I have enabled 'rownumber' property. So it is displaying row numbers by inserting a row in left most . So first column is displaying row numbers. ButI want to display row numbers in between i.e. in 3rd column.
Is there is any way to change column position?
Some columns of jqGrid have special meaning and will be created by jqGrid depend of the options which you use. It's "rn" (rownumbers: true), "cb" (multiselect: true) and subgrid (subGrid: true). See for example the line of code. Many parts of jqGrid code just test the options and then uses the indexes of columns based on the assumption that the columns are the first columns in jqGrid. So it's probably theoretically possible to write the code which could move the original "rn" column to another position, but the code will be very tricky and probably long. Look at the demo of the answer for example.
So I would recommend you just add your custom column to the grid which looks like "rn" column and fill it with the corresponding data. The column definition could be like below
{ name: "myRowNumbern", width:25, sortable: false, resizable: false, hidedlg: true,
search: false, align: "center", fixed: true,
classes: "ui-state-default jqgrid-rownum" }
UPDATED: I created the demo which demonstrates the approach. The most important parts of the corresponding code is below:
$("#list").jqGrid({
curRowNum: 1, // current row number parameter used in custom formatter
colModel: [
...
{ name: "myRowNumbern", width: 25, sortable: false, resizable: false,
hidedlg: true, search: false, align: "center", fixed: true,
classes: "ui-state-default jqgrid-rownum",
formatter: function () {
var p = $(this).jqGrid("getGridParam"),
rn = p.curRowNum +
(parseInt(p.page, 10) - 1)*parseInt(p.rowNum, 10);
p.curRowNum++;
return rn.toString();
} },
...
],
loadComplete: function () {
var p = $(this).jqGrid("getGridParam");
p.curRowNum = 1; // reset curRowNum
}
});
Related
I have 3 editable dropdown columns in jqgrid, 2 dropdowns are populated dynamically onedit. Because values of those dropdown depends on selection of first. Grid row becomes editable as soon as row is selected & its configured to update the values after user clicks enter. Issue is after saving row, dropdown column 3 & 4 are not showing selected text in the grid. Cells are empty.
This is how both columns are setup
{
name: "FirstDrop", index: "FirstDrop", width: 180, align: "left", editable: true, formatter: 'select', edittype: "select", editoptions: {
value:"0:select;1:first,2:second", valuesToSelect: "0"
, dataEvents: [
{
type: 'change',
fn: function (e) {
//function that reload SecDrop based on value selected in this
}
}
]
}
},
{
name: "SecDrop", index: "SecDrop", width: 180, align: "left", editable: true, edittype:'select', formatter: 'select', editoptions: {
dataInit: function (elem) {
// function that appends options to "elem"
//$(elem).val(""); Select initial value here
}
}
}
So now, when I change selection in first dropdown, second dropdown is reloaded with new value. After selecting new value in second dropdown, upon saving, the column is empty. Instead it should show text value of selected option.
I get the value of the second column when I loop through all rows.
Not sure what I am missing here.
Any suggestions, is something missing, can same result be achieved using custom format option.
Not sure if this is a known issue in jqgrid v.4.7 But I figured out a hack, now handling the aftersave function to get the value of selected option and retrieving its corresponding text from the source json & assigning it to cell.
I have jqGrid table with many columns. Searching in grid is made using filter toolbar. For most of them search is just simple default operator. For one datetime column I want different kind of operators and datepicker selector.
I have added dataInit datepicker initialization to searchoptions, necessary operators to searchoptions.sopt. To show this operators I have set searchOperators to true. So for this column all is ok. I have datepicker with operator selector popup. But for all other columns default operator icon is shown on the left of it. It is annoying as operator is default and user couldn't change it. So is there is some possibility to hide them using jqGrid API? As far as I could see I could hide this only using my custom code:
I need to check my column model and after rendering of grid (may be in loadComplete) for all columns that have empty sopt or sopt.length == 0 to remove operator selector. Or add CSS class that hide it. Not sure which of these solution is better (hide or remove) because removing could broke some logic, and hiding could affect width calculation. Here is sample of what I mean on fiddle
function fixSearchOperators()
{
var columns = jQuery("#grid").jqGrid ('getGridParam', 'colModel');
var gridContainer = $("#grid").parents(".ui-jqgrid");
var filterToolbar = $("tr.ui-search-toolbar", gridContainer);
filterToolbar.find("th").each(function()
{
var index = $(this).index();
if(!(columns[index].searchoptions &&
columns[index].searchoptions.sopt &&
columns[index].searchoptions.sopt.length>1))
{
$(this).find(".ui-search-oper").hide();
}
});
}
Does anybody have some better ideas?
