I have a form which has a tab and in this tab is a quickview form. On the quickview form, I have a subgrid and a text field.
The tab has a default state of 'collapsed'. When I open the form, only the text field is displayed. It seems as if the subgrid in no rendering at all.
If I change the tab default state to 'expanded', then when I open the form, the
subgrid is rendering correctly.
I have tried to refresh the quickform view outlined here
https://msdn.microsoft.com/en-us/library/mt736908.aspx
But it does not seem to work.
UPDATE:
I have tried the following, but still no success.
FIRST VERSION
// Triggering when the tab is expanded
function onChange(){
console.log('on change');
// get quick view form
var qv = Xrm.Page.ui.quickForms.get("myquickformview");
qv.refresh();
// get subgrid
try {
qv.getControl(0).refresh();
}
catch (e)
{
console.log(e);
}
}
SECOND VERSION
function onLoad(){
console.log('onload');
Xrm.Page.getAttribute('new_person').addOnChange(refresh);
}
function onChange(){
Xrm.Page.getAttribute('new_person').fireOnChange();
}
function refresh(){
console.log('on change');
// get quick view form
var qv = Xrm.Page.ui.quickForms.get("myquickformview");
// get subgrid
try {
qv.getControl(0).setVisible(false);
qv.getControl(0).setVisible(true);
qv.getControl(0).refresh();
}
catch (e)
{
console.log(e);
}
qv.refresh();
}
Any advice appreciated. Thanks in advance.
1.Add onchange event handler for the lookup (on which Quick view form is rendered) to have the code to refresh the quick view control.
Xrm.Page.getAttribute("lookup_fieldname").addOnChange(function);
Keep the below code in function.
var quickViewControl = Xrm.Page.ui.quickForms.get(“your quick view form name”);
if (quickViewControl != undefined) {
if (quickViewControl.isLoaded()) {
quickViewControl.refresh();
}
}
2.Trigger fireOnChange() of lookup on tab expanded handler, so that onchange will refresh QVform totally.
Xrm.Page.getAttribute("lookup_fieldname").fireOnChange();
Got a hint from this. I just answered here (in mobile without testing) to unblock you.
I am using Chosen with jquery valodation.
you can see this example: http://jsfiddle.net/hfdBF/9/
if you click submit, you see that the validation is working for combo and input.
if you put one letter in the input box you will get an alert that the letter un-highlighted, as it suppose to be.
BUT, if you choose in the select box a value, it wont be alert as un-highlighted.
do you have any idea how to get that to work?
$(function() {
var $form = $("#form1");
$(".chzn-select").chosen({no_results_text: "No results matched"});
$form.validate({
errorLabelContainer: $("#form1 div.error"),
wrapper: 'div',
});
var settings = $.data($form[0], 'validator').settings;
settings.ignore += ':not(.chzn-done)';
settings.unhighlight= function(el){
alert(el.name + " hit unhighlight")
}
});
Thanks
You don't need to do the validation manually. You can bind a function to the change event handler that will check if the specific select is valid. This will also clear the error if the select is valid.
$('.chzn-select').change(function () {
$(this).valid();
});
You need to reset the jQuery validation manually:
$(".chzn-select").chosen().change(function () {
var value = $('#UserCompanyIds').val(); // the select element
if (value != "") {
$('.field-validation-error').each(function () {
if ($(this).attr('data-valmsg-for') == 'UserCompanyIds') {
$(this).html("");
$(this).removeClass("field-validation-error").addClass("field-validation-valid");
}
});
}
});
I hope this helps!
I have buttons that trigger jQuery validation. If the validation fails, the button is faded to help draw attention away from the button to the validation messages.
$('#prev,#next').click(function (e)
{
var qform = $('form');
$.validator.unobtrusive.parse(qform);
if (qform.valid())
{
// Do stuff then submit the form
}
else
{
$('#prev').fadeTo(500, 0.6);
$('#next').fadeTo(500, 0.6);
}
That part works fine.
However, I would like to unfade the buttons once the invalid conditions have been cleared.
Is it possible to hook into jQuery Validation to get an appropriate event (without requiring the user to click a button)? How?
