I have a hidden iframe where the submission of a form is handled. It goes like this:
<iframe name="foo" style="display:none;"></iframe>
So, I was wondering, if it is possible that after the stuff has happened that needs to be within the iframe, I can use javascript or something to print out data on the parent page? Thanks
EDIT: here is my form code.
<form id="bar" name="bar" method="post" target="foo" action="include/database.php">
<input type="text" name="betamount">
<input type='text' name="multipler">
<input type="checkbox" name="hilo" value="High" checked>
<input type="submit" name="submit" value="Bet">
</form>
<iframe name="foo" style="display:none;"></iframe>
Database.php handles these POST requests inside the iframe. Now, there is this one thing inside database.php which goes like this
$betamount = $_POST['betamount'];
$multiplier = $_POST['multiplier'];
$payout = (int)$betamount*(int)$multiplier;
What I want to do is, I want to use AJAX or something to echo out the 'payout' variables inside a div present on index.php
For the purposes of my answer, I'm assuming that the actions you are doing in server side cannot be replaced by a simple client-side one (using javascript).
If you are expecting a return, why don't you use AJAX directly, without iframes? Simply post the data to your php page, and return it asynchronously.
JsFiddle: http://jsfiddle.net/ericwu91/28U8n/
HTML Code:
<input type="text" id="amount" name="betamount">
<input type='text' id="multiplier" name="multipler">
<input type="checkbox" name="hilo" value="High" checked>
<button onclick="submit();return false;">Submit</button>
JS Code:
var yourData = {multiplier:$("#multiplier").val(),betamount:$("#amount").val()};
$.post("yourUrl.php",yourData,function(result){
//Success: Use the "result" parameter to retrieve the data returned from server
alert(result);
});
I'm using jQuery's ajax post method. Documentation here: http://api.jquery.com/jquery.post/
The perks of doing it this way is that it does exactly what you wanted to, but simplifies it by using an almost-native javascript property (asynchronous responses).
EDIT: I forgot to put the real jsfiddle link... And after I pasted all the HTML and JS code, I realized how useless the fiddle is, as it won't return any respose at all... xD
Related
I have a form to submit and send data to 2 pages via POST.
I have tried the code with javascript. One form submit is working but other submit is not working
<form id="add">
<input type="text" name="test">
<input type="submit" onclick="return Submit();">
</form>
javascript
function SubmitForm()
{
document.forms['add'].action='filecreate.php';
document.forms['add'].submit();
document.forms['add'].action='filecreate.fr.php';
document.forms['add'].submit();
return true;
}
The first submission is not working but 2nd submission is working.
Since you appear to send the exact same data to two different handlers, you can flip the coin - and say that you just submit one form, and process them both in filecreate.php.
As you are sending a form, you cannot send two separate forms in the same HTTP request - so you can either do them both through asynchronous methods, or process them both backend after the submission of one form.
Since you haven't shown any PHP code, I'm making some assumptions and writing some pseudo-code, but it should be enough to get you started.
So first off, set a static action-property to your form.
<form id="add" action="filecreate.php">
<input type="text" name="test">
<input type="submit">
</form>
If you are sending it over POST, then you need to specify the method as well,
<form id="add" action="filecreate.php" method="POST">
Then, in PHP, you can get both files executed if you include it to the other. Meaning, in your filecreate.php, you include the filecreate.fr.php. Once you do that, the contents of that file will be executed as well.
<?php
// Once you require the file, it will be executed in place
require "filecreate.fr.php";
// .. handle the rest of your normal execution here.
That said, if you are doing the very similar thing multiple times, just with different data, you may want to create functions for it instead - going with the DRY principle ("Don't Repeat Yourself"), you can probably create a function that handles the structure and processing, then send the data separately through that function.
Try this :
<form id="add">
<input type="text" name="test">
<input type="button" onclick="return SubmitForm();">
</form>
function SubmitForm()
{
if(document.forms['add'].onsubmit())
{
document.forms['add'].action='filecreate.php';
document.forms['add'].submit();
document.forms['add'].action='filecreate.fr.php';
document.forms['add'].submit();
}
return true;
}
I have a form inside a form and this makes the top form unresponsive. When I take off the second form (which is inside the first form), the first form works. This is the second form I have:
<form action="imgupload.php" method="post" enctype="multipart/form-data">
<h3>Upload a new image:</h3>
<input type="file" name="fileToUpload" id="fileToUpload">
<br>
<input type="hidden" value="<?php echo $row['Gallery_Id']; ?>" name="gid">
<input type="hidden" value="User" name="user">
<input type="submit" value="Upload Image" name="imgup">
</form>
Since this makes the first form not work, I was wondering if I can take off the form fields and then the submit button can send the form data to the imgupload.php like this.