I find the idea to define visibility of searching operations in every column very good idea. +1 from me.
I would only suggest you to change a little the criteria for choosing which columns of searching toolbar will get the searching operations. It seems to me more native to include some new property inside of searchoptions. So that you can write something like
searchoptions: {
searchOperators: true,
sopt: ["gt", "eq"],
dataInit: function(elem) {
$(elem).datepicker();
}
}
I think that some columns, like the columns with stype: "select", could still need to have sopt (at least sopt: ["eq"]), but one don't want to see search operators for such columns. Specifying of visibility of searching operations on the column level would be very practical in such cases.
The modified fiddle demo you can find here. I included in the demo CSS from the fix (see the answer and the corresponding bug report). The full code is below
var dataArr = [
{id:1, name: 'steven', surname: "sanderson", startdate:'06/30/2013'},
{id:2, name: "valery", surname: "vitko", startdate: '07/27/2013'},
{id:3, name: "John", surname: "Smith", startdate: '12/30/2012'}];
function fixSearchOperators() {
var $grid = $("#grid"),
columns = $grid.jqGrid ('getGridParam', 'colModel'),
filterToolbar = $($grid[0].grid.hDiv).find("tr.ui-search-toolbar");
filterToolbar.find("th").each(function(index) {
var $searchOper = $(this).find(".ui-search-oper");
if (!(columns[index].searchoptions && columns[index].searchoptions.searchOperators)) {
$searchOper.hide();
}
});
}
$("#grid").jqGrid({
data: dataArr,
datatype: "local",
gridview: true,
height: 'auto',
hoverrows: false,
colModel: [
{ name: 'id', width: 60, sorttype: "int"},
{ name: 'name', width: 70},
{ name: 'surname', width: 100},
{ name: 'startdate', sorttype: "date", width: 90,
searchoptions: {
searchOperators: true,
sopt: ['gt', 'eq'],
dataInit: function(elem) {
$(elem).datepicker();
}
},
formatoptions: {
srcformat:'m/d/Y',
newformat:'m/d/Y'
}
}
]
});
$("#grid").jqGrid('filterToolbar', {
searchOnEnter: false,
ignoreCase: true,
searchOperators: true
});
fixSearchOperators();
It displays the same result like youth:
Each row has an id in our db different to the jqgrid row id. How can I send this lineid when saving a row?
Also, is there a way to delete a row?
This is my code so far:
var mydata = [
{
lineItemId: "785",
productSku:"n123",
productName:"hello there",
pieces:"123",
value:"23.00",
line:"123"
}
,
{
lineItemId: "803",
productSku:"n1234",
productName:"hello there",
pieces:"123",
value:"23.00",
line:"123"
}
];
var colNames = ['SKU','Product Name', 'Pieces','Total Value','Line Number'];
var colModel = [
{name:'productSku', index:'productSku', width:10, sorttype: 'text', editable:true},
{name:'productName', index:'productName', width:60, editable:true},
{name:'pieces', index:'pieces', width:10, sorttype: 'int', editable:true, formatter: 'integer'},
{name:'value', index:'value', width:10, sorttype: 'int', editable:true, formatter: 'number'},
{name:'line', index:'line', width:10, sorttype: 'int', editable:true, formatter: 'integer', formatoptions:{thousandsSeparator: ""}}
];
initOrdersJqGrid("orderContent", mydata, '<xsl:value-of select="$datapath/OrderId"/>', colNames, colModel, "sku", "desc");
var orderLineOptions = {
keys: true,
aftersavefunc: function (rowid, response, options) {
// only update page if orderis is nil i.e. a new order
if($('#orderidlabel').text() == "") {
log('saving order line item from order with no id yet.');
var dummy = $('<div />').html(response.responseText);
var id = dummy.find('#orderId').val();
$('#orderidlabel').text(id);
$('#orderId').val(id);
$('button[value="Save Order"]').trigger('click');
}
}
}
function initOrdersJqGrid(id, data, orderid, colNames, colModel, defaultSortColumn, defaultSortOrder) {
$("#" + id + "Table")
.jqGrid({
datatype: "local",
data: data,
colNames: colNames,
colModel: colModel,
localReader: { id: "lineItemId"},
pager: '#' + id + 'Pager',
autowidth: true,
gridview: true,
autoencode: true,
height: "auto",
forceFit: true,
