I'm trying to pass a variable from an html file, using jquery and ajax to a php file, where I want to process it. I have two test files:
ajax.html
<html>
<head>
<title>ajax</title>
<script type="text/javascript" src="jquery.js"></script>
<link href="ajax.css" rel="stylesheet" type="text/css" />
<script type="text/javascript">
$(document).ready(function () {
$("#call_back").click(function () {
$.post(
"ajax.php", {
value: $("#input_text").val()
},
function (data) {
$("#response_text").val(data);
}
);
});
});
</script>
</head>
<body>
<div id="wrapper">
<h3>Ajax</h3>
<div class="entry-wrapper">
<h4>Data to send to the server</h4>
<input type="text" size="50" id="input_text" />
<input type="button" value="Ajax Callback" id="call_back" />
</div>
<div id="response_wrapper">
<textarea id="response_text"></textarea>
</div>
</div>
</body>
</html>
and ajax.php
<?php
$value = $_POST['value'];
echo "Returned from the server: $value";
?>
The html works fine. If I enter a value in the text field, it's passed to the php file, processed and sent back to be displayed in the textarea.
I've looked in firebug and the post tag in the console shows the values that I enter.
The problem is that I want to be able to use the variable in the php file, for arithmetic operations or running an sql query with it, but it shows up as undefined index when I open the php file.
I've looked through multiple threads on passing variables from jquery to php but still found no answer. I have no idea what the problem is. Suggestions are much appreciated. Thanks.
If I understand you question correctly, you need to define the variable in your url.
Example:
http://yoursite.com/ajax.php?value=yourdata
This way when you open the file you are defining the value which is all you are doing with your ajax request. If this is not what you are trying to accomplish, try to rephrase your question.
Revised after 1st comment:
<?php
if(isset($_POST['value'])){
//do whatever in here
dowhatever();
}
function dowhatever() {
$value = $_POST['value'];
echo "Returned from the server: $value";
if(value == 'data') {
//run query
}//end if
}
?>
I assume that when you say you can confirm it's working, but when you directly open the .php file, it has undefined index, then I would say that makes perfect sense. If you are browsing directly to the .php file, then no POST data is being sent. But if you are able to type in some text, fire off the submit, and get text back as planned, then it's working.
Related
I'm learning AJAX and want to create a really simple web app to use my knowledge in the "real world".
I'm trying to calculte different percentages of a user input value, and make it appears on the webpage, without refreshing, thanks to AJAX.
Here is my HTML form:
<form id="warmupForm" class="form">
<label for="userWorkLoad">Work load (in kgs)</label><br>
<input type="text" name="userWorkLoad" id="userWorkLoad">
<button type="submit">Calculate</button>
</form>
<div id="#output">This is where I want the result to be shown with AJAX</div>
Here is some of my PHP code, for you to get the idea:
# Get the user input (work load in kgs)
if (isset($_POST['userWorkLoad'])) {
$workload = $_POST['userWorkLoad'];
# Avoid JS hacking
$workload = htmlspecialchars($workload);
}
# CALCULATION #
# Calculate 55% of the work load (1st warm up set)
$FirstWarmupSet = ($workload * 0.55);
# Calculate 70% of the work load (2nd warm up set)
$SecondWarmupSet = ($workload * 0.7);
# First Warmup set #
echo "<li>Do 8 reps with " . $FirstWarmupSet . " kgs, then take 1 minute rest.</li>";
echo "<br>";
# Second Warmup set #
echo "<li>Do 5 reps with " . $SecondWarmupSet . " kgs, then take 1 minute rest.</li>";
echo "<br>";
// etc etc...
I'd like the different variables values from PHP to be shown in my "#output" div when the user click on the submit button.
I've tried a lot of different things (AJAX without jQuery, AJAX with jQuery), but didn't manage to get what I want.
I'm sure I'm doing something wrong, but I don't know what. I'm sure my PHP script is working, since I used it without AJAX without any problem.
I would be very grateful if someone could help me on that.
