqli_query($conexion, $sql)) die();
$clients = array();
$num_result = mysqli_num_rows($result);
if ($num_result == 0) {
$clients = array("error" => "true", "msg" => "We can't found this user", "data" => $username);
} else {
while($row = mysqli_fetch_array($result))
{
$id=$row['ID'];
$Name=$row['Name'];
if ($row['A_Login'] == $username && $row['A_Password'] == $password){
$clients[] = array('id'=> $id, 'Name'=> $Name);
} else {
$clients[] = array('error'=> "true", "msg" => "Incorrect data");
}
}
}
$close = mysqli_close($conexion)
or die("Unespected error with DB");
}
else {
$clients = array("error" => "true", "msg" => "You must fill all fields", "username" => $username);
}
//We build the JSON
$json_string = json_encode($clients);
echo $json_string;
?>
In a wordpress page I have this code, I build a form where if the user click the submit button call doLogin()
<script type="text/javascript"> function doLogin(){
data = {username: jQuery("#user").val(), password: jQuery("#pass").val()}
console.log(data);
jQuery.ajax({
type: "POST",
url: "Mywebsiteurl.php",
data: data,
beforeSend: function(){
},
success: function(data){
console.log(data);
//var arr = JSON.parse(data);
//$('#forma').html(data);
},
error: function(XMLHttpRequest, textStatus, errorThrown) {
alert("Error");
console.log(textStatus);
console.log(errorThrown);
}
});
} </script>
I need to show in <div id="forma"> a kind of list usign html, for example:
Id: VALUE ID
Name: VALUE NAME
and more information...
When i try to print in my website the required information using $('#forma').html(data); I obtain error or just an empty space.
How can I fix it? thanks.
In WordPress we need to hook the ajax hook to your check_user function here.
add_action('wp_ajax_your_action_from_js', 'your_function');
//Using ajax for non-logged users as well (PUBLIC)
add_action('wp_ajax_nopriv_your_action_from_js', 'your_function');
Check below code for how it is done regarding your context.
In functions.php
function check_user() {
$username = $_REQUEST['username'];
$password = $_REQUEST['password'];
if (isset($username) && isset($password)) {
//CONEXION
$host="localhost";
$user="DB_Username";
$pass="DB_Password";
$dbname="DB_Name";
//Conexion
$conexion = mysqli_connect($host, $user, $pass,$dbname)
or die("unexpected error");
//gWe made the search
$sql = "SELECT * FROM Column WHERE A_Login='$username'";
mysqli_set_charset($conexion, "utf8");
if(!$result = mysqli_query($conexion, $sql)) die();
$clients = array();
$num_result = mysqli_num_rows($result);
if ($num_result == 0) {
$clients = array("error" => "true", "msg" => "We can't found this user", "data" => $username);
} else {
while($row = mysqli_fetch_array($result))
{
$id=$row['ID'];
$Name=$row['Name'];
if ($row['A_Login'] == $username && $row['A_Password'] == $password){
$clients[] = array('id'=> $id, 'Name'=> $Name);
} else {
$clients[] = array('error'=> "true", "msg" => "Incorrect data");
}
}
}
$close = mysqli_close($conexion)
or die("Unespected error with DB");
}
else {
$clients = array("error" => "true", "msg" => "You must fill all fields", "username" => $username);
}
//We build the JSON
$json_string = json_encode($clients);
echo $json_string;
}
add_action('wp_ajax_check_user', 'check_user');
//Using ajax for non-logged users as well (PUBLIC)
add_action('wp_ajax_nopriv_check_user', 'check_user');
In your JS called file.
In the script the action is related to your _your_action_from_js. So action is needed for knowing where the ajax has to hit. In our case it executes our check_user and returns the appropriate values.
