This is the answer, enjoy: .grow { transition: all .2s ease-in-out; } .grow:hover { transform: scale(1.5,0); } where grow is the class of Yours divs 1 solved how to make hover effect like this site( grow gride on hover ) [closed]
You don’t need to get data from external API in client side and post them again to your server. PHP can send get request and receive response without jQuery. You can write php script to get product price list from external API like the following: $url = “https://www.notimportant/etc/”; $header = [‘GUID’ => ‘something’] echo “Sending … Read more
function sb_save(){ alert(‘saving’); } $(‘button’).on(‘click’, function(){ if (typeof sb_save===”function”){ sb_save(); } }); <script src=”https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js”></script> <button>Click</button> 2 solved how to check if a function exists [duplicate]
The remainder operator is really useful for wrapping things like your index around: index = (index + 1) % length …where length is the length of what you’re indexing into (this assumes 0-based indexing). So your Slide function can do something along these lines: function Slide(car) { var urls = bs.url[car]; var index = 0; … Read more
function buttonClicked() { alert(“clicked”); } button1.addEventListener(‘click’, buttonClicked); button2.addEventListener(‘click’, buttonClicked); Or, for IE8 (which doesn’t have support for addEventListener: button1.onclick = button2.onclick = buttonClicked; Or, with jQuery: $button1.click(buttonClicked); $button2.click(buttonClicked); solved how to do function if one of two buttons clicked in jquery? [duplicate]
Try this code, which is not completely coded. please, refer Link for further queries. $(foo).bar({ data:{ $.ajax({ url: “complete_path.php”, success: function(result){ console.log(result); // Process your data as per the requirement. }}); }, }); 1 solved Need to put dynamically generated (PHP) data into javascript [closed]
You can try using JQUERY, put this in your head tag : <head> <script src=”https://stackoverflow.com/questions/23381099/jquery-1.11.0.min.js”></script> </head> Then below your head, to set Checked: function setChecked(string checkbox) { $(“#” + checkbox).prop(‘checked’, true); } to Uncheck: function setChecked(string checkbox) { $(“#” + checkbox).prop(‘checked’, false); } Call these methods in your PHP. Good luck. 1 solved How to … Read more
You can get a reference to your div like this: var $div = $(‘#hiddenChart’); To clone it, var $clonedDiv = $div.clone(); Then, in the $cloneDiv object, you make the changes: $clonedDiv.find(–selector of a node–).attr(atributeName, atributeValue); //change/add attribute $clonedDiv.find(–selector of a node–).removeAttr(atributeName); //remove attribute And so on. I won’t explain how jQuery works, Lekhnath gave you … Read more
Ok so I seem to have gotten it right now… Here the JS <script> function myFormFunction() { if (document.myForm.Name.value == “”) { var el1 = document.getElementById(“name”); el1.style.display = “block”; document.myForm.Name.focus(); return false; } if (document.myForm.Surname.value == “”) { var el2 = document.getElementById(“surname”); el2.style.display = “block”; document.myForm.Surname.focus(); return false; } if (document.myForm.Income.value == “”) { var … Read more
try $json=json_encode($_POST[‘myData’]); $decoded = json_decode($json,true); if (is_array($decoded)) { for($i = 0; $i <= count($decoded); $i++) { echo $decoded[$i][“name”]; } } 1 solved php Json_decode correct way to get individual data
http://malsup.com/jquery/form/ needs to be separately included. its not part of jquery core. 1 solved Object doesn’t support property or method ‘ajaxSubmit’ [closed]
button1 $( “#button1” ).click(function() { $( “#whatever” ).show(); }); button2 $( “#button2” ).click(function() { $( “#whatever” ).hide();$( “#whatever2” ).show(); }); and so on solved jQuery show/hide select and button [closed]
EDIT: I cleaned up my answer a little bit, should be a bit more readable. You just need to break up your string and concatenate it with the returned value. jQuery(document).ready(function() { var $parent = jQuery(parent.document); // cache this so jquery doesnt have to instantiate the same object twice var theFormID = $parent .find(‘.theformclass’) .find(‘input[name=”form_id”]’).val(); … Read more
The following should work for your use case. $(‘img’).attr(‘src’) 0 solved Kindly help me in getting source of image tag
You can use .has() selector: if(!$(‘ul’).has(‘li’)){//ul is empty } 4 solved jquery – get number of li inside a ul element [duplicate]