I am using Ajax to send information about the screenwidth to a PHP page and I am also wanting the user to type a value into a text box however this is currently not working as only the textbox value is passed and I am told that the other value is an undefined index. Any help would be appreciated.
HTML/JS page
<form method="post" action = "insert.php">
<input type = "text" name ="carName" id ="carName"/>
<div id ="screenWidth" name = "screenWidth" method="post"></div>
<input type="submit" id= "submit" name="submit" value="submit"/>
<script type="text/javascript">
var screenWidth = screen.width + "px";
document.getElementById("screenWidth").innerHTML = screenWidth;
console.log(screenWidth)
var TestVal = ("test");
$.ajax({
type: 'POST',
url: 'insert.php',
data: {'screenWidth': screenWidth},
success: function(data){
console.log(data);
}
});
PHP Page
<?php
include 'db.php';
$screenWidth = $_POST['screenWidth'];
$screenHeight = $_POST['screenHeight'];
if(isset($_POST['submit']))
{
$screenWidth = $_POST['screenWidth'];
$phoneType = $_POST['carName'];
echo 'hello';
$sql = "INSERT INTO deviceInfo (screenWidth, carType, )
VALUES ('$screenWidth','$carType',)";
if (sqlsrv_query($conn, $sql)) {
echo "New record has been added successfully !";
} else {
echo "Error: " . $sql . ":-" . sqlsrv_errors($conn);
}
sqlsrv_close($conn);
}
?>
There are several issues with your setup.
1) only form inputs are submitted with the form. When you click the submit button, the form will submit through a browser HTML post and your JS will not be run. The form submit will only submit the carName variable in the post data as the screenWidth is not in a form input.
2) the JS you have written will only submit the screenWidth over ajax - and that will run as soon as the page is loaded (assuming this is only an extract and that this JS is in side a jQuery(document).ready(function($){ }) block.
3) your carName becomes carType and then phoneName in your code, so $carType is not set when you insert your SQL.
You have 2 options here..
1) change you div containing the screenWidth to a text input that is readonly. This way you can populate your screenWidth on page load and then have the user submit the form over HTML post when they are finished updating the carName. This will not use AJAX, or
2) get rid of the form and create a "click" function for submitting your AJAX that includes both vars... something like this (not tested).
jQuery(document).ready(function($){
$("#submit").click(function(){
e.preventDefault(); // stop the button from posting
$.ajax({
type: 'POST',
url: 'insert.php',
data: {'screenWidth': $("#screenWidth").text(),
'carName': $("#carName").val()},
success: function(data){
console.log(data);
}
});
});
});
Related
I am adding a text area on click of a particular div. It has <form> with textarea. I want to send the jquery variable to my php page when this submit button is pressed. How can this be achievable. I am confused alot with this . Being new to jquery dizzes me for now. Here is my code,
`
<script type="text/javascript">
$(document).ready(function(){
$('.click_notes').on('click',function(){
var tid = $(this).data('question-id');
$(this).closest('ul').find('.demo').html("<div class='comment_form'><form action='submit.php' method='post'><textarea cols ='50' class='span10' name='notes' rows='6'></textarea><br><input class='btn btn-primary' name= 'submit_notes' type='submit' value='Add Notes'><input type='hidden' name='submitValue' value='"+tid+"' /></form><br></div>");
});
});
</script>`
Your code works fine in the fiddle I created here -> https://jsfiddle.net/xe2Lhkpc/
use the name of the inputs as key of $_POST array to get their values.
if(isset($_POST['submitValue'])) { $qid = $_POST['submitValue']; }
if(isset($_POST['notes'])) { $notes = $_POST['notes']; }
You should send your data after form submitted, something like this
:
$(".comment_form form").submit(function(e) {
var form = $(this);
var url = form.attr('action');
$.ajax({
type: "POST",
url: url,
data: form.serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
e.preventDefault(); // avoid to execute the actual submit of the form.
});
you can assign event after insert your form.
// handling with the promise
$(this).closest('ul').find('.demo').html("<div class='comment_form'><form action='submit.php' method='post'></form><br></div>").promise().done(function () {
// your ajax call
});;
This is my Fiddle code:
$("form.signupform").submit(function(e) {
e.preventDefault();
var data = $(this).serialize();
var url = $(this).attr("action");
var form = $(this); // Add this line
$.post(url, data, function(data) {
$(form).children(".signupresult").html(data.signupresult);
$(form).children(".signupresult").css("opacity", "1");
});
return false;
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<form class="signupform" method="post" action="admin/signupinsert.php">
<p class="signupresult"></p>
<input type="text" name="firstname" />
<input type="submit" value="Sign Up"/>
</form>
Signupinsert.php page code:
// Code to insert data into Database
$signupresult = "Some value here";
$response = new \stdClass();
$response->signupresult = $signupresult;
header('Content-Type: application/json');
print json_encode($response);
Expected Result:
When user clicks on submit form button, the code runs in background. And submits the form without reloading the page.