Update
Based on #Darin's answer, I have opened the following ticket with the jquery-validation project
https://github.com/jzaefferer/jquery-validation/issues/459
It might sound you strange but the jQuery.validate plugin doesn't have a global success handler. It does have a success handler but this one is invoked per-field basis. Take a look at the following thread which allows you to modify the plugin and add such handler. So here's how the plugin looks after the modification:
numberOfInvalids: function () {
/*
* Modification starts here...
* Nirmal R Poudyal aka nicholasnet
*/
if (this.objectLength(this.invalid) === 0) {
if (this.validTrack === false) {
if (this.settings.validHandler) {
this.settings.validHandler();
}
this.validTrack = true;
} else {
this.validTrack = false;
}
}
//End of modification
return this.objectLength(this.invalid);
},
and now it's trivial in your code to subscribe to this event:
$(function () {
$('form').data('validator').settings.validHandler = function () {
// the form is valid => do your fade ins here
};
});
By the way I see that you are calling the $.validator.unobtrusive.parse(qform); method which might overwrite the validator data attached to the form and kill the validHandler we have subscribed to. In this case after calling the .parse method you might need to reattach the validHandler as well (I haven't tested it but I feel it might be necessary).
I ran into a similar issue. If you are hesitant to change the source as I am, another option is to hook into the jQuery.fn.addClass method. jQuery Validate uses that method to add the class "valid" to the element whenever it is successfully validated.
(function () {
var originalAddClass = jQuery.fn.addClass;
jQuery.fn.addClass = function () {
var result = originalAddClass.apply(this, arguments);
if (arguments[0] == "valid") {
// Check if form is valid, and if it is fade in buttons.
// this contains the element validated.
}
return result;
};
})();
I found a much better solution, but I am not sure if it will work in your scenario because I do not now if the same options are available with the unobtrusive variant. But this is how i did it in the end with the standard variant.
$("#form").validate({
unhighlight: function (element) {
// Check if form is valid, and if it is fade in buttons.
}
});
I have 3 different tabs where i am displaying data using jQGrids(each tab contain one grid).
But i just thought that my grids are completely the same, only the difference is that they using different url to get data.
So I have three similar girds on each tab only with different urls:
First: url: '/Home/GetData?id=1' Second: url: '/Home/GetData?id=2' and Third: url: '/Home/GetData?id=3'
So i was thinking that may be i may declare grid only once and than on each tab click a can pass the url to load data? So on each tab click jQGrid will be populating from the new url.
May be some one may have any ideas about that?
Or may be some one may have better ideas how to reduce "jQGrid copy-paste" in that case?
UPDATE 0:
Nearly get it work i mean it is working but there is one small problem,
When i am switching tabs the header of the grid getting lost...and some jqgrid formatting as well.
here is my code:
$("#tabs").tabs({
show: function (event, ui) {
if (ui.index == 0) {
//$("#Grid1").appendTo("#tab1");
//$("#Grid1Pager").appendTo("#tab1");
//When Appending only pager and grid div, header getting lost so i've append the whole grid html instead
$("#gbox_Grid1").appendTo("#tab1");
changeGrid("#Grid1", 1);
}
else if (ui.index == 1) {
//$("#Grid1").appendTo("#tab2");
//$("#Grid1Pager").appendTo("#tab2");
$("#gbox_Grid1").appendTo("#tab2");
changeGrid("#Grid1", 2);
}
else if (ui.index == 2) {
//$("#Grid1").appendTo("#tab3");
//$("#Grid1Pager").appendTo("#tab3");
$("#gbox_Grid1").appendTo("#tab3");
changeGrid("#Grid1", 3);
}
}
});
function changeGrid(grid, id) {
$(grid).jqGrid('setGridParam', {
url: '/Home/GetData?id=' + id
});
$(grid).trigger('reloadGrid');
}
UPDATE 1
All right, i've changed the code to append the whole grid instead of appending grid div and pager only. So it is working like that.
You can basically make the tabs as regular buttons that will call some function which sets new URL parameter to the grid and reloads it.
The function should be something like this:
function changeGrid(grid, id) {
$(grid).jqGrid('setGridParam', {
url: '/Home/GetData?id=' + id, page: 1
});
$(grid).trigger('reloadGrid');
}
Note that I set the page to 1. Keep in mind that you might need to set the default sorting column or something similar depending on your solution.
