So, when i was dealing with forms before what i would do is find web element that is part of the form input, set value of that form input by element.setattribute("value", "infoi want to add")
then simply when i set value of form elements i click submit button of form.
Is this possible to do with web-form input type="file" and how can i do it?
Thanks
The simplest and the most straightforward way to do exactly what you've described is to get form's target address with QWebFrame::hitTestContent and then to post a file with QNetworkAccessManager::post. You also could inject some javascript code into a loaded page (see this for more details), but that would be an overkill for such a simple taks. And besides in the latter case all your injected javascript will be gone after submitting the form (i.e. after the page is reloaded).
Related
I have a list of objects on my page. I need to edit an object in a popup.
There are many objects, and generating many forms for each object is not correct.
What can I do, can an iframe do in a popup?
You don't need to use iframe.
Build your form in your controller and render it in the html. The fact that it's in popup doesn't change anything. It doesn't matter if it's in popup or not, the final result will be POST call to your action.
You should have an action that renders a form, callable from an ajax in the view where you have all those entities.
Just change the entity id that will be received in the action as argument by changing the ajax url using js depending on the clicked entity you want to edit.
Then return with the ajax the form already rendered so with only one form you can edit as many entities (of the same class) as you want, without even need to render one form before they click which one they want to edit.
I have a form with a drop-down 'select' element that the user will sometimes need to add to. So, I added a link to open a Modal Frame form, created the Modal Frame form, and made its _submit() function add the new data to the table that holds options for the select element. Then it "returns" the new ID and name to the Javascript callback (the way Modal Frames do), and the JS callback adds the new element to the dropdown, and makes it the currently selected element in the browser, with JQuery. This all works great.
Finally, the user submits the original form, and gets the error "An illegal choice has been detected. Please contact the site administrator."
The form building function creates the option list from the database, which (I checked) DOES include the new option, so I'm guessing Drupal is using a cached version of the form rather than re-building it before it does the automatic validation. If you go back one page, then forward, the newly-added choice is there in the list.
I tried adding:
global $GLOBALS;
$GLOBALS['conf']['cache'] = FALSE;
to the function that builds the page with the problem, but it made no difference.
FYI: this is part of a multi-page form, if it matters.
I also thought about trying to add the option to the cached version of the "parent" form in the "child" form _submit() function, but don't know how to get the form_build_id of the "parent" form. Besides, that seems like way too much of a kludge (though I could be wrong about that).
Ideas?
You are submitting the form which is different then the time of form render so Drupal consider it as malicious input. If you want to skip that malicious check then add below line as property of your form in your_module_form() function.
'#DANGEROUS_SKIP_CHECK' => TRUE,
This will solve your problem but make sure that you are adding manual validation to protect from malicious inputs :). Hope this will help you.
I think you'd be better off using Drupal's AHAH functions to request an updated select element from a callback function. This way the cached form is rebuilt on the server side and you don't need to add the dreaded DANGEROUS_SKIP_CHECK. Here's a tutorial for Drupal 6:
http://randyfay.com/ahah
I need to pass variables from one ASP(Classic ASP) to another ASP page. I am using javascript (window.open()) to open the second ASP page. I should not pass the variables in the URL like window.open('second.asp?first_name=sdf&last_name=asdas');
I have number of buttons in first ASP page,on click different buttons-different pages will get opened thro java script functions. So, I cant use "POST" method all the time.
Please let me know the ways of passing variables from one page to other page with out passing explicitely in the URL.
One thought m having is : using sessions/hidden/viewstate variables.
Thanks in Advance
Rupa
I think the solution can be founnd be working round the assumption regarding "POST". You could modify the containing HTML form elements Action property in the buttons onclick. Have an expando attribute attached to the button which contains the real action value:-
btn_onclick() { this.form.action = this.getAttribute("action"); }
Now you can give each input type="submit" its own action attribute and use onclick="btn_onclick".
