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298
ViewsHow to validate HTML elements when submitting through JQuery?
I have an HTML form as shown below with some form fields and a submit and a delete button:
There is also a floating component which appears whenever there are changes in the form as shown in the same diagram with text: You have unsaved changes. This is a common component which appears for all the forms in my website.
When I submit the form using the form's Submit button, it validates all the fields as per the validations.
(for example: <input type="number" min="0"> will check that the number should be positive)
But if I submit the form from the Save button on the floating element, it does not checks for any validation, and just posts the request.
I tried using the following code, but the reportValidity() function doesn't do anything.
if (!form.checkValidity()) {
form.reportValidity();
}
form.checkValidity() and form.reportValidity() both are returning false when I do a console.log.
What am I missing here, and how can I fix this?
P.S. I tried this on chrome v98.
Edit: Adding HTML code:
<form method="post" action="/products/manage/{{.Product.ID}}/submit/">
<div class="form-group col-md-5">
<label>Product Quantity</label>
<input type="number" min="0" name="ProductQuantity" value="{{if .Product}}{{.Product.Qunatity}}{{else}}10{{end}}">
</div>
<button type="submit" class="btn btn-success" value="Submit">Submit</button>
<button type="submit" class="btn btn-danger" value="Delete" formaction="/products/manage/{{.Product.ID}}/delete/">Delete</button>
</form>
Save button calls this function:
function submitForm(form, url) {
const form = $(form)[0]
if (!form.checkValidity()) { //<- Added the code here
form.reportValidity();
}
var serialized = serializeForm(form);
// Do some more things then use HTTP to request the API
}
1 answers
Answer question0
You can easily use the jQuery event handlers do the work for you. A simplified example below:
Give your form some identifier (example: id="form1")
Catch the click event on button click
Trigger the submit event to submit that form
$('.unsavedChangesBtn').on('click',function(){ $('#form1').submit(); });
In below example you can submit the form either by clicking the submit button or the save button.
$('document').ready(function(){
$('#form1').on('submit',function(){
alert("SUBMITTED");
});
$('.unsavedChangesBtn').on('click',function(){
$('#form1').submit();
});
});<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<form method="post" id="form1" action="javascript:void(0);">
<div class="form-group col-md-5">
<label>Product Quantity</label>
<input type="number" min="0" name="ProductQuantity" value="{{if .Product}}{{.Product.Qunatity}}{{else}}10{{end}}">
</div>
<button type="submit" class="btn btn-success" value="Submit">Submit</button>
<button type="submit" class="btn btn-danger" value="Delete" formaction="/products/manage/{{.Product.ID}}/delete/">Delete</button>
</form>
<button class="unsavedChangesBtn"> SAVE </button>