As mentioned above, the easiest way to make an AJAX request for you is probably to try jQuery:
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<!-- Add jQuery on your HTML page -->
<script src="https://code.jquery.com/jquery-3.5.1.min.js"></script>
<!-- Add some custom JavaScript file -->
<script type="text/javascript" src="script.js"></script>
</head>
<body>
<form id="warmupForm" class="form">
<label for="userWorkLoad">Work load (in kgs)</label><br>
<input type="text" name="userWorkLoad" id="userWorkLoad">
<button id="btn" type="submit">Calculate</button>
</form>
<div id="output">This is where I want the result to be shown with AJAX</div>
</body>
</html>
The script.js content:
$(function() {
// Process a button click
$("#btn").click(function() {
event.preventDefault();
// Get input field
var userWorkLoadInput = $("#userWorkLoad");
// Build some request parameters
var params = {
userWorkLoad: userWorkLoadInput.val()
};
// Let's name your PHP script file as "server.php"
// And send POST request with those parameters
$.post("server.php", params, function(response) {
// Response text we're going to put into the `output`
$("#output").html(response);
});
});
});
You can simply do it using Jquery instead of using Ajax (using PHP you should add method="POST" to the form).
Here's an example:
$(document).ready(function(){
$("#send").click(function(){
// your calculates
$("#output").html(...);
});
});
...
<button type="submit" id="send">Calculate</button>
I am having a bit of trouble figuring out how exactly to make a certain connection. It's a project I thought might be simple enough for me to do on my own without help, but I've hit a wall unfortunately. It's an 'exercise generator' that basically asks a few basic questions, and based on your answers, it outputs a recommended workout routine. I've constructed the MySQL database with plenty of exercises, have successfully connected the database, and have made the form.
What I am having trouble though, however, is storing those form results into variables, and based on those variables (if workout days is 3, for example, there would only be 3 groups of workouts printed instead of 5) output a routine into the same div as the form, effectively replacing it with the answer instead of placing it underneath the submitted form.
Index.php
<!DOCTYPE html>
<html>
<?php $page_title = "Workout Generator"; ?>
<link rel="stylesheet" type="text/css" href="style.css">
<head>
<script type="text/JavaScript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js" ></script>
</head>
<body>
<?php include("header.php"); ?>
<?php include("connect.php"); ?>
<div id="appWindow">
<h3>Please answer the following questions and click 'Submit' to generate your workout routine.</h3>
<form id="homeForm" method="post" >
<label for="name">Name: </label>
<input type="text" name="name"><br>
<label for="age">Age: </label>
<input type="number" name="age"><br>
<label for="workoutdays">How many days a week can you workout?</label>
<select name="workoutdays">>
<option value="1">3</option>
<option value="2">5</option>
</select><br>
<label for="workoutstyle">Do you prefer a gym, or bodyweight exercises?</label>
<select name="workoutstyle">>
<option value="1">Gym</option>
<option value="2">Bodyweight</option>
</select><br>
<button type="submit" name="submit">Submit</button>
<button type="reset" name="reset">Reset</button>
<div class="form_result"></div>
</form>
<?php
$age = $_POST['age'];
$workoutdays = $_POST['workoutdays'];
$workoutstyle = $_POST['workoutstyle'];
?>
</div>
<br><br><br>
<?php include("footer.php"); ?>
</body>
</html>
I don't necessarily want an answer giving me the exact code to enter, but would appreciate being pointed in the right direction to get that form pulling data from MySQL, and using AJAX to print in the same window without refreshing.
THANK YOU
First of all, you need to post your request. With $.ajax this
//Do this thing on page load
$(function() {
//handle submit
$("#homeForm").submit(function(e) {
//customize your submit
e.preventDefault();
$.ajax({
type: "POST",
url: youURL, //maybe an url pointing to index.php
data: yourData, //attach everything you want to pass
success: function(response) {
$("#appWindow").html(response);
}
});
});
});
code should help. You need to make sure you pass the necessary elements and you provide the correct url. On server side, generate the desired html and send back as response.
In the script file
$(document).ready(function(){
$(document).on('click','#submit_btn',function(){
var url = '/xyz.php'; //full path of the function where you have written the db queries.
var data = '';//any data that you would like to send to the function.
$.post(url,{data: data}, function(result){
var arr = JSON.parse(result);
});
});
});
in your php file once your database query has been executed and you get the result. for example
$result = //result of your mysql query.
echo json_encode($result);
exit;
I don't know if something like this has already been asked and answered, but since no matter what search query I make, nothing seems to be close to what I am looking to do. I am working on a project where the user will upload a file. Once the file has been uploaded it will show the user a success message as well as some file info. I am trying to keep this all within one page, if possible, but can't seem to get it to work. File gets uploaded, but the info does not show.
