I am trying to show data from the database in my textbox. But when I start the script I am getting no results. I tested the script in different ways and i figured out that the variable: $product1 is empty. Does anybody know how I can fix this?
index.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "database";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM forms";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "<select class='form-control select2' id='product1' name='product1' onChange='getPrice(this.value)' style='width: 100%;'>";
echo "<option selected disabled hidden value=''></option>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<option value='" . $row["id"]. "'>" . $row["name"]. "</option>";
}
echo "</select>";
} else {
echo "0 results";
}
$conn->close();
?>
<html>
<body>
<!-- Your text input -->
<input id="product_name" type="text">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
function getPrice() {
// getting the selected id in combo
var selectedItem = jQuery('.product1 option:selected').val();
// Do an Ajax request to retrieve the product price
jQuery.ajax({
url: 'get.php',
method: 'POST',
data: 'id=' + selectedItem,
success: function(response){
// and put the price in text field
jQuery('#product_name').val(response);
},
error: function (request, status, error) {
alert(request.responseText);
},
});
}
</script>
</body>
</html>
get.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "database";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname) ;
// Check connection
if ($conn->connect_error)
{
die('Connection failed: ' . $conn->connect_error) ;
}
else
{
$product1 = filter_input(INPUT_POST, 'id', FILTER_SANITIZE_NUMBER_INT) ;
$query = 'SELECT price FROM forms WHERE id=" . $product1 . " ' ;
$res = mysqli_query($conn, $query) ;
if (mysqli_num_rows($res) > 0)
{
$result = mysqli_fetch_assoc($res) ;
echo $result['price'];
}else{
echo 'no results';
}
}
?>
Change
var selectedItem = jQuery('.product1 option:selected').val();
To
var selectedItem = jQuery('#product1 option:selected').val();
You are selecting a class with name product1, but you set only an ID with this name. Id's are specified with # and classes with .
Update on your script, because you used getPrice(this.value);
<script>
function getPrice(selectedItem) {
// Do an Ajax request to retrieve the product price
jQuery.ajax({
url: 'get.php',
method: 'POST',
data: 'id=' + selectedItem,
success: function(response){
// and put the price in text field
jQuery('#product_name').val(response);
},
error: function (request, status, error) {
alert(request.responseText);
},
});
}
</script>
TIP:
Did you know that you can use jQuery.ajax and jQuery('selector') also like this: $.ajax and $('selector') :-)
You have not a form tag in your HTML. The default form Method is GET.
In Your get.php you try to get a POST Variable with filter_input
The function filter_input returns null if the Variable is not set.
Two possible solutions:
1. Add a form to your html with method="post"
2. Change your php code to search for a GET variable
Related
After i searched for this solution on this site, nothing i found. Here is basic php code, just for testing.
index.php
<input type="checkbox" name="chk" id="chk" value="1" />
<button type="button" name="" id="submit">TEST</button>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
$('#submit').click(function(){
var check = $('#chk').val();
$.ajax({
url:"qry.php",
method:"POST",
data: {check:check},
success:function(data)
{
//alert(data);
window.location.reload();
}
});
});
});
</script>
qry.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "check";
// Create connection
$connect = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($connect->connect_error) {
die("Connection failed: " . $connect->connect_error);
}
if($_REQUEST['chk'] == true){
$stok = '1';
}
elseif ($_REQUEST['chk'] == false) {
$stok = '0';
}
$query = "INSERT INTO test(checked) VALUES('$stok')";
$result = mysqli_query($connect, $query);
if ($result === TRUE) {
echo "Zahtev je uspešno poslat!";
} else {
echo "Error: " . $query . "<br>" . $connect->error;
}
?>
How to set checkbox checked true or false into mysql and then echo if in html? It's always set to 0 in mysql boolen.
<input type="hidden" name="chk" id="chk" value="1" <?php if ($checked == '1') {echo 'checked';} else {} ?>/>
I tried everything from this site and nothing works.
Try this Jquery. This will get rid of the always value 1 problem you're having. What this code does is when you click on the "submit" button it check the status of your check box. If the check box is checked then the code will take the checked check box value and send that in the ajax function if it's not checked then the value 0 get assigned and that will be sent using the ajax.
Doing this will reduce the work has to be done by the back end PHP. I also made some changes to your PHP code as well.
$(document).ready(function() {
$(document).on('click', '#submit', function() {
if ($("#chk").is(":checked")) {
var chk = $('#chk').val();
}else{
chk = 0;
}
$.ajax({
url: "qry.php",
method: "POST",
data: {
check: chk
},
success: function(data) {
//alert(data);
window.location.reload();
}
});
});
})
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="checkbox" name="chk" id="chk" value="1" />
<button type="button" name="" id="submit">TEST</button>
You will see that the way I'm inserting the data is a bit different from the way you have done. I'm using mysqli_ prepared statements which makes SQL injection a hard to do.