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="hidden" value="<?php echo $row['Gallery_Id']; ?>" name="gid">
<input type="hidden" value="User" name="user">
<input type="submit" value="Upload Image" name="imgup" action="imgupload.php" method="post" enctype="multipart/form-data">
This does not work now. Is there a way I can get this working? If not, what's an alternative way to send this data to the other php?
Since you are uploading files, have a look at Ravi Kusuma's Hayageek jQuery File Upload plugin. It's simple, it's a Swiss Army Knife, and it works.
Study the examples.
http://hayageek.com/docs/jquery-upload-file.php
Ravi breaks down the process into three simple steps, that basically look like this:
<head>
<link href="http://hayageek.github.io/jQuery-Upload-File/uploadfile.min.css" rel="stylesheet"> // (1)
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script src="http://hayageek.github.io/jQuery-Upload-File/jquery.uploadfile.min.js"></script> // (1)
</head>
<body>
<div id="fileuploader">Upload</div> // (2)
<script>
$(document).ready(function(){
$("#fileuploader").uploadFile({ // (3)
url:"my_php_processor.php",
fileName:"myfile"
});
});
</script>
</body>
The final step is to have the PHP file specified in the jQuery code (in this case my_php_processor.php) to receive and process the file:
my_php_processor.php:
<?php
$output_dir = "uploads/";
$theFile = $_FILES["myfile"]["name"];
move_uploaded_file($_FILES["myfile"]["tmp_name"],$output_dir.$fileName);
Note the relationship between myfile in the PHP ($_FILES["myfile"]), and the filename specified in the jQuery code block.
Don't forget to check out the server-side code from the Server Side tab -- you need both parts (js and php).
Looking at your question again, you will probably want to use this functionality as well:
dynamicFormData: function()
{
var data ={ location:"INDIA"}
return data;
}
or
dynamicFormData: function(){
return {
newID: $("#newNID").val(),
newSubj: $("#newSubj").val(),
newBody: $("#newBody").val(),
formRole: $('#formRole').val()
};
These will appear on the PHP side, thus:
$newID = $_POST['newID'];
$subj = $_POST['newSubj'];
etc
As with any plugin, resist the temptation to just plop it into your code. Do a couple of quick-and-dirty tests with it first. Kick its tires. Fifteen minutes will save you two hours.
And don't forget to verify what was uploaded. You never know when a developing country black hat might be trying to get a new account.
I have a beginner question. What is the easiest way to take data from a form on one html page and display it on another when the user clicks submit? I have two functions, a Submit() that calls the display() function (the display function displays the data on the page). I first displayed the result on the index.html page but realized it was too cluttered so I opted to print the results to a separate html page. However, I cannot recall the proper way of doing this. I tried putting location.href='results.html' inside my display() function by it didn't work.
You can use just HTML + Javascript to achieve this.
Just create a form with method="get". So the values will be passed by querystring to the another page.
Example:
index.html
<html>
<form method="get" action="results.html">
<input type="text" name="age" />
<input type="submit" value="Send" />
</form>
</html>
results.html
<html>
<h1></h1>
<script>
document.querySelector("h1").innerHTML = window.location.search.substring(1);
</script>
</html>
Whilst technically this is possible using HTML5 local storage, the best solution to your question is to use a server side language such as PHP, which you can read up on here as a beginners tutorial, or in more detail on the PHP Manual
Hope this helps
Here is an example. Write your html page (e.g. "index.html") like
<html>
<head>
<title>form with output</title>
</head>
<body>
<form target="out" action="tst.php">
<input type="text" name="a">
<input type="text" name="b">
<input type="submit" name="sub" value="OK">
</form>
</body>
</html>
and, assuming you have PHP available on your webserver you can write a second (php) script (filename: "tst.php") like this
<?php
echo json_encode($_REQUEST);
?>
(The php script simply outputs all passed variables as a JSON string). The important thing that will redirect your form's output into a separate window is the target="out" part in the <form> tag.
Hello and thank you for viewing my question. I am a complete beginner and am looking for simple ways to do the following...
