jqgrid-asp.net

问题 I've added the following code for my jqgrid: changeTextFormat = function (data) { return "Activity or one from the same price group already used"; }; jQuery.extend(jQuery.jgrid.edit, {errorTextFormat: changeTextFormat }) It works great for insert and I get the error message appearing in the top of the dialog. However for edits it's not working so well. Instead a massive dialog is appearing with the error message. I'm currently thinking that it's because I'm using inline editing so it has to

问题 I've added the following code for my jqgrid: changeTextFormat = function (data) { return "Activity or one from the same price group already used"; }; jQuery.extend(jQuery.jgrid.edit, {errorTextFormat: changeTextFormat }) It works great for insert and I get the error message appearing in the top of the dialog. However for edits it's not working so well. Instead a massive dialog is appearing with the error message. I'm currently thinking that it's because I'm using inline editing so it has to

问题 I have constructed a jqgrid dynamically, i am unable to invoke the function onclick of the button. My Code: function buildButtons(cellvalue, options, rowObject) { var optionsRowId = options.rowId; var editDelButtons = "<input style=\"height:22px;width:40px;\" type=\"button\" value=\"Edit\" onclick=\"javascript:testRow('" + optionsRowId + "')\" />"; return editDelButtons; } function testRow(rowID) { alert(rowID); } } The error i get always when i click on the buton in each row of jqgrid is

问题 I have a jqGrid and there are several pages of items. I have the Id of a row which may be on page one or may be buried in the other pages somewhere. Given the ID of the row, How do I programmatically select such a row ? I am using the click event of a button as follows .on("click", function(){ var myId = $(this).attr("id"); $("#studentGrid").jqGrid.setSelection(myId, true); }); When I click on the button I get the following th the firebug console. TypeError: this.each is not a function Any

问题 I am using Asp.net Web Forms , earlier I used this jqGrid sample which is not free , this I realized later , now that the confusion was cleared , I will now have to use jqGrid open source thing as I wont opt for commercial one , Can anyone point me to links for getting started with jqGrid and Asp.Net Web Forms , it would be really great if someone gives an example which will show free columns from a sample table in jqGrid , I found some articles online , but couldnot understand how it works.

问题 I'm adding some rows data to my JQGrid on client side with javascript : var grid = jQuery("#<%= JQGridMembers.ClientID %>"); var rowKey = grid.getGridParam("selrow"); var newRow = [{ ID: memberId, FullName: memberFullName, Percent: parseInt(percent)}]; grid.addRowData(memberId, newRow); above code works well , but How can I get all inserted rows data (in JQGrid) in code-behind? 回答1: You can get all rows from the grid by var myData = grid.jqGrid('getRowData'); or with respect of var myData =

问题 I m facing a issue in Jqgrid with row selection. I have duplicated rows in the grid. whenever i select the duplicate or it select the first record of that duplicate row. 回答1: It's common problem for many people who start to use jqGrid. Such strange selection behavior exist if you fill the grid with rows having id duplicates . So it's extremely important to understand how the ids from your input will be used by jqGrid . jqGrid use internally HTML markup to display the grid. It uses <table> ,

问题 I wanted to change the grid column sequence dynamically. For e.g. By default the grid will be loaded in LoginId, FirstName and LastName sequence. Based on some condition, I need to change the FirstName and LastName sequence. Is there any way I can do this? I tried doing like: {name:'UserName',index:'UserName',width:82,sortable:false}, if(true) { {name:'FirstName',index:'FirstName',width:65,sortable:false}, {name:'LastName',index:'LastName',width:65,sortable:false}, } else { {name:'LastName'

问题 I wanted to change the grid column sequence dynamically. For e.g. By default the grid will be loaded in LoginId, FirstName and LastName sequence. Based on some condition, I need to change the FirstName and LastName sequence. Is there any way I can do this? I tried doing like: {name:'UserName',index:'UserName',width:82,sortable:false}, if(true) { {name:'FirstName',index:'FirstName',width:65,sortable:false}, {name:'LastName',index:'LastName',width:65,sortable:false}, } else { {name:'LastName'

问题 I generate my jqgrid from model class which I pass into view. I get constructed and working jqgrid. However, I want to set postData on one view, where I use jqGrid, from script in that view, after I call helper for creating jqgrid, without having to change whole partial view which creates jqgrid. I tried running $("#@Model.Id").jqGrid('setGridParam', { postData: { test: 233} }); and $("#@Model.Id").setGridParam({ postData: { test: 233} }); but without error or any result. If I set postData in