And the signupinsert.php page return some text, and its text display on a paragraph with class signupresult.
And the form can be submitted unlimited times, without reloading the page.
Problem:
The form only gets submitted once. If I try to submit it twice, "Nothing Happens" (No values inserted into database, no value returned in paragraph with class signupresult.
Where is the problem?
You have to tell your request that you expect JSON as return. Else data.signupresult doesn't make sense; data is seen as a string.
I always use $.ajax, never $.post; I find it easier to add options.
$.ajax({
url: $(this).attr("action"),
dataType: 'JSON',
type: 'post',
data: $(this).serialize(),
success: function(data) {
...
}
})
I have a form with an input field for a userID. Based on the entered UID I want to load data on the same page related to that userID when the user clicks btnLoad. The data is stored in a MySQL database. I tried several approaches, but I can't manage to make it work. The problem is not fetching the data from the database, but getting the value from the input field into my php script to use in my statement/query.
What I did so far:
I have a form with input field txtTest and a button btnLoad to trigger an ajax call that launches the php script and pass the value of txtTest.
I have a div on the same page in which the result of the php script will be echoed.
When I click the button, nothing happens...
Test.html
<script type="text/javascript" src="http://code.jquery.com/jquery-1.4.2.min.js"></script>
<script type="text/javascript" src="http://ajax.microsoft.com/ajax/jquery.validate/1.7/jquery.validate.min.js"></script>
<script>
//AJAX CALL
function fireAjax(){
$.ajax({
url:"testpassvariable.php",
type:"POST",
data:{userID:$("#txtTest").val(),},
success: function (response){
$('#testDiv').html(response);
}
});
}
</script>
</head>
<body>
<form name="testForm" id="testForm" action="" method="post" enctype="application/x-www-form-urlencoded">
<input type="text" name="txtTest" id="txtTest"/>
<input type="button" id="btnLoad" name="btnLoad" onclick="fireAjax();"
<input type="submit" name="SubmitButton" id="SubmitButton" value="TEST"/>
</form>
<div id="testDiv" name="testDiv">
</div>
</body>
The submit button is to insert updated data into the DB. I know I have to add the "action". But I leave it out at this point to focus on my current problem.
testpassvariable.php
<?php
$player = $_POST['userID'];
echo $player;
?>
For the purpose of this script (testing if I can pass a value to php and return it in the current page), I left all script related to fetching data from the DB out.
As the documentation says 'A page can't be manipulated safely until the document is ready.' Try this:
<script>
$(document).ready(function(){
//AJAX CALL
function fireAjax(){
$.ajax({
url:"testpassvariable.php",
type:"POST",
data:{userID:$("#txtTest").val(),},
success: function (response){
$('#testDiv').html(response);
}
});
}
});
</script>
You need to correct two things:
1) Need to add $(document).ready().
When you include jQuery in your page, it automatically traverses through all HTML elements (forms, form elements, images, etc...) and binds them.
So that we can fire any event of them further.
If you do not include $(document).ready(), this traversing will not be done, thus no events will be fired.
Corrected Code:
<script>
$(document).ready(function(){
//AJAX CALL
function fireAjax(){
$.ajax({
url:"testpassvariable.php",
type:"POST",
data:{userID:$("#txtTest").val(),},
success: function (response){
$('#testDiv').html(response);
}
});
}
});
</script>
$(document).ready() can also be written as:
$(function(){
// Your code
});
2) The button's HTML is improper:
Change:
<input type="button" id="btnLoad" name="btnLoad" onclick="fireAjax();"
To:
<input type="button" id="btnLoad" name="btnLoad" onclick="fireAjax();"/>
$.ajax({
url: "testpassvariable.php",
type: "POST",
data: {
userID: $("#txtTest").val(),
},
dataType: text, //<-add
success: function (response) {
$('#testDiv').html(response);
}
});
add dataType:text, you should be ok.
You need to specify the response from the php page since you are returning a string you should expect a string. Adding dataType: text tells ajax that you are expecting text response from php
This is very basic but should see you through.
Change
<input type="button" id="btnLoad" name="btnLoad" onclick="fireAjax();"/>
Change AJAX to pass JSON Array.
data = $(this).serialize() + "&" + $.param(data);
$.ajax({
type: "POST",
dataType: "json",
url: "action.php",
data: data,
....
// action.php
header('Content-type: application/json; charset=utf-8');
echo json_encode(array(
'a' => $b[5]
));
//Connect to DB
$db = mysql_connect("localhst","user","pass") or die("Database Error");
mysql_select_db("db_name",$db);
//Get ID from request
$id = isset($_GET['id']) ? (int)$_GET['id'] : 0;
//Check id is valid
if($id > 0)
{
//Query the DB
$resource = mysql_query("SELECT * FROM table WHERE id = " . $id);
if($resource === false)
{
die("Database Error");
}
if(mysql_num_rows($resource) == 0)
{
die("No User Exists");
}
$user = mysql_fetch_assoc($resource);
echo "Hello User, your number is" . $user['number'];
}
try this:- for more info go here
$(document).ready(function(){
$("#btnLoad").click(function(){
$.post({"testpassvariable.php",{{'userID':$("#txtTest").val()},function(response){
$('#testDiv').html(response);
}
});
});
});
and i think that the error is here:-(you wrote it like this)
data:{userID:$("#txtTest").val(),}
but it should be like this:-
data:{userID:$("#txtTest").val()}
happy coding :-)
How can i submit a hidden form to php using ajax when the page loads?