$query = $connect -> prepare("INSERT INTO test(checked) VALUES (?)";
$query -> bind_param("i", $_REQUEST['check']);
if ($query -> execute()) {
echo "Zahtev je uspešno poslat!";
} else {
echo "Error: " . $query . "<br>" . $connect->error;
}
jsFiddle if you want to test it.
You write the value into check, but read it back from $_REQUEST['chk']. That won't work. Change that to $_REQUEST['check'].
You are using val() to get the state of the checkbox, you should use checked.
Also, you are possibly open to SQL injection, start using prepared statements.
Add another line while sending the AJAX request.
While sending the value of checkbox, send 1 or 0.
It will reduce our work at PHP end.
So, the code should be:
var check = $('#chk').is(":checked");
check = (check) ? 1 : 0;
Final code should be:
<input type="checkbox" name="chk" id="chk" value="1" />
<button type="button" name="" id="submit">TEST</button>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
$('#submit').click(function(){
var check = $('#chk').val();
check = (check) ? 1 : 0;
$.ajax({
url:"qry.php",
method:"POST",
data: {check:check},
success:function(data) {
//alert(data);
window.location.reload();
}
});
});
});
</script>
And PHP:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "check";
// Create connection
$connect = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($connect->connect_error) {
die("Connection failed: " . $connect->connect_error);
}
$stok = $_REQUEST['chk'];
$query = "INSERT INTO test(checked) VALUES('$stok')";
$result = mysqli_query($connect, $query);
if ($result === TRUE) {
echo "Zahtev je uspešno poslat!";
}
else {
echo "Error: " . $query . "<br>" . $connect->error;
}
?>
I am trying to select multiple columns with concat an put the returned data into one textbox.
I think there is something wrong with my definition for the variables. But I could not figured out what is wrong. Here are the variables:
$id = isset($_POST['id'])?$_POST['id']:'';
$name = isset($_POST['firstname'])?$_POST['firstname']:'';
$name .= isset($_POST['insertion'])?$_POST['insertion']:'';
$name .= isset($_POST['lastname'])?$_POST['lastname']:'';
When I define just one variable for $name the script works. But that is not what I want.
Does someone know what is wrong?
Here is the other part of my script.
First I have a textbox. The data needs to be send to this textbox:
<input type="text" class="form-control" id="name" name="name" placeholder="Name">
The button calls sends '5' as the ID and runs the script getName():
<button type="button" rel="5" onclick="getName();"
<script type="text/javascript">
$('body').on('click', '.selectClass', function () {
var id = $(this).attr('rel');
$("#id").val(id);
modal.style.display = "none";
});
</script>
After clicking on the button the id is deployed here:
<input type="text" class="form-control" id="id" name="id">
The onClick event runs the following script:
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
function getName(value) { // Do an Ajax request to retrieve the product price
console.log("getName before ajax", jQuery('#id').val());
jQuery.ajax({
url: './get/getname5.php',
method: 'POST',
data: {'id' : jQuery('#id').val()},
success: function(response){
console.log("getName after ajax", jQuery('#id').val());
jQuery('#name').val(response);
},
error: function (request, status, error) {
alert(request.responseText);
},
});
}
</script>
The jquery script calls the PHP, which is not working with the multiple variables for $name
<?php
session_start();
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "db";
$conn = new mysqli($servername, $username, $password, $dbname) ;
if ($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['insertion'])?$_POST['insertion']:'';
$name .= isset($_POST['lastname'])?$_POST['lastname']:'';
$query = 'SELECT concat(firstname, ' ', insertion, ' ', lastname) as name FROM users WHERE id="' . mysqli_real_escape_string($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'];
}
}
?>
I want to POST data from form. It works fine.
In the other functionality i want to get data from database.
I don't know where is mistake. I suspect that AJAX call is fine.
My PHP code:
<?php
$uuid = $_POST['uuid'];
$minor = $_POST['minor'];
$mayor = $_POST['mayor'];
$lokalizacja = $_POST['lokalizacja'];
$servername = "";
$username = "";
$password = "";
$dbname = "";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} else{
echo "Polaczono";
}
$sql = "INSERT INTO beacons (uuid, major, minor, lokalizacja)
VALUES ('$uuid', '$minor', '$mayor', '$lokalizacja')";
if ($conn->query($sql) === TRUE) {
echo "Dane dodano prawidłowo";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$sqlget = "SELECT uuid, major, minor, lokalizacja FROM beacons";
$result = $conn->query($sqlget);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo json_encode(array("value" => "UUID: " . $row["uuid"]));
}
} else {
echo "Brak rekordów w bazie";
}
$conn->close();
?>
AJAX call:
$('#admin').submit(function(e){
e.preventDefault();
if( ($("input[name=uuid]").val().length) > 40 || ($("input[name=minor]").val().length) > 5 || ($("input[name=mayor]").val().length) > 5 || ($("input[name=lokalizacja]").val().length) > 20){
$(".error-pola").show();
} else{
$.post('administrator-connect.php', $(this).serialize() )
.done(function(){
$(".success-wyslanie").show();
})
.fail(function(){
$(".error-wyslanie").show();
});
}
});
$(document).ready(function() {
$.ajax({
type: "GET",
url: 'administrator-connect.php',
dataType: 'json',
success: function(data)
{
alert("fsdfsd"+ data);
},
error: function(){
alert("not");
}
});
});
I am using:
echo json_encode(array("UUID" => $row["uuid"]));
and in ajax:
var jqxhr = $.get( "administrator-get.php", function(data) {
var jsonx = JSON.parse(JSON.stringify(data));
$( "#data-listing" ).html(jsonx);
});
But I get response:
{"UUID":"B9407F30-F5F8-466E-AFF9-25556B57FE6D"}
How to get only string ?
