I'm currently using wordpress 3.5 and creating a page template of my own. I have my own form in that template and when I click the submit button, it is successfully saved the data to my database. Unfortunately, when I click F5 or refresh button on my browser, It prompts an alert which says there will be a double form submission if I continue.
Usually I prevent this by using "redirect to the same page after submitting" technique. But I can't use header("location: ") to redirect because it generates error: Warning: Cannot modify header information - headers already sent by. Probably there are echos on other wordpress file that prevent redirecting.
Does anyone know how to solve this? Or does anybody know other technique to prevent double form submission beside redirecting?
I've always done this with Javascript:
<?php if(isset($_POST['submit_flag'])){ ?>
<script type='text/javascript'>
window.location='URL';
</script>
<?php } ?>
But now that I think about it, you could easily create another PHP page somewhere in your theme that's not included by the rest of your theme to handle the form data and re-direct back to your form.
I'm also about 98% sure that you can include $wpdb without sending headers by simply requiring "wp-blog-header.php".
There is no output besides in your template file. What would that be? Look at the source code of your website, it's only exactly what you tell Wordpress to create.
So that's first: you can use header("Location: ") at the top of the template file that's called on first. Usually header.php.
Secondly, you can (and usually should) use hooks to handle forms. For example:
add_action( 'init', function() {
// Handle stuff
} );
But perhaps with a different hook (I don't recall any best practice). Tutorials here and there will give you suggestions. In this case you will definitely call header("Location: " ) before there is any output. Your theme hasn't even been involved yet.
Related
I am creating reports using Jasper right now and almost everything goes well. We set it in a way that if the user will preview a report, the report(pdf) will be shown on a new Tab. The problem is if an error occurs, a new Tab would still be opened instead of just showing the Feedback Panel on the original page.
How can the form be setup in such a way that the feedback panel will be shown on the original page instead of the newly opened Tab?
Form<?> form = new Form<Void>("form");
form.add(new AttributeAppender("target", Model.of("_blank")));
SubmitLink view= new SubmitLink("view") {
#Override
public void onSubmit() {
//code inside a try-catch to generate the report using Jasper}
};
CptiDownloadButton download = new CptiDownloadButton("download", new AbstractReadOnlyModel<File>(){
//CptiDownloadButton extends SubmitLink button and is a modification of Mr Ivaynberg's DownloadLink
};
<form wicket:id="form">
<input type="button" wicket:message="value:search"/>
<input type="button" wicket:message="value:download"/>
</form>
Thanks in advance to anyone who'll answer. ^^
If you do any form submission to a form with target="_blank", the browser will automatically open a new tab to render the response from the form submission. It is the intended behavior, and trying to prevent it is breaking the standard target="_blank" behavior. I guess what I'm saying is you should really think whether breaking this standard behavior is something you want to do.
If it is, here's how I would go about it. Warning: ain't gonna be clean.
Use Ajax (AjaxButton or AjaxFormSubmitBehavior) to submit the form. Since it is done via ajax, the browser will not invoke default form submission behavior, hence not opening a new tab.
Ajax then invokes the form processing. On error, re-render the feedback panel and return. On success append JavaScript to invoke the default form submission on the respective link when the request returns. This will perform standard form submission behavior, hence performing the target="_blank". It will once more validate the form, but then it will proceed to perform the originally intended behavior.
The way you invoke the default form submission on the link you desire can be done in a few different ways and is entirely up to you. As a quick and dirty way you can hide the buttons that you have right now (visually) and perform javascript to click the button. Or you can use a hidden form field to identify which button has been clicked if you don't want ugly hidden clicking behavior.
You'll have to do a form (Ajax)-submit without target, and then initiate the actual download after checking possible errors.
See https://cwiki.apache.org/confluence/display/WICKET/AJAX+update+and+file+download+in+one+blow for an example with Ajax and an attachment content disposition.
I'd like to integrate dokuwiki as part of a Bootstrap site, within a div. Is this possible? I've tried a php include:
<?php include('dokuwiki/index.php'); ?>
but this effectively redirects - it just generates a completely new page, replacing the existing html. Note that the Bootstrap plugin doesn't do the job.
For a quick solution, it might be the time to use an HTML Frame.
I have a textarea that I first print some values coming from a request.getParameter("some_textarea_name"). The user is allowed to modify those values and then I want to get the new values and replace the old ones with the new ones so as to query the new ones and get the results from my database tables. Can I do that without redirecting the user to a new page e.g without using the <form method> and the request.getParameter()
Can I do that without redirecting the user
Yes, you can implement an ajax call that will submit the form without redirecting the user and you can do something with the response (perhaps add it to the page).
If you need help using ajax follow this tutorial, but be aware that it implements AJAX in pure javascript (its a bit more bloated / complicated). If you want to keep it simple look into jQuery ajax, and here is a tutorial too.
without using the
No, you need to use the form to be submitted, however if you use ajax you wont need to redirect the user.
My application has "survived" many Facebook API changes, to the point where it's quite messy.
Anyway: I'm no longer able to POST a simple HTML form. The entire page reloads and no data is saved.
I've tried several things. Changing the "action" url to include several Facebook parameters, changing the target (canvas_iframe) but to no avail.
Has anyone else encountered problems with this?
try to write absolute path in "action". Its not important include Facebook parameter
Give site url("www.example.com/") as the value of the action attribute rather the Canvas URL(http://appifylabs.com/facebook/yourapp/processform.php)
e.g
if processform.php is the script that will manipulate your form so give the value
site_url = "www.example.com/";
action="<?php echo site_url; ?>processform.php"
I've been searching the net for an answer to my question but I just can't seem to find one, even though it's probably pretty simple.
I have a joomla module that signs up users to a newsletter, when clicking the submit button I navigate to submitsignup.php file. I do this using the form action value like so:
form action="modules/mod_cmsnewslettersignup/otherfiles/submitsignup.php" method="post" id="subForm"
Within this submitsignup.php file I can not access any joomla classes, such as:
$db = JFactory::getDBO();
I know that I can't access any joomla classes because I made direct access to the submitsignup.php file, but I was wondering how do I access this file so that I can have access to all the Joomla classes?
Thanks.
If you are reloading the page when you submit the form, then there is a simple solution that solves both the problem of using a direct URL and of having to load the Joomla framework in that file. Basically change your module code to something like this -
if ($_POST["formsubmitted"]){
the code you run when the form is submitted
echo success or failure message
} else {
the code you run to display the form
<form action="<?php echo JURI::current(); ?>" method="post">
<input type="hidden" value="true" name="formsubmitted">
}
Basically, you submit the form to the page that displays it. Then in your module you add a hook to either process the submitted form or display the form depending on what you find in $_POST. You can add some simple security to make sure that the form is being submitted from your site.