When I used datepicker with trigger icon so that users could choose date from clicking this icon or type directly in textbox (txtDate), I also used jquery validation to require textbox must be not empty.
But when a user submit the form with empty textbox (txtDate.Text=""), the error message of validation push trigger icon to the right. Could you tell me the solution? Thank you very much!
$('#some_form').validate({
rules: {...},
messages: {...},
errorPlacement: function(error, element) { //solve you problem
var trigger = element.next('.ui-datepicker-trigger');
error.insertAfter(trigger.length > 0 ? trigger : element);
}
});
The errorPlacement function receives two parameters - the error message and the validated element. Use the latter to decide whether or not to customize the placement for your fields (for example, by adding a class):
$('form').validate({
rules: {...},
errorPlacement: function(error, element) {
if (element.hasClass('customError')) {
// custom error placement
}
else {
element.after(error); // default error placement
}
}
});
Related
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'm implementing a simple (at least ,that was the goal) Kendo UI grid that displays two columns: one holding a checkbox, bound to a boolean, and one holding a display name for the item. The checkbox column has a simple template, and the change() event of the checkbox is handled so that the model in the datasource gets updated. I have verified this, and it works.
The data source has been configured for batch, and defines a transport for read and update. Both call a function that perform the ajax call. As I said before, the read function is handled as expected. However, the update function defined on the transport is not. The sync() on the datasource is triggered with a simple button whose click event is hooked to a function that calls datasource.sync() (or grid.saveChanges()).
transport: {
read: function(options) {
return loadStuff(options);
},
update: function (options) {
return updateStuff(options);
}
}
When debugging in the Kendo UI code, it looks like the models attribute on the ModelSet is always empty, and therefore the sync() decides that there's nothing to sync. Anyone got a clue what is happening here?
UPDATE:
Looks like something may be wrong when handling the checkbox check / uncheck. Apparently I should use something like
$('#divGrid').on('click', '.chkbx', function() {
var checked = $(this).is(':checked');
var grid = $('#divGrid').data().kendoGrid;
var dataItem = grid.dataItem($(this).closest('tr'));
dataItem.set("Selected", checked);
});
Unfortunately, it looks like the set() method is not defined on the data item. When debugging, it only contains the data, and no Model object having the set() method.
UPDATE 2:
Tried wrapping the data returned from the ajax call in a model defined with Model.define(). That seems to solve the issue of the model not being dirty, as the _modified property on the model returns true. However, the models array in the ModelSet remains empty. Is this a bug in Kendo UI, or am I going the wrong way?
You don't actually need to bind to click event on the checkboxes.
I´ve posted an example on using it in JSFiddle where you can see it running. This example displays in a grid two columns: first text (tick) and second boolean rendered as a checkbox (selected); the update is batch (so, it's pretty close to what you have).
Questions to keep in mind are:
For displaying the checkbox while not in edit mode, you should define a template, something like this. You might realize that the checkbox is in disabled state by default since you want to edit it as other fields (selecting the cell first). This also guarantees that the model is correctly updated:
{
field : "selected",
title : "Selected",
template: "<input type='checkbox' name='selected' #= selected ? 'checked' : '' # disabled/>"
}
Define in the model that this field is boolean:
schema : {
id : "id",
model: {
fields: {
symbol : { type: "string" },
selected: { type: "boolean" }
}
}
},
Define the transport.update function, something like:
transport: {
read : function (operation) {
// Your function for reading
},
update: function (operation) {
// Display modified data in an alert
alert("update" + JSON.stringify(operation.data.models, null, 4));
// Invoke updating function
// that should ends with an operation.success(the_new_data)
// In this example just say ok
operation.success(operation.data.models)
}
}
EDIT: If you want to be able to modify the checkbox state without having to enter in edit mode first, you should:
Remove the disabled from the template:
{
field : "selected",
title : "Selected",
template : "<input type='checkbox' name='selected' #= selected ? 'checked' : '' #/>"
},
Then bind the click event on checkboxes to the following handler function:
$("#stocks_tbl").on("click", "input:checkbox", function(ev) {
var dataItem = grid.dataItem($(this).closest('tr'));
dataItem.set("selected", this.checked);
});
Where #stocks_tbl is the id of the div that contains the grid. You might see it running here.
NOTE: It's important the on with the three parameters for making it live
I want to change the default behavior for the jquery.validate.unobtrusive.js
I want to be able to change options like the error message placement or highlight a field when an error is happening , etc
all that I could do using the jquery.validate plugin alone. Just change some of the validate method options like highlight or errorPlacement
so I want to override some of the functionality of the unobtrusive validation , is it possible without changing the jquery.validate.unobtrusive.js file
You could fetch the native validator from the form data and then subscribe to any standard option. For example errorPlacement:
$(function() {
var validator = $('form').data('validator');
validator.settings.errorPlacement = function(error, element) {
// do your custom error placement
};
validator.settings.highlight = function(element, errorClass) {
// do your custom error highlight
};
});
I have a simple MVC3 form that has a date field and client side validation enabled (jquery.validate / jquery.validate.unobtrusive). I've added code to attach a datepicker (jQuery UI) to the date field per the documentation. However, if the datepicker is the last thing I click prior to clicking the submit button and the date field is invalid, it causes the datepicker for that field to automatically show itself. I don't want this. How can I disable?
