To check whether a certain field is valid, use: $(‘#myField’)[0].checkValidity(); // returns true|false To check if the form is valid, use: $(‘#myForm’)[0].checkValidity(); // returns true|false If you want to display the native error messages that some browsers have (such as Chrome), unfortunately the only way to do that is by submitting the form, like this: … Read more
The first is the most concise, I would go with that. The first two are the same, but the first is just that little bit shorter, so you’ll save on bytes. The third is plain wrong, because that condition will always evaluate true because the object will never be null or falsy for that matter.
children() only looks at the immediate children of the node, while find() traverses the entire DOM below the node, so children() should be faster given equivalent implementations. However, find() uses native browser methods, while children() uses JavaScript interpreted in the browser. In my experiments there isn’t much performance difference in typical cases. Which to use … Read more
The best you can achieve with HTML only (documentation): <input type=”number” min=”0″ step=”1″/>
You need to find the selected option: $(this).find(‘:selected’).data(‘id’) or $(this).find(‘:selected’).attr(‘data-id’) although the first method is preferred.
Getting the element type the jQuery way: var elementType = $(this).prev().prop(‘nodeName’); doing the same without jQuery var elementType = this.previousSibling.nodeName; Checking for specific element type: var is_element_input = $(this).prev().is(“input”); //true or false
Neither of the answers here helped me. The problem was: I was using the slim build of jQuery, which had some things removed, ajax being one of them. The solution: Just download the regular (compressed or not) version of jQuery here and include it in your project.
The following will return true if the element is an input: $(“#elementId”).is(“input”) or you can use the following to get the name of the tag: $(“#elementId”).get(0).tagName
A better and more generic solution IMO: <!– Insert this line above script imports –> <script>if (typeof module === ‘object’) {window.module = module; module = undefined;}</script> <!– normal script imports etc –> <script src=”https://stackoverflow.com/questions/32621988/scripts/jquery.min.js”></script> <script src=”scripts/vendor.js”></script> <!– Insert this line after script imports –> <script>if (window.module) module = window.module;</script> Benefits Works for both browser and … Read more
$.ajax({ url:url, type:”POST”, data:data, contentType:”application/json; charset=utf-8″, dataType:”json”, success: function(){ … } }) See : jQuery.ajax()