UPDATE
If you really want to go with the tabs, the 'show' event handler can be simplified.
Try this:
$("#tabs").tabs({
show: function (event, ui) {
var id = ui.index + 1;
$("#gbox_Grid1").appendTo("#tab" + id);
$("#Grid1").jqGrid('setGridParam', {
url: '/Home/GetData?id=' + id
});
$("#Grid1").trigger('reloadGrid');
}
});
I need to avoid the double click submitting behavior. I'm using the client validation with the unobtrusive library. I have the following code for avoiding the double clic:
jQuery.fn.preventDoubleSubmit = function () {
var alreadySubmitted = false;
return jQuery(this).submit(function () {
if (alreadySubmitted)
return false;
else {
alreadySubmitted = true;
}
});
};
jQuery('form').preventDoubleSubmit();
Unfortunately, if my form has some validable fields (for example, a required field), the code above is still being fired, hence, even if I correct any mistakes on the form, I won't be able to submit it again.
How can I fire the double click code after the validation has been succesfully done?
You can also use the JQuery One event.
I have found that I could get past most guards against double-clicks by double-clicking fast. Using the one event is the only true way to make sure the event is only fired once. I don't think this technique will work "out of the box" with an input type=submit tag. Instead, you can simply use an input type=button or JQueryUI's .button().
$("#submitButton").one("click", function(event) {
$('#theForm').submit();
});
If you need to re-wire the event on a validation error (or other circumstance), I recommend that you create a function for the event handler. The function isn't necessary in this example because all the event handler does is submit the form, but in more complicated scenarios you may want to avoid repeating yourself.
function submitClick(event) {
$('#theForm').submit();
}
$("#submitButton").one('click', function(event) {
submitClick(event);
});
// This handler will re-wire the event when the form is invalid.
$('#theForm').submit(function(event) {
if (!$(this).valid()) {
event.preventDefault();
$('#submitButton').one('click', function(event) { submitClick(event); });
}
});
You could obviously add the disabling code here if you wanted to give feedback to the user that the button doesn't work anymore. One great side-effect of using the One event is that you don't actually have to make the button disabled, you can use a style of your own.
function submitClick(event) {
$('#submitButton').addClass('disabledButton');
$('#theForm').submit();
}
$("#submitButton").one('click', function(event) {
submitClick(event);
});
// This handler will re-wire the event when the form is invalid.
$('#theForm').submit(function(event) {
if (!$(this).valid()) {
event.preventDefault();
$('#submitButton').one('click', function(event) { submitClick(event); });
$('#submitButton').removeClass('disabledButton');
}
});
JQuery One Event: http://api.jquery.com/one/
I solved it with the following code:
var tryNumber = 0;
jQuery('input[type=submit]').click(function (event) {
var self = $(this);
if (self.closest('form').valid()) {
if (tryNumber > 0) {
tryNumber++;
alert('Your form has been already submited. wait please');
return false;
}
else {
tryNumber++;
}
};
});
NOTE: You can also replace the:
return false;
line, for:
self.attr('disabled', true);
BUT, if you use the name of your submit buttons on your controller for extra logic, they will be sent as null. (you can use an additional hidden field to charge them before submitting)
that's it, hope it helps
Rodrigo
EDIT: Thanks to these posts:
jquery newbie: combine validate with hidding submit button
Why not just use:
function disableButtons() {
var form = $(this);
var btns = $("input:submit", form);
if (!form.valid()) {
// allow user to correct validation errors and re-submit
btns.removeAttr("disabled");
} else {
btns.attr("disabled", "disabled");
}
}
to disable your buttons and activate it using:
$("form").bind("submit", disableButtons);
Based on Ryan P's popular answer I created the following generic solution that also works with my ajax form.
decorate your custom submit button with the following class:
<button type="button" class="one-click-submit-button">Submit</button>
Add the following to your javascript file:
function OneClickSubmitButton() {
$('.one-click-submit-button').each(function () {
var $theButton = $(this);
var $theForm = $theButton.closest('form');
//hide the button and submit the form
function tieButtonToForm() {
$theButton.one('click', function () {
$theButton.hide();
$theForm.submit();
});
}
tieButtonToForm();
// This handler will re-wire the event when the form is invalid.