What I have in seperate linked documents:
HTML, CSS, Javascript, PHP
What I am having trouble with:
I need to use something like JSON (although I would also accept XML requests or Ajax at this point if they work) to transfer variables from Javascript to PHP. I need the variables to search in a database, so they need to be literally available within PHP (not only seen on a pop-up message or something).
I have seen a LOT of different ways to do this, I have even watched tutorials on YouTube, but nothing has worked for me yet. The things I am having the biggest problem with is that when I add a submit button to my form it doesn't submit my form and I don't know why.
Form code snippet:
<form id="form" name="input" method="post" action="javascript:proofLength();">
<input id="userinput" type="text" autofocus />
<input id="submit" type="button" value="submit" onsubmit="post();">
</form>
The second to last line there doesn't work. Do I need javascript to submit the form? Because I really thought that in this case it was part of the functionality of the form just like method="post"...
The other thing is that for JSON, I have no idea what to do because my variables are determined by user input. Therefore, I cannot define them myself. They are only defined by document.getElement... and that doesn't fit the syntax of JSON.
Those are really my main problems at the moment. So if anyone could show me a simple way to get this variable transfer done, that would be amazing.
After this I will need to search/compare in my database with some php/sql (it's already connecting fine), and I need to be able to return information back to a in HTML based on what I find to be true. I saw one example, but I am not sure that was very applicable to what I am doing, so if you are able to explain how to do that, that would be great also.
Thank you very, very much.
April
You don't need ajax to submit this form. You don't even need javscript. Just do this:
<form id="form" name="input" method="post" action="mytarget.php">
<input id="userinput" name="userinput" type="text" autofocus />
<input id="submit" type="submit" value="submit" />
</form>
This will send the form data to mytarget.php (can be changed of course)
See that i have added the name attribute to your text-field in the form and i changed the type of the button to submit.
Now you can work the Data in mytarget.php like this:
<?
$username = $_POST['userinput'];
echo "Your name is: ".$username;
?>
You wanted to have a check for length in the submit. There are two ways to this:
Before the input is send (the server is not bothered)
Let the server Check the input
for 1 you will have to append a event listener, like this:
var form = document.getElementById("form");
form.addEventListener("submit", function(event){
console.log("test");
var name = form.elements['userinput'].value;
if(name.length < 3){
alert("boy your name is short!");
event.preventDefault();
}
});
Enter a name with less then 3 characters and the form will not be submitted. test here: http://jsfiddle.net/NicoO/c47cr/
Test it Serverside
In your mytarget.php:
<?
$username = $_POST['userinput'];
if(strlen($username) > 3)
echo "Your name is: ".$username;
else
echo "your name was too short!";
?>
You may also do all this with ajax. You will find a lot of good content here. But I'd recommend a framework like jQuery to do so.
The problem is in this line
<form id="form" name="input" method="post" action="javascript:proofLength();">
The action should be a PHP page (or any other type of server script) that will process the form.
Or the proofLength function must call submit() on the form
In the php page you can obtain variable values using $_GET["name"] or $_POST["name"]
To summarize; your code should look like this
<form id="form" name="input" method="post" action="yourpage.php">
<input id="userinput" type="text" autofocus />
<input id="submit" type="button" value="submit">
</form>
and for your php page:
<?php
$userinput = $_POST["userinput"];
//Do what ever you need here
?>
If you want to do something in your javascript before submitting the form, refer to this answer
I'm using Google App Engine Go SDK and I want to put some basic javascript code into my HTML templates that will use parameters passed from the application. The template looks like this:
<script type="text/javascript">
function CopyToClipboard()
{
CopiedTxt = document.selection.createRange();
CopiedTxt.execCommand("Copy");
}
</script>
[...]
<form name="Form1">
<input type="hidden" name="link" value="{{.Link}}">
<input type="button" onClick="CopyToClipboard()" value="Copy to clipboard" />
</form>
What the code is supposed to do is copy the {{.Link}} value into the clipboard. But instead of getting things like http://example.com in the clipboard, I get {{.Link}}, even though the page source of the executed template clearly reads
<input type="hidden" name="link" value="http://example.com">
How can I make the javascript work properly with the GAE Golang template?
This has nothing to do with app engine, templates, or go. The problem is that .execCommand() will not generally work. Clipboard access is not something that can be done successfully through javascript. You must use a flash plugin.