form_data | Utility-belt to build form data request bodies | Form library
kandi X-RAY | form_data Summary
kandi X-RAY | form_data Summary
Utility-belt to build form data request bodies.
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QUESTION
I have a form on my website where in some cases I need to validate a phone number and in other cases I don't have to (based on changes in the input field.
The form is populated with data from an database and if the input "phone_Number" has been changed I need to validate the number when submit is pressed.
If the input "phone_number" hasn't changed the form has other inputs that needs to be validated.
My issue is: If the phone number has changed I need to make 2 calls with ajax instead of 1, and I can't figure out what the best way to do that is. I have read that async: false was deprecated, so I need help how to arrange my code in order for the submit to work.
I have looked at some different jquery when/wait examples, but I do not understand jquery good enough to change the examples results into my question.
...ANSWER
Answered 2022-Apr-17 at 15:33You can put the second AJAX call in the success method of the previous AJAX function. Like this:
QUESTION
I am making a blogging application with Laravel 8 and Bootstrap 5.
I run into a problem trying to validate my "Add new article" form: when the form fails validation, the valid fields do not keep their values, despite ne using Blade's old() method.
In the controller, I have:
...ANSWER
Answered 2022-Mar-26 at 18:57You're redirecting back to the previous page without telling Laravel to pass the old inputs. Use the method withInput() with the route return.
Change
QUESTION
I am making a blogging application with Laravel 8 and Bootstrap 5.
I run into a problem with validating the image field of the "Add new article" form.
In the controller, I have:
...ANSWER
Answered 2022-Mar-26 at 19:53First of all, when you are dealing with an image/file you must have enctype=" multipart/form-data " in your form. Otherwise, you cannot submit the form. and validation will not be working.
Next validation rule for image or file will be like this :
QUESTION
I want to render a StreamingResponse image on React from FastAPI. The image is in the form of numpy array which is of cv2 type of object
...ANSWER
Answered 2022-Mar-22 at 20:04You could either encode the image data to base64 format and return it, which can then be used to display the image in HTML as shown here (e.g., btoa(), String.fromCharCode() and Uint8Array) or Blob object (and then call URL.createObjectURL() to convert the blob into a URL.) on client side. The below examples show how to achieve the last two methods, using Axios library, which you seem to be using in your project, as well as Fetch API. Make sure to set responseType: "arraybuffer" in axios.
Using Axios
Option 1 - Convert raw image bytes into Blob
QUESTION
I have created an API using FastApi that processes the image uploaded and return the processed image.
Here is the image upload endpoint
...ANSWER
Answered 2022-Mar-01 at 16:34You should add your image to the FormData object, using the same key defined in the endpoint for the file upload (in your case, that is, file). Thus, you should do:
QUESTION
I'm trying to develop a simple Django app of a contact form and a thanks page. I'm not using Django 'admin' at all; no database, either. Django 3.2.12. I'm working on localhost using python manage.py runserver
I can't get the actual form to display at http://127.0.0.1:8000/contact/contact; all I see is the submit button from /contact/contactform/templates/contact.html:
Static files load OK: http://127.0.0.1:8000/static/css/bootstrap.css
The thanks.html page loads OK: http://127.0.0.1:8000/contact/thanks
This is the directory structure:
/contact/contact/settings.py
...ANSWER
Answered 2022-Feb-17 at 03:06The form does not display as you are not passing it into your template. You can do this instead in the contact view:
QUESTION
In my form.php
...ANSWER
Answered 2022-Feb-20 at 17:26Try passing the form element to FormData. No need to call the append method afterwards.
QUESTION
I am at beginner level for using Jquery.
Problem : so the problem is that, I have to add "choose file" for each of the element's inside a tag. And display the image after selecting it inside the tag. But Has this is this list of elements using for each loop, it cannot different between the id property. please see the images and code for reference and help me out, Thank You !!!
...ANSWER
Answered 2022-Feb-19 at 09:29HTML id attribute should be unique. In your code, only the first element with id=productImage is being replaced.
What you could do is use $(this).find(selector). This will find the element inside the current element.
In the code below, the current element is .uploadFile and .find() will search inside that element.
QUESTION
I'm trying edit a code from this tutorial: [https://www.youtube.com/watch?v=fSIg5y5QXbc&ab_channel=Webslesson][1]. The problem is a sytuation when two or more countires have the same state name (imagination only). For example, USA and Russia have the same "xyz" state, and when I select USA and state: "xyz", the program sees no difference (as if it just forgot which country we are in) and displays all the cities in the "xyz" state from both country. Can you help me resolve this problem?
EDIT: I think i'm close, but i need to know, how to save for example: $select_country=(selected country) in fetch.php file
index.php
...ANSWER
Answered 2022-Feb-13 at 03:05For situations like this, the countries, states and cities should all be store in separate tables. Then you link them with the help of foreign keys.
But since all the records live in one table in your case, you can add the selected country to your queries to ensure that only records belonging to a particular country and/or state are affected.
As you can see below, var country = $('#country').val(); gets the selected country and was added to the ajax request. And then, included in the select query on the server.
index.php
QUESTION
I am trying to open a tippy tooltip when a button is clicked, not append the tooltip to this button but open it in it's own spot.
In my example I have an add to cart button and a shoppingcart icon in my menu that shows the shopping cart inside a tippy tooltip. I want this tooltip to also show when clicking the add to cart button.
I've tried to create an instance of my tippy and use this with the show() method but I have no luck.
So as a simple example: there is element1 and element2. Element1 has the tippy code which works fine, but now I also want to trigger the tippy on element1 when clicking element2.
My code:
...ANSWER
Answered 2022-Jan-26 at 08:11I think you are looking for triggerTarget property:
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