Edit:
After reviewing the code for the validation plugin, it looks like it tries to manually focus on the last active input control using the focusInvalid() function below:
focusInvalid: function() {
if( this.settings.focusInvalid ) {
try {
$(this.findLastActive() || this.errorList.length && this.errorList[0].element || [])
.filter(":visible")
.focus()
// manually trigger focusin event; without it, focusin handler isn't called, findLastActive won't have anything to find
.trigger("focusin");
} catch(e) {
// ignore IE throwing errors when focusing hidden elements
}
}
},
There appear to be 2 options for dealing with this. One is to set the focusInvalid setting on the validator itself to false. I opted to monkey patch the focusInvalid function instead because it allows me to focus on the FIRST invalid element in the form, not necessarily the last active element.
$('form').data('validator').focusInvalid = function () {
$(this.currentForm).find('.input-validation-error').first().focus();
};
I'd be interested to hear any other approaches to this problem, however.
If acceptable in your case, you could show the datepicker only using a button, using the following datepicker options:
showOn: 'button', showButtonPanel: true, buttonImage: '<your image file>',
buttonImageOnly: true,
buttonText: 'Choose a Date',
If you do it this way the datepicker won't automatically display in case validation fails when submitting the form.
Does anyone know how to trigger the stock jqGrid "Loading..." overlay that gets displayed when the grid is loading? I know that I can use a jquery plugin without much effort but I'd like to be able to keep the look-n-feel of my application consistent with that of what is already used in jqGrid.
The closes thing I've found is this:
jqGrid display default "loading" message when updating a table / on custom update
n8
If you are searching for something like DisplayLoadingMessage() function. It does not exist in jqGrid. You can only set the loadui option of jqGrid to enable (default), disable or block. I personally prefer block. (see http://www.trirand.com/jqgridwiki/doku.php?id=wiki:options). But I think it is not what you wanted.
The only thing which you can do, if you like the "Loading..." message from jqGrid, is to make the same one. I'll explain here what jqGrid does to display this message: Two hidden divs will be created. If you have a grid with id=list, this divs will look like following:
<div style="display: none" id="lui_list"
class="ui-widget-overlay jqgrid-overlay"></div>
<div style="display: none" id="load_list"
class="loading ui-state-default ui-state-active">Loading...</div>
where the text "Loading..." or "Lädt..." (in German) comes from $.jgrid.defaults.loadtext. The ids of divs will be constructed from the "lui_" or "load_" prefix and grid id ("list"). Before sending ajax request jqGrid makes one or two of this divs visible. It calls jQuery.show() function for the second div (id="load_list") if loadui option is enable. If loadui option is block, however, then both divs (id="lui_list" and id="load_list") will be shown with respect of .show() function. After the end of ajax request .hide() jQuery function will be called for one or two divs. It's all.
You will find the definition of all css classes in ui.jqgrid.css or jquery-ui-1.8.custom.css.
Now you have enough information to reproduce jqGrid "Loading..." message, but if I were you I would think one more time whether you really want to do this or whether the jQuery blockUI plugin is better for your goals.
I use
$('.loading').show();
$('.loading').hide();
It works fine without creating any new divs
Simple, to show it:
$("#myGrid").closest(".ui-jqgrid").find('.loading').show();
Then to hide it again
$("#myGrid").closest(".ui-jqgrid").find('.loading').hide();
I just placed below line in onSelectRow event of JQ grid it worked.
$('.loading').show();
The style to override is [.ui-jqgrid .loading].
You can call $("#load_").show() and .hide() where is the id of your grid.
its is worling with $('div.loading').show();
This is also useful even other components
$('#editDiv').dialog({
modal : true,
width : 'auto',
height : 'auto',
buttons : {
Ok : function() {
//Call Action to read wo and
**$('div.loading').show();**
var status = call(...)
if(status){
$.ajax({
type : "POST",
url : "./test",
data : {
...
},
async : false,
success : function(data) {
retVal = true;
},
error : function(xhr, status) {
retVal = false;
}
});
}
if (retVal == true) {
retVal = true;
$(this).dialog('close');
}
**$('div.loading').hide();**
},
Cancel : function() {
retVal = false;
$(this).dialog('close');
}
}
});
As mentioned by #Oleg the jQuery Block UI have lots of good features during developing an ajax base applications. With it you can block whole UI or a specific element called element Block
For the jqGrid you can put your grid in a div (sampleGrid) and then block the grid as:
$.extend($.jgrid.defaults, {
ajaxGridOptions : {
beforeSend: function(xhr) {
$("#sampleGrid").block();
},
complete: function(xhr) {
$("#sampleGrid").unblock();
},
error: function(jqXHR, textStatus, errorThrown) {
$("#sampleGrid").unblock();
}
}
});
If you want to not block and not make use of the builtin ajax call to get the data
datatype="local"
you can extend the jqgrid functions like so:
$.jgrid.extend({
// Loading function
loading: function (show) {
if (show === undefined) {
show = true;
}
// All elements of the jQuery object
this.each(function () {
if (!this.grid) return;
// Find the main parent container at level 4
// and display the loading element
$(this).parents().eq(3).find(".loading").toggle(show);
});
return show;
}
});
and then simple call
$("#myGrid").loading();
or
$("#myGrid").loading(true);
to show loading on all your grids (of course changing the grid id per grid) or
$("#myGrid").loading(false);
to hide the loading element, targeting specific grid in case you have multiple grids on the same page
In my issues I used
$('.jsgrid-load-panel').hide()
Then
$('.jsgrid-load-panel').show()