I want to display the array data that I have called in getword.php into the text field in index.php using AJAX. but how ?
this is the index.php
<body>
<div class="container" >
<h2>View data</h2>
<h4>Word List : </h4>
<div class="form-group">
<input id="wordlist" type="text" class="form-control" name="wordlist">
</div><br>
<button id="display" title="Generate Word">Generate</button>
<div class="input-single">
</div>
<script type="text/javascript">
$(document).ready(function() {
$("#display").click(function() {
$.ajax({
type: "POST",
url: "getword.php",
dataType: "json",
success: function($result){
$('#wordlist').val('');
}
});
});
})
</script>
and this is the sql query to get data from database into array (getword.php)
<?php
$host = "localhost";
$user = "root";
$password = "";
$dbname = "wordlist";
$con = mysqli_connect($host, $user, $password, $dbname);
if (!$con) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "select kata from word";
$res = mysqli_query($con,$sql);
$result = array();
while($row = mysqli_fetch_array($res)){
$result[]= $row[0];
}
echo json_encode(array('kata'=>$result));
mysqli_close($con);
?>
I would suggest you write PHP code as below it will be lot more simple
<?php
$host = "localhost";
$user = "root";
$password = "";
$dbname = "wordlist";
$con = mysqli_connect($host, $user, $password, $dbname);
if (!$con) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT kata FROM word";
$res = $con->query($sql);
while($row = mysqli_fetch_array($res)){
$dataToSend = $row[0];
}
mysqli_close($con);
echo json_encode($dataToSend);
?>
Now the script be like this to place the result in the text box
$.ajax({
type: "POST",
url: "getword.php",
dataType: "json",
success: function($result) {
var val = JSON.parse($result);
console.log(val); // see console in browser to check what is sent back
var txtBox = docuent.getElementById('wordlist');
txtBox.value = val;
}
});
Here I guess the data from the PHP file will be a single value since you are using a single text box. if multiple values are there then use an array to store data in PHP and send it in json_encode() itself.
hope that this becomes helpful to you.
Related
I have button which pressed update row in database based on variable:
<input type="submit" id="wyslij" name="przycisk" value="<?php echo $checkboxstatus;?>">
and I try to make ajax connection in wordpress but its not working, i tried diffrent ways but still the same results. It's even not responding while clicking the button and not receving error.
Ajax:
jQuery(document).ready( function() {
jQuery("#wyslij").click( function(e) {
e.preventDefault();
checkboxstatus = <?php echo $checkboxstatus; ?>
jQuery.ajax({
type : "post",
url : czekboks.php,
data : {checkboxstatus},
})
})
})
</script>
czekboks.php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "obejrzaneodcinki";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
$query = " UPDATE uzytkownik
SET status = '$checkboxstatus'
WHERE id = '2115' ";
if(mysqli_query($conn,$query))
{
echo "good job";
echo "<br />";
}
I'm developing a website using php. I have some problems. I want to know know how to get modified table values while running the php page without refreshing the page.
<html>
<?php
function fun_get_user_name() {
$host_name = "localhost";
$db_user_name = "root";
$password = "";
$database_name = "database_name";
$connect = mysqli_connect($host_name, $db_user_name, $password, $database_name);
$query = "SELECT * FROM `users` ";
$result = mysqli_query($connect, $query);
$output = "";
while ($row = mysqli_fetch_array($result)) {
$output = $output."<br/>"..$row[0];
}
}
?>
<script>
function js_function() {
result = "<?php echo fun_get_user_name; ?>";
document.getElementById('div_body_users').innerHTML = result;
}
window.setInterval(function() {
js_function();
}, 1000);
</script>
<body>
<div id="div_body_users">
</div>
</body>
</html>
when I made a change in phpmyadmin table the change didn't affect the page. But I expected the updated table.
So move the fun_get_user_name to another file and then in a setInterval do a n ajax call to that file.
$.get( "users.php", function( data ) {
$( ".result" ).html( data );
alert( "Load was performed." );
});
For more info on ajax request look at this link
https://api.jquery.com/jquery.get/
On user.php you just need to add fun_get_user_name
You can do it with using ajax, here changed with your ajax url and database connection details.