<script type="text/javascript">
function doLogin(){
data = {action: 'check_user', username: jQuery("#user").val(), password: jQuery("#pass").val()}
console.log(data);
jQuery.ajax({
type: "POST",
url: ajax_url,
data: data,
beforeSend: function(){
},
success: function(data){
console.log(data);
//var arr = JSON.parse(data);
//$('#forma').html(data);
},
error: function(XMLHttpRequest, textStatus, errorThrown) {
alert("Error");
console.log(textStatus);
console.log(errorThrown);
}
});
}
</script>
Reference Simple AJAX Form: http://wptheming.com/2013/07/simple-ajax-example/
CODEX Reference: https://codex.wordpress.org/AJAX_in_Plugins
WordPress has specific methods to enable ajax requests.
// registering ajax request for Logged users
add_action( 'wp_ajax_my_action', 'my_action_callback' );
// registering ajax request also for public area
add_action( 'wp_ajax_nopriv_my_action', 'my_action_callback' );
function my_action_callback()
{
// Your code here
wp_die(); // this is required to terminate immediately and return a proper response
}
To call it:
jQuery(document).ready(function($) {
var data = {action: "my_action", username: jQuery("#user").val(), password: jQuery("#pass").val()}
jQuery.ajax({
url: '/wp-admin/admin-ajax.php',
data: data,
method: 'POST',
success: function(response) {
console.log(response);
},
error: function(a,b,c) {
}
});
});
Source: https://codex.wordpress.org/AJAX_in_Plugins
I'm currently using Slim and Ajax to develop a mobile application. In my database I have a column which stores session codes for logins. I need to be able to once logged in, compare the username entered to the database and retrieve the sessionCode based on that.
My current PHP is this:
$app->post('/logIn/', 'logIn');
function logIn()
{
$request = \Slim\Slim::getInstance()->request();
$q = json_decode($request->getBody());
//$hashedPassword = password_hash($q->password, PASSWORD_BCRYPT);
$sql = "SELECT * FROM users where username=:username AND password=:password";
try {
$db = getConnection();
$stmt = $db->prepare($sql);
$stmt->bindParam("username", $q->username);
$stmt->bindParam("password", $q->password);
$stmt->execute();
//$row=$stmt->fetch(PDO::FETCH_ASSOC);
$row=$stmt->fetch(PDO::FETCH_OBJ);
//$verify = password_verify($q->password, $row['password']);
$db = null;
echo json_encode($row);
} catch (PDOException $e) {
echo $e->getMessage();
}
} //PHP 5.4 LogIn
$app->get('/getSessionCode/:username','getSessionCode');
function getSessionCode($username)
{
$request = \Slim\Slim::getInstance()->request();
$q = json_decode($request->getBody());
$sql = "SELECT * FROM users WHERE username=:username";
try{
$db = getConnection();
$stmt = $db->prepare($sql);
$stmt->bindParam("username", $username);
$stmt->execute();
$row=$stmt->fetchALL(PDO::FETCH_OBJ);
$dbh=null;
echo json_encode($row);
}
catch(PDOException $e){
if(db != null) $db = null;
echo $e->getMessage();
}
}
Ajax Code:
$("#checkLogIn").click(function()
{
username = $("#enterUser").val();
$.ajax({
type: 'POST',
contentType: 'application/json',
url: rootURL + '/logIn/',
dataType: "json",
data: checkLogIn(),
success: function(data)
{
if(data != false)
{
$.ajax({
type: 'GET',
url: rootURL + "/getSessionCode/" + username,
dataType: "json",
success: sessionData
});
}
else
{
alert("Username and/or Password was incorrect");
}
}
})
});
function checkLogIn(data)
{
return JSON.stringify({
"username": $("#enterUser").val(),
"password": $("#enterPass").val(),
});
}
function sessionData(data){
session = data.sessionCode;
alert(username);
alert(session);
$.mobile.changePage("#mainMenu");
}
When I run the application and log in. It runs though no problem but when it reaches alert(session) it returns undefined. I have both session and username set globally as variables so that isn't an issue and when it reaches alert(username), it returns the username I entered.
I was able to solve my issue by adding in a for loop into the sessionData function
function sessionData(data){
for(var i = 0; i < 1; i++)
{
sessionStorage.code = data[i].sessionCode;
}
$.mobile.changePage("#mainMenu");
}