I have a form with one hidden value which i want to submit without refreshing the page or any response message from the server. How can implement this in ajax? This is my form. I also have another form in the same page.
<form id = "ID_form" action = "validate.php" method = "post">
<input type = "hidden" name = "task_id" id = "task_id" value = <?php echo $_GET['task_id'];?>>
</form>
similar to Zafar's answer using jQuery
actually one of the examples on the jquery site https://api.jquery.com/jquery.post/
$(document).ready(function() {
$.post("validate.php", $("#ID_form").serialize());
});
you can .done(), .fail(), and .always() if you want to do anything with the response which you said you did not want.
in pure javascript
body.onload = function() {
var xmlhttp = new XMLHttpRequest();
xmlhttp.open("POST","validate.php",true);
xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
xmlhttp.send("task_id=" + document.getElementById("task_id").value);
};
I think you have doubts invoking ajax submit at page load. Try doing this -
<script type="text/javascript">
$(document).ready(function(){
$.ajax({
"url": "validate.php",
"type": "post"
"data": {"task_id": $("#task_id").val();},
"success": function(){
// do some action here
}
})
})
</script>
If you're using jQuery you should be able to get the form and then call submit() on it.
E.g.:
var $idForm = $('#ID_form');
$idForm.submit();
Simple solution - jQuery AJAX post the value as others have suggested, but embed the PHP value directly. If you have multiple forms, you can add more key:value pairs as needed. Add a success/error handler if needed.
<script type="text/javascript">
$(document).ready(function(){
$.post( "validate.php", { task_id: "<?=$_GET['task_id']?>" } );
})
</script>
As others have said, no need for a form if you want to send the data in the background.
validate.php
<?php
$task_id = $_POST['task_id'];
//perform tasks//
$send = ['received:' => $task_id]; //json format//
echo json_encode($send);
JQuery/AJAX:
$(function() { //execute code when DOM is ready (page load)//
var $task = $("#task_id").val(); //store hidden value//
$.ajax({
url: "validate.php", //location to send data//
type: "post",
data: {task_id: $task},
dataType: "json", //specify json format//
success: function(data){
console.log(data.received); //use data received from PHP//
}
});
});
HTML:
<input type="hidden" name="task_id" id="task_id" value=<?= $_GET['task_id'] ?>>
What I am trying to do is create a "save" button for my website which saves specific posts and comments, exactly like the "save" button on Reddit. For now I am trying to self teach jQuery AJAX and attempting to figure out how to submit data to the database without having to reload the whole page. What I am attempting to do here is save a string by submitting it to a table called "Saved" when I click on "save".
HTML
<div id="message1">
<div id="pmessage"><p><?php echo $myComment;?></p></div>
Save
Edit
Hide
</div>
<form action="ajaxexample.php" method="post" style="display: none" id="1234">
<input type="hidden" name="message" id="message" value="<?php echo $myComment; ?>">
</form>
jQuery
$('a.Save').click(function () {
if ($(this).text() == "Save") {
$("#1234").ajax({ url: 'ajaxexample.php', type: 'post', data: 'message' });
$("a.Save").text("Unsave");
} else {
$("a.Save").text("Save");
}
});
PHP5.3
$message = $_POST['message'];
$query = "INSERT INTO saved (comment) VALUES (?)";
$statement = $databaseConnection->prepare($query);
$statement->bind_param('s', $message);
$statement->execute();
$statement->store_result();
$submissionWasSuccessful = $statement->affected_rows == 1 ? true : false;
if ($submissionWasSuccessful)
{
header ("Location: index.php");
}
$myComment = "This is my message!";
As of now all I am trying to do is submit the message "This is my message!" into the database table "Saved". What is wrong with my code? Why can I not submit the data to the table and how can I fix it? Thanks in advance!
Submit form when someone clicks on a.Save
$('a.Save').click(function (e) {
e.preventDefault();
$("#1234").submit();
});
submit handler on form#1234
$("#1234").submit(function(e) {
e.preventDefault();
$.ajax({
type: "POST",
url: 'ajaxexample.php',
data: $("#1234").serialize(),
success: function(data)
{
// data stores the response from ajaxexample.php
// Change the html of save by using $("a.Save").html("Unsave");
}
});
});
Serialize automatically makes a query string.
$(".save").bind("click",function(e){e.preventDefault();
$.ajax({
url : $("#1234").attr("action"),
type : "POST",
data : $("#1234").serialize(),
success : function(data){},
fail : function(data){},
});
});