Here is something like what I am working with:
<?php
if(isset($_POST['uploadFile']) && isset($_FILES['file'])) {
move_uploaded_file($_FILES['file']['tmp_name'], "files/" . $_FILES['file']['name']);
$message = "\"" . $_FILES['file']['name'] . "\" uploaded successfully...";
}
?>
<html>
<head>
<title>Upload File</title>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
$(document).ready(function() {
$(".uploaded-file-info").hide();
$(".uploadForm").submit(function() {
$(".upload-form").hide();
$(".uploaded-file-info").show();
});
});
</script>
</head>
<body>
<div class="upload-form">
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" enctype="multipart/form-data" class="uploadForm" >
<input type="file" name="file" /><br />
<input type="submit" name="uploadFile" value="Upload File" />
</form>
</div>
<div class="uploaded-file-info">
<p><?php echo $message; ?></p>
</div>
</body>
</html>
Like I said, the file gets uploaded, but the form doesn't hide and the file info ($message) doesn't show. Any suggestions?
The problem is the JQuery part :
$(".uploadForm").submit(function() {
$(".upload-form").show();
$(".uploaded-file-info").show();
});
coupled with this line :
<form method="post"
The JQuery part is saying : As soon as the form on the page is submitted, Show the information DIV.
The "Form" part just say : Submit the form.
So, when you click the button, the form is submitted and at the same time, the JQuery is executed. But then the form that you just posted needs to "refresh" the page to get the response back from the server. Basically, the JQuery you wrote display your div while you submit it. Meaning that it will work for a fraction of a second but will display an empty div because the response of the server is not there yet.
What you probably want to do is something like :
When the page loads
And there is content in the uploaded-file-info
Show the info and hide the form.
Add a Style tag with the following :
<style>
.uploaded-file-info {
display: none;
}
</style>
It will always hide the uploaded-file-info when the page loads.
Then, change your JavaScript code with :
<script>
$(document).ready(function() {
if ($('.uploaded-file-info > p').html() != "") {
$(".uploaded-file-info").show();
}
});
</script>
It says that when the page loads, if something is present inside the children of the element "uploaded-file-info", then show it. Otherwise, do nothing.
An easier solution would be to display the block, with php (so on the server side), only if a file was uploaded. No need for JQuery (client side) code.
Remove all the JQuery code and code within "<style>" tags and replace surround your "div class="uploaded-file-info" with an IF like this :
<?php if ($message != '') { ?>
<div class="uploaded-file-info">
<p><?php echo $message; ?></p>
</div>
<?php } ?>
Here's what will happen then:
you post (from your browser) the form
the server receives your post
if there is a file uploaded, it will initiate your "message" variable
and if the message variable exists, the server will put the "div uploaded-file-info" into the response. If not, everything surrounded by the "if" won't be put into the response.
your browser will receive the response and display it on screen.
My code is as follows:
<html>
<input type="datetime" name="datepicker" id="datepicker" />
<P id=firstp></P>
<button id="button" onClick="test_function();" type="button">View</button>
<script type="text/javascript">
function test_function(){
$("#firstp").load("view_calls_sort.inc.php?datepicker="+ $("#datepicker").val());}
</script>
</html>
view_calls_sort.inc.php:
$datepicker=$_GET['datepicker'];
echo $datepicker;
im trying to get the datepicker value to view_calls_sort.inc.php, Above code doesn't seem to work,as an example when I input datetime like this
2005-04-19 12:00:00.000 it is not shown in view_calls_sort.inc.php, Can any one help me with this. Thanks
Try encoding it:
var value = $("#datepicker").val();
$("#firstp").load("view_calls_sort.inc.php", { datepicker: value });
or:
$("#firstp").load("view_calls_sort.inc.php?datepicker=" + encodeURIComponent(value));
Also make sure that your script tag is inside your <html> tag and not as it is currently shown in your question.
First, make sure that the function is called.
<script type="text/javascript">
function test_function(){
alert($("#datepicker").val());
$("#firstp").load(
"view_calls_sort.inc.php?datepicker="+$("#datepicker").val());
}
</script>
If the alert shows up, the method is called. And you get to see which value is sent to the server.
You also could change the php script and have it return a fixed value (echo 'Foo';) so you can see if the script is called.
I want to submit search query form & get search result without redirecting/reloading/refreshing on the same page.
My content is dynamic so can not use those "submit contact form without refreshing page which replies back on success".