$theForm.submit(function (event) {
if (!$(this).valid()) {
$theButton.show();
event.preventDefault();
tieButtonToForm();
}
});
});
}
OneClickSubmitButton();
since this is an ajax form we want to reload the handlers if we fail server validation.
function MyForm_OnSuccess() {
if (true if your form passed validation logic) {
//do something since your form submitted successfully
} else { //validation failed on server
OneClickSubmitButton(); //reinitialize the button logic
}
}
Obviously if you don't have ajax forms you can omit the whole OneClickSubmitButton function business and run $('.one-click-submit-button').each(... directly.
I have a form that uses MVC3 unobtrusive validation, and a viewmodel with a [RemoteAttribute].
It looks to me like the form's submit event only fires after all validation has passed. I'm currently using this, and it seems to work:
<input type="submit" value="Submit the Form"
data-app-disable-on-submit="true" />
$('form').live('submit', function() {
$(this).find('input[type="submit"][data-app-disable-on-submit="true"]')
.attr('disabled', 'disabled');
})
;
I set breakpoints on both the remote attribute validation action method and the HttpPost action method. Clicking the submit button the first time hits the breakpoint on the validation action method. At this point, the button is still enabled. I can click it multiple times, and after resuming the validation method, the HttpPost is hit only once. When the HttpPost is hit, the submit button is disabled.
Update
Right you are Alex. So an updated version of the above would look like this:
$('form').on('submit', function() {
$(this).find('input[type="submit"][data-app-disable-on-submit="true"]')
.attr('disabled', 'disabled');
})
$('form').submit(function () {
$('input[type="submit"]', this).attr('disabled', 'disabled');
});
I use a different approach to this. Not wiring to the click event of the button, but to the submit event of the form. Works like a charm to prevent multiple simultaneous submits of forms.
function initFormsToPreventSimultaneousSubmits(selector) {
if (!selector) {
selector = 'form'; // No selector supplied, apply to all forms on the page
}
// Make sure all forms that conform to selector are marked as not submitting
$(selector).each(function()
{
var $form = $(this);
$form.data('submitting', false);
});
// Attach to submit event of all forms that conform to selector
$(selector).off('submit').on('submit', function (e) {
var $form = $(this);
if (!$form.valid || $form.valid()) { // Make sure to only process when the form is valid or jquery validation is not used
if ($form.data('submitting')) {
// form is already submitting. Classic case of double click on one of the submit buttons of the form. Stop the submit
e.preventDefault();
return false;
} else {
// All ok, mark the form as submitting and let the form perform the submit
$form.data('submitting', true);
return true;
}
}
});
}
On document ready i call initFormsToPreventSimultaneousSubmits() to init all forms on the page.
Only thing to remember is that when u use a ajax form post is to call the initFormsToPreventSimultaneousSubmits('#formId') on the OnComplete event of the AjaxOptions settings. Because otherwise the form will still be marked as submitting when its done. When a 'normal' form post is used this is not an issue.
Extends answers by Alex and Ryan P to accounts for situations where jQuery Validation might be missing and where multiple submit buttons exist in a single form.
oneClickSubmitButton = function () {
$('input[type=submit], button[type=submit], input[type=image]').each(function () {
var $theButton = $(this);
var $theForm = $theButton.closest('form');
//hide the button and submit the form
function tieButtonToForm() {
$theButton.one('click', function () {
$theButton.addClass('ui-state-disabled');
});
}
tieButtonToForm();
$theForm.submit(function (event) {
// Only proceed for the clicked button
if (!$theButton.hasClass("ui-state-disabled"))
return;
// If jQuery Validation is not present or the form is valid, the form is valid
if (!$theForm.valid || $theForm.valid())
return;
// Re-wire the event
$theButton.removeClass('ui-state-disabled');
event.preventDefault();
tieButtonToForm();
});
});
};
I was able to fix a similar issue with a couple of lines of code. I prefer this if you don't want to "alert" to user that they double clicked and just silently ignore the second click.
I just made a global javascript variable that I toggled when my function was executing during a critical section. This kept subsequent function calls from re-executing the same section.
var criticalSection = false;
SomeOnClickEventFired = function () {
if (!criticalSection)
{
criticalSection = true;
//Ajax Time
criticalSection = false;
}
}