If you write this
dataType: 'json',
It expect for JSON value not string be sure to return only JSON.
You returns string value not JSON.
With like this code
echo "Polaczono";
Any echo would be the return value for ajax
At last you should return only one value like this.
echo json_encode($result);//an array result
You can check by string return. By removing dataType
I am working on a chrome extension (freshment) and have a little problem.
I have a button, and I want that when button is clicked, to show my information from my database on my extension page.
HTML :
<button class="button" id="show" style="vertical-align:middle" onclick="myAjax()"><span>Show my purchaes</span></button>
<div id="showhere">
//this is where i want to show the info
</div>
Java Script :
$(document).ready(function(){
function myAjax() {
$.ajax({
url:"http://127.0.0.1/show.php",
data:{ action:'showhere' },
method:"POST",
success:function(data) {
('#showhere').html(data);
}
});
}
});
PHP :
<?php
if($_POST['action'] == 'showhere') {
$servername = "localhost";
$username = "root";
$password = "********";
$dbname = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT ProductName, Amount, Date, WebStore FROM budget";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "<table><tr><th>ID</th><th>Name</th></tr>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>".$row["ProductName"]."</td><td>".$row["Amount"]."</td><td>".$row["Date"]."</td><td>".$row["WebStore"]."</td></tr>";
}
echo "</table>";
} else {
echo "0 results";
}
$conn->close();
}
?>
What I want it to do is pretty simple : I have a button and below I have a div called : "showhere", and in this div I want to take mysql info and write it.
your write i didnt write the exact problem, the problem is that the button doesnt do anything.
agian , thx!
I suggest you set it this way:
$(document).ready(function() {
$('#show').on('click', function(e) {
e.preventDefault();
$.ajax({
url: "http://127.0.0.1/show.php",
data: {
action: 'showhere'
},
method: "POST",
success: function(data) {
('#showhere').html(data);
}
});
});
});
Here i want to call the php function with some parameter that will be held in some jquery function.
I have php function that takes one parameter and executes the sql query.
and also i have a jquery function where i am getting different values when select item from dropdown box.
now i want to pass the value of dropdown box to php function, how can i do this?
here is what i have done
$("#product_category").change(function(){
var category = $(this).val(); //getting dropdown value
// here i want to pass this variable value to php function
});
php
function productManagement($procat){
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT product_name,product_category,product_image_url FROM product_list where product_category='$procat'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>". $row["product_name"]."</td><td><a href=''><span class='glyphicon glyphicon-remove'></span>Remove</a></td><td><a href='edit-product.php'><span class='glyphicon glyphicon-edit'></span>Edit</a></td></tr>";
}
} else {
echo "No results found";
}
$conn->close();
}
How can i do this?
Another way is to use Ajax like :
Jquery code:
$("#product_category").change(function(){
var category = $(this).val(); //getting dropdown value
// here i want to pass this variable value to php function
-----------------------------------------------------
$.ajax({
url: 'newFunctionFile.php',
type: 'POST',
data: 'category='+category,
success: function(data) {
//you can add the code to show success message
},
error: function(e) {
//called when there is an error
//console.log(e.message);
}
});
});
and you have to create a file for php function let say file name is newFunctionFile.php as i have mention in ajax call:
if(isset($_POST['category'])) {
productManagement($_POST['category']);
}
function productManagement($procat){
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT product_name,product_category,product_image_url FROM product_list where product_category='$procat'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>". $row["product_name"]."</td><td><a href=''><span class='glyphicon glyphicon-remove'></span>Remove</a></td><td><a href='edit-product.php'><span class='glyphicon glyphicon-edit'></span>Edit</a></td></tr>";
}
} else {
echo "No results found";
}
$conn->close();
}
How about something along these lines.
In your javascript:
$("#product_category").change(function(){
var category = $(this).val(); //getting dropdown value
location.href = location.href + "?val="+category; //reload the page with a parameter
});
And then in your PHP, add this code as your function call
if(isset($_GET['val'])) { //if 'val' parameter is set (i.e. is in the url)
productManagement($_GET['val']); //call function, with that parameter
}