I have a main.php page
the following code is hardcoded in main form
<form action="test.php" method="POST">
the following is the code generated dynamically using AJAX
<input type="checkbox" value="test" name="test[]"/>
<input type="checkbox" value="test1" name="test[]"/>
<input type="submit" value="go">
ideally speaking on clicking the go button the page should submit to test.php page with the post value of the check elements
but now i find no action is taken by the browser. Also i find no error message in error console.
this used to work upto firefox 3.5 and IE 8. However in Firefox 3.6 the dynamically generated form elements are not recognized at all
Is there a mistake in the code and is a there a work around
Make sure that the dynamic content does not replace the form .. but gets appended to it (or replaces the contents and not the actual form tag..)
Related
So I have a form that requires a user to submit their website to a form. Here is the html line:
<input type='url' name='link'>
And I'm using <input type="submit" value="submit" formmethod="post"> to submit the form to a php
And I'm trying to retrieve the values in my php file with:
$link = $_POST['link'];
Why isn't this working? At first I thought it was because I had htmlspecialchars() but it's not coming through without it either. I can't find anything in any google search that even mentions anything related to this kind of problem (with a type="url" form)
What do I need to do to process form data with type of "url" in PHP with a $_POST?
Get your form method to be set to post e.g
<form method=post>,
if you submit the form and in the url in your browser u can see some more inf then be sure 2 check your form method
I think this is wrong,
method="post"
Its only method, not formmethod
Also make sure, you dont have one more for element name with link.
I'm trying to write a simple script to write some data to mysql then read it. My code is working without a problem alone but I'm trying to use it inside a WordPress page, this is the point problem starts.
I have created a template file for WordPress, and using this template to create the page. Page shows up without a problem but whenever I try to submit the form inside it (my custom php form) it forwards me to index.php .
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>" method="post">
<span>Enter Your Code : </span><br/>
<input type="text" name="sha256"><br/>
<p align = right><input type="submit" name="shaSubmit" value="Submit" /></p>
</form>
this is my form (inside custom php), and as you can see it posts the data to itself. At the the start of my custom php code I have
if(isset($_POST['Submit']))
But it doesn't matter, as soon as I click on button, it forward me to domain.com/index.php
Btw, this custom php is on a page with such url domain.com/custompage/
How can make this form work ?
ps. Code above is for reading from mysql.
You're using the following conditional statement if(isset($_POST['Submit'])) along with the submit button's named element name="shaSubmit".
You need to make those match.
Either by changing the name of your submit button to name="Submit"
or by changing your conditional statement to if(isset($_POST['shaSubmit']))
which is why your code is failing because of the conditional statement you've set is relying on a form element named "shaSubmit".
You need to change 2 things.
(1)make action=""
don't need to use action=" echo htmlspecialchars($_SERVER["PHP_SELF"])"
(2)if(isset($_POST['shaSubmit'])){
the code ....
}
I have a PHP page that loads several parts of a form using AJAX. For instance, first check if the user is already registered, if so the script loads (with AJAX) the rest of the form. The form will not be submited using AJAX what can be a problem when the user submits the form (without AJAX) - imagine there are some errors - the form will loose all values.
I'm wondering if CSS hiding part of the form and after the successful login use JS to display the rest of the form, would be better.
Here some code:
<form action="some_action.php">
Email: <input type="text" name="email" id="email"> <br />
Password: <input type="password" name="password" id="password"> <br />
<button id="vrf_login">Verificar</button>
<div id="rest_form">
</div>
</form>
AJAX:
- CHECK login: if email and password matches then
- LOAD the form for div with id "rest_form"
(it is in another file, for instance:
<input type="text" name="place" id="place">
<input type="text" name="age" id="age">
<input type="submit" name="submit" value="submit">
)
The problem is if I submit the form (without AJAX) and there are errors I will loose the form loaded with AJAX
EDIT (again)
Thank you all for your constructive suggestions:
The solution I adopted is close to the first Alkis's suggestion:
almost all the form is hidden (CSS)
after some logic choices the (part of the) form is turned visible (jQuery) - to "remember" what parts should be visible in case of submission failed (server side validation) some session variables hold the information (AJAX) - and then, after the submission (failed) use jQuery to restore the prior form structure (get the session variables with JS this way: var xpto = "<?php echo $_SESSION['prior_xpto']; ?>" ; )
the fields of the form will remember theirs values (with PHP)
You have 3 options.