<html>
<?php
function fun_get_user_name()
{
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "test";
// Create connection
$conn = new mysqli($servername, $username, $password,$dbname);
$sql = "SELECT * FROM users";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["name"]. "<br>";
}
} else {
echo "0 results";
}
}
if($_GET['ajax']==1){
$data=fun_get_user_name();
echo $data;
exit(0);
}
?>
<script src="https://code.jquery.com/jquery-2.2.4.min.js"></script>
<script>
$(document).ready(function(){
setInterval(js_function,1000);
function js_function()
{
$.ajax({
url: "http://localhost/test2/test.php?ajax=1",
data: '',
cache: false,
processData: false,
contentType: false,
type: 'POST',
success: function (result) {
document.getElementById('div_body_users').innerHTML=result;
}
});
}
});
</script>
<body>
<div id="div_body_users">
</div>
</body>
</html>
You for update page without page refresh you have to use AJAX along with setInterval function.
Please check below link
https://www.w3schools.com/asp/asp_ajax_intro.asp
I have my php file on a server that retrieves data from my database.
<?php
$servername = "myHosting";
$username = "myUserName";
$password = "MyPassword";
$dbname = "myDbName";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, name, description FROM tableName;";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
$row_number = 0;
while($row = $result->fetch_assoc()) {
$row_number++;
echo $_GET[$row_number. ";". $row["id"]. ";". $row["name"]. ";". $row["description"]. "<br>"];
}
} else {
echo "0 results";
}
$conn->close();
?>
Unfortunately, I do not know how to receive data from a php file using javascript.
I would like the script in javascript to display the received data in the console in browser.
The script written in javascript is Userscript in my browser extension(tampermonkey) and php file is on my server.
I've tried to use ajax, unfortunately without positive results.
(the php script works as expected).
JS(not working):
$.ajax({
url: 'https://myserver.com/file.php',
type: 'POST',
success: function(response) {
console.log(response);
}
});
The code within the loop is a little screwy
$_GET[$row_number. ";". $row["id"]. ";". $row["name"]. ";". $row["description"]. "<br>"]
that suggests a very oddly named querystring parameter which is not, I think, what was intended.
Instead, perhaps try like this:
<?php
$servername = 'myHosting';
$username = 'myUserName';
$password = 'MyPassword';
$dbname = 'myDbName';
$conn = new mysqli($servername, $username, $password, $dbname);
if( $conn->connect_error ) {
die( 'Connection failed: ' . $conn->connect_error );
}
$sql = 'select `id`, `name`, `description` from `tablename`;';
$result = $conn->query($sql);
if( $result->num_rows > 0 ) {
$row_number = 0;
while( $row = $result->fetch_assoc() ) {
$row_number++;
/* print out row number and recordset details using a pre-defined format */
printf(
'%d;%d;%s;%s<br />',
$row_number,
$row['id'],
$row['name'],
$row['description']
);
}
} else {
echo '0 results';
}
$conn->close();
?>
A full example to illustrate how your ajax code can interact with the db. The php code at the top of the example is to emulate your remote script - the query is more or less the same as your own and the javascript is only slightly modified... if you were to change the sql query for your own it ought to work...
<?php
error_reporting( E_ALL );
ini_set( 'display_errors', 1 );
if( $_SERVER['REQUEST_METHOD']=='POST' ){
ob_clean();
/* emulate the remote script */
$dbport = 3306;
$dbhost = 'localhost';
$dbuser = 'root';
$dbpwd = 'xxx';
$dbname = 'xxx';
$db = new mysqli( $dbhost, $dbuser, $dbpwd, $dbname );
$sql= 'select `id`, `address` as `name`, `suburb` as `description` from `wishlist`';
$res=$db->query( $sql );
$row_number=0;
while( $row=$res->fetch_assoc() ){
$row_number++;
/* print out row number and recordset details using a pre-defined format */
printf(
'%d;%d;%s;%s<br />',
$row_number,
$row['id'],
$row['name'],
$row['description']
);
}
exit();
}
?>
<!DOCTYPE html>
<html lang='en'>
<head>
<meta charset='utf-8' />
<script src='//code.jquery.com/jquery-latest.js'></script>
<title>Basic Ajax & db interaction</title>
<script>
$( document ).ready( function(){
$.ajax({
url: location.href,
type: 'POST',
success: function( response ) {
console.log( response );
document.getElementById('out').innerHTML=response;
}
});
} );
</script>
</head>
<body>
<div id='out'></div>
</body>
</html>
Hi you can do it this way:
your php script:
if (isset($_POST["action"])) {
$action = $_POST["action"];
switch ($action) {
case 'SLC':
if (isset($_POST["id"])) {
$id = $_POST["id"];
if (is_int($id)) {
$query = "select * from alumni_users where userId = '$id' ";
$update = mysqli_query($mysqli, $query);
$response = array();
while($row = mysqli_fetch_array($update)){
.......