In order to submit a form, collect the results from the database and present them to the user without a page refresh, redirect or reloading, you need to:
Use Ajax to Post the data from your form to a php file;
That file in background will query the database and obtain the results for the data that he as received;
With the query result, you will need to inject it inside an html element in your page that is ready to present the results to the user;
At last, you need to set some controlling stuff to let styles and document workflow run smoothly.
So, having said that, here's an working example:
We have a table "persons" with a field "age" and a field "name" and we are going to search for persons with an age of 32. Next we will present their names and age inside a div with a table with pink background and a very large text.
To properly test this, we will have an header, body and footer with gray colors!
index.php
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html lang="pt" dir="ltr">
<head>
<title>Search And Show Without Refresh</title>
<meta HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=ISO-8859-1">
<meta http-equiv="Content-Style-Type" content="text/css">
<!-- JQUERY FROM GOOGLE API -->
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript">
$(function() {
$("#lets_search").bind('submit',function() {
var value = $('#str').val();
$.post('db_query.php',{value:value}, function(data){
$("#search_results").html(data);
});
return false;
});
});
</script>
</head>
<body style="margin:0;padding:0px;width:100%;height:100%;background-color:#FFFFFF;">
<div style="width:1024px;margin:0 auto;height:100px;background-color:#f0f0f0;text-align:center;">
HEADER
</div>
<div style="width:1024px;margin:0 auto;height:568px;background-color:#f0f0f0;text-align:center;">
<form id="lets_search" action="" style="width:400px;margin:0 auto;text-align:left;">
Search:<input type="text" name="str" id="str">
<input type="submit" value="send" name="send" id="send">
</form>
<div id="search_results"></div>
</div>
<div style="width:1024px;margin:0 auto;height:100px;background-color:#f0f0f0;text-align:center;">
FOOTER
</div>
</body>
</html>
db_query.php
<?php
define("HOST", "localhost");
// Database user
define("DBUSER", "username");
// Database password
define("PASS", "password");
// Database name
define("DB", "database_name");
// Database Error - User Message
define("DB_MSG_ERROR", 'Could not connect!<br />Please contact the site\'s administrator.');
############## Make the mysql connection ###########
$conn = mysql_connect(HOST, DBUSER, PASS) or die(DB_MSG_ERROR);
$db = mysql_select_db(DB) or die(DB_MSG_ERROR);
$query = mysql_query("
SELECT *
FROM persons
WHERE age='".$_POST['value']."'
");
echo '<table>';
while ($data = mysql_fetch_array($query)) {
echo '
<tr style="background-color:pink;">
<td style="font-size:18px;">'.$data["name"].'</td>
<td style="font-size:18px;">'.$data["age"].'</td>
</tr>';
}
echo '</table>';
?>
The controlling stuff depends from what you want, but use that code, place those two files in the same directory, and you should be fine!
Any problems or a more explicative code, please let us know ;)
You'll probably want to start with any of the thousands of "AJAX for beginners" tutorials you can find on the net. A Google search with that term should get you going.
Try this for starters:
http://www.destraynor.com/serendipity/index.php?/archives/29-AJAX-for-the-beginner.html
After you've read through that, keep in mind that you really don't need to be writing any XHR handling code. As pointed out by Jamie, jQuery or any of the other multitudes of Javascript libraries out there, can greatly simplify your client-side AJAX code.
This is what AJAX is for.
In jQuery (apologies if you're looking for a different library)
$("form#search").bind('submit',function() {
$.post("search.php",this.serialize(),function(data) {
// Put the code to deal with the response data here
});
return false;
});
It's good if you can get some basics of Ajax before straight away going to the code.
Ajax , is the exact solution for your problem. It asynchronously makes a request to the server, get the result and the data in the page can be modified with the result . It's all done in JavaScript.
Suppose you have an html like this:
<html>
<body>
<div id="myDiv"> your content area</div>
<button type="button" onclick="loadByAjax()">Change Content</button>
</body>
</html>
Now, your javascipr code will be like this:
<script type="text/javascript">
function loadByAjax()
{
$.ajax({
type: "POST",
url: "yourserverpage.php",
data: "searchkey=data_from_user_input",
success: function(response_data){
$('myDiv').html(response_data)
}
});
}
</script>
so, basically upon click of the button, the JavaScript will be executed. It wil call the php serverside script, pass the parameters it got from user input and retrieve the response data and place it inside the div.
So your page is updated without full refresh.
Also, please understand that, i used jquery library here for the Ajax function.