Stop loading the whole form by ajax. Hide it with css and show it if the the conditions are met. If the page is shown after some validation error, just show it (change the css inline or give it a different class)
Have a condition and every time the page loads check if it is a first load or if the page is shown after some validation error occured. If the latter is true then load again the form with ajax. This condition can be a hidden field that takes its value from the server and you check it on the client every time you serve the page.
The second solution can be done on the server too. Have the condition be checked on the server. If it's a first load, then don't populate the form and let it be populated from ajax as you do now. If it's after a validation error then pre-populate the form. It's just an if/else clause.
Please provide some codes for your question, but i guess your problem is sending result using a button with "submit" type !
if you have a form like this:
<form>
<inputs ...>
<input type="submit" value="Send data" onclick="SendDataUsingAjax()" >
</form>
after clicking on submit all values on input will reset regardless of what your ajax function is doing. to fix this problem you only need to change type="submit" to type="button".
I am looking for a bit of code to do the following:
A form containing a single text field and a submit button, must send the value of the text field to a landing page that automatically counts how many html tags that this page contains.
E.g. if the text field states stackoverflow.com, the landing page should say (H1 tags = 20) with many more parameters to come.
How is this done? I know how to make a form, but I do not know how to make it send its value to the landing page.
<form action="landingpage.php/" method="post">
The URL
<input type="text" name="cf_name">
<input type="submit" value="Submit">
</form>
This piece of code is a perfect answer to your question.
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="GET">
Type In Something: <input name="random-info" type="text" size="25">
<input type="submit" value="Submit">
</form> <br>
<?php
echo "You Typed: " . $_GET['random-info'];
?>
you get the method into the url, then you can use them on another page.
To access data from a form it depends on the method. Since your code shows a post message you simply access it in the php on the landing page by user $POST_['cf_name'].
To learn more you can check out:
http://w3schools.com/php/php_post.asp about the post method and http://w3schools.com/php/php_get.asp about the get method.
Also an invaluable source is php manual itself.
As far as counting the tags, not really sure what you are trying to achieve.
If you are counting the tags in the page you create, just make a variable and add to it each time you put that specific tag on the page.
Then you can put those values in a hidden field of the form to be passed into your landing page.
I try to create an upload progress bar with PHP and jQuery. However, I have a problem when I bring it to the form data. The code is similar like this:
<form method="post" action="upload.php" enctype="multipart/form-data" id="upload-form" target="upload-frame">
Suburb:<input type="text" name="txtSuburb" id="txtSuburb">
Picture:
<input type="hidden" id="uid" name="UPLOAD_IDENTIFIER" value="<?php echo $uid; ?>">
<input type="file" name="file">
<input type="button" name="submit" value="Upload!">
<iframe id="upload-frame" name="upload-frame">
</iframe>
<input type="submit" name="DataSubmit" value="Submit Data"/>
</form>
As you can see, I got 2 submit buttons. If I keep the form like this then the form can't submit data to server. It just submits the file to iFrame. If I change the action and target of the form then the upload progress function will not work.
Could anyone please help me to find the solution for this?
I want the user can click on upload button to upload their file. Then they can take the rest to fill the form. When everything is done, they can click on another submit data button to submit their data (included the file) to the server.
Make sure that you have only one input element of type submit within your form.
If you want the first button to trigger some Javascript, use a regular input element or even a styled link and attach a Javascript event to it's onclick event, then prevent it's default behavior, e.g. by returning false.
Like this only the second button will actually submit your form which should do what you're describing.
In general I'd second #Treffynnon's suggestion to use a existing library for this purpose. These hacks have a tendency to get pretty nasty, especially when it comes to crossbrowser compatibility.