fill your response here
}
echo json_encode($response);
}
}
break;
}
}
Where action is a command you want to do SLC, UPD, DEL etc and id is a parameter
then in your ajax:
function getdetails() {
var value = $('#userId').val();
return $.ajax({
type: "POST",
url: "getInfo.php",
data: {id: value}
})
}
call it like this:
getdetails().done(function(response){
var data=JSON.parse(response);
if (data != null) {
//fill your forms using your data
}
})
Hope it helps
I have a textbox with the ID and NAME called "name". In my database I have firstname, preposition, lastname. In SQL I am using concat to combine these to a "name".
When I am trying to echo the result $result['name'] I get a browser error.
I guess something is wrong with the multiple variable $name = in my code. But I could not fix it.
Does someone know what is wrong with my code?
<?php
session_start();
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "db";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname) ;
// Check connection
if ($conn->connect_error) {
die('Connection failed: ' . $conn->connect_error) ;
}else {
$id = isset($_POST['id']) ? $_POST['id'] : '';
$name = isset($_POST['firstname']) ? $_POST['firstname'] : '';
$name .= isset($_POST['preposition']) ? $_POST['preposition'] : '';
$name .= isset($_POST['lastname']) ? $_POST['lastname'] : '';
$query = 'SELECT concat(firstname, ' ', preposition, ' ', lastname) as name FROM users WHERE id="' . mysqli_real_escape_string($conn, $id) . '"';
$res = mysqli_query($conn, $query) ;
if (mysqli_num_rows($res) > 0) {
$result = mysqli_fetch_assoc($res) ;
echo $result['name'];
}else{
$result = mysqli_fetch_assoc($res) ;
echo $result['name'];
}
}
?>
Edit 1:
Textbox:
<input type="text" class="form-control" id="name" name="name">
Javascript:
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
function getUser(value) { // Do an Ajax request to retrieve the product price
console.log("getUser before ajax", jQuery('#id').val());
jQuery.ajax({
url: './get/get5.php',
method: 'POST',
data: {'id' : jQuery('#id').val()},
success: function(response){
console.log("getUser after ajax", jQuery('#id').val());
jQuery('#name').val(response);
},
error: function (request, status, error) {
alert(request.responseText);
},
});
}
</script>
All I want to do is print 'win!' if they log in with their details in the Database (working correctly) and 'loss' if for some reason their info was not found in the DB.
So my issue is that for some reason my line of code 'echo $email;' doesn't work. It seems be set to NULL.
At the moment it only ever prints 'loss' regardless what i enter, but, if I add a row in the database that has a blank email and password (email = "", password="") then the php script returns 'win!'.
PHP CODE:
<?php
// echo "php test";
//server info
$servername = "localhost";
$username = "root";
$dbpassword = "root";
$dbname = "personal_data";
//Establish server connection
$conn = new mysqli($servername, $username, $dbpassword, $dbname);
//Check connection for failure
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
//Read in email & password
echo "reading in email & password...";
$email = mysqli_real_escape_string($conn, $_POST['email1']);
$password = mysqli_real_escape_string($conn, $_POST['password1']);
echo $email; //this prints blank
echo $password; //this also prints blank
$sql = "SELECT Name FROM personal_data WHERE Email='$email' AND Password='$password' LIMIT 1";
$result = mysqli_query($conn, $sql);
if(mysqli_num_rows($result) > 0){
echo "win!!";
} else {
echo "loss";
}
mysqli_close($conn);
?>
JS CODE:
$(document).ready(function(){
// alert("js working");
$('#login_button').click(function(){
var email = $('#email').val(); //prints the correct value
var password = $('#password').val(); //prints the correct value
var dataString = 'email1=' + email
+ '&password1=' + password;
$.ajax({
type: "POST",
url: "http://localhost:8888/php/login.php",
data: dataString, //posts to PHP script
success: success()
});
});//eo login_button
function success(){
alert("success");
}
});//eof
Apart from the fact that that is completely, insanely useless and with no security whatsoever, you can just exchange $.ajax() for $.post() and do like this:
var loginEmail = $('#email').val();
var loginPassword = $('#password').val();
$.post('login.php',{email:loginEmail,password1:loginPassword},function(data) {
console.log(data);
})