I need to supply an html form (not a server form) for searching within an asp.net web-forms website. The form will post to another website where an indexed search is performed. Seeing as nested forms don't work well in asp.net, what is the simplest approach for this?
The nested form is a simple html form that performs a "get" against the search website, which I do not have programmatic control over.
Update: I resolved my issue by simply moving the server form to the appropriate place on the page, rather than having it surround the entire page. However, I'm still wondering how this would be handled if the html form needed to be placed physically between server controls (which require the server form).
However, I'm still wondering how this would be handled if the html form needed to be placed physically between server controls (which require the server form).
You can place controls on your page without requiring an HtmlForm.
In your case there's no issue declaring another form markup, but you could also just use some search control on your main form and make it issue a GET to that website.
Not only do nested forms "not work well," you basically can't have >1 form per page at all. The simplest approach is the approach you are forced to go with: write a page that only uses one <form runat="server"></form>. Since you need search functionality, is there no ASP.NET search box control that you could use?
Have a read here.
There are 4 workarounds:
Use an IFRAME
Force Submission to Navigate Using a GET Request
Dynamically Change the Form Action
Use a 3rd Party Form Handler
More details on http://www.revindex.com/Blogs/tabid/65/EntryID/21/Default.aspx
Nested forms don't work well in HTML full stop! You should never do it.
Perhaps you mean more than one form on page? Whilst it's true you can only have one form with runat="server", I can't see any reason why you couldn't have a standard form (not server form) that posted to another site at the same level (ie. not nested).
Try adding your HTML input elements to wherever you want the nested form to be. Then use JQuery to change the page form action to point to the external Website. The entire form is submitted, but with a different external Url. The minor downside is the values for all input elements on the page are posted, but most times that is not big deal.
(I only tried this using a POST, not a GET. This is basically Roman O's #3 workaround in more detail)
<div id="nested-form">
<input type="text" name="q">
<input name="searchbtn" value="Go" type="submit" class="search-button">
</div>
<script>
$(function() {
$("input.search-button").click(function() {
$('form').get(0).setAttribute('action', 'http://external.com');
});
});
</script>
maybe you try Server.Transfer() to your target page that do the search from a button for example!
I have a form in asp.net with a FileUpload control inside. Next to this control I have "Upload" button which is used to upload a file to a list of files. The problem is that I also have "Submi"t button used for submiting whole form.
Now when somebody selects a file through browse button and presses on "Submit" not "Upload" file is being uploaded and I wouldn't want that.
I'm hoping for asp.net or general HTML answer
Sounds like the FileUpload control and Upload button should be inside it's own form. With ASP.NET and the 1 server-side form restriction, you may have to resort to regular HTML controls to do this.
Alternatively, you may be able to use JavaScript to clear the FileUpload if Upload isn't clicked. Browsers can be picky about access to file inputs, though (for security reasons) so it may not be accessible.
For the general HTML answer, have your file upload button be part of a separate form from the other form, meaning they are encapsulated within separate Form tags.
For ASP.NET, you're going to have more pain, since ASP.NET by default has only one form per page.
Ultimately, however, I think you're better off rethinking your design. Try to put that file upload on a separate page; if you have a page that does more than one thing, you're inviting confusion on the part of the user.
Submit button will send all of the form's information.
Try putting the FileUpload control outside, on a diferent form.
Your FileUpload control needs to be in its own form if you want to to be submitted separately. When you click the Upload button, the other parts of your form are being submitted as well. That's just how it works.
I would separate the two forms. Using JavaScript or something like that could get to be a pain. The problem with separating the forms is you have to do it in a way where you don't lose your original form, meaning if you have other form information around the file upload you might have a bit of a nightmare on your hands but if you can put the file upload either above or below the form that submit controls then it's easy.
<form id="A">
<button name="submit>
</form>
<form id="B">
<fileupload>
</form>
I use this method on one of my sites and it works quite well.