shrinkToFit: true, //Width of columns should be expressed in integers which add to 100
sortname: defaultSortColumn,
sortorder: defaultSortOrder,
url: "fs/servlet/CS",
editurl: "CS?action=com.agistix.webinterface.controllers.OrderIC,saveLineItems&orderId=" + orderid
})
.jqGrid('navGrid',"#" + id + "Pager",{edit:false,add:false,del:false,search: false, refresh: false})
.jqGrid('inlineNav',"#" + id + "Pager", { addParams: { addRowParams: orderLineOptions }, editParams: orderLineOptions});
}
If you use datatype: "local" then the items from the array of input data specified by data parameters should have additional property id which specify the value of id attribute of every row (<tr>) of the grid. If you prefer to have another name of the rowsid property you can use localReader to specify it. For example localReader: { id: "Id" } option inform jqGrid to get value of id attribute of rows (rowids) from the Id property. In the case the items of your data should bi like below
{
Id: 76453
productSku:"n123",
productName:"hello there",
pieces:"123",
value:"23.00",
line:"123"
}
The value of id property need be unique.
If you have already some column in the grid which contains id from some database table you don't need to add the same value with id property. Instead of that you can just key: true in the column. jqGrid allows to place key: true in only one item of colModel.
One more common problem with ids of rows it's important to understand. If you need to place more as one grid on a page or if you need to use Subgrid as Grid then you can still have one problem. The ids in database are unique in a table, but one can have the same ids in multiple tables. On the other side the ids of HTML elements (inclusive <tr> elements used for rows) must be unique over the whole page.
To solve the problem one can use idPrefix option of jqGrid. For example you have INT IDENTITY column in the database for the primary key of the table in the database. In the case you will have integers as native ids for rowids. The values can be for example 3, 5, 40 in the first grid. By usage idPrefix: "g1_" the ids assigned to the rows (to <tr> elements) will be "g1_3", "g1_5", "g1_40". So usage of idPrefix: "g1_" for the first grid and another value like idPrefix: "g2_" can solve the problem with potential id duplicates. It's important that jqGrid automatically strip the prefix idPrefix from rowid if it sends some data to the server (if you use editing in the grid for example). One can distinguish "id" and "rowid" names. The "rowids" will be always with prefix. You can use $.jgrid.stripPref function to cut the prefix.
I have a jqgrid that has main rows and a footer row (with userdata loaded) and then a formatter that alters the data in the cells to be linkable. The cells in the main body can be clicked and the onCellSelect event will capture the click. However, clicking on data in the footer row does not seem to fire off the onCellSelect event. How do I capture a select/click event in the footer row? Below is the script for the jqgrid.
$('#jqgSummaryResults').jqGrid({
datatype: 'json',
mtype: 'GET',
url: 'some action',
postData: { 'criteria': function () {
some function}},
rowNum: 100,
rowList: [],
pager: '#jqgpSummaryResults',
viewrecords: true,
sortorder: 'asc',
sortname: 'DateField',
width: 1250,
height: 350,
shrinkToFit: true,
gridview: true,
footerrow: true,
userDataOnFooter: true,
onCellSelect: function (rowid, iCol, cellcontent, e) {
var selectedDate = rowid;
savedMailDueDateString = rowid;
var selectedColumn = iCol;
...
},
loadComplete: function (data) {
...
},
colNames: ['DateField',
'Total Jobs',
...
'% Not Mailed'],
colModel: [
{ name: 'DateField', index: 'DateField', align: 'left' },
{ name: 'TotalJobs', index: 'TotalJobs', align: 'left', formatter: hyperlinkColumnFormatter },
...
{ name: 'PercentNotMailed', index: 'PercentNotMailed', align: 'left', formatter: hyperlinkColumnFormatter },
]
}).navGrid('#jqgpSummaryResults', {
excel: false,
edit: false,
add: false,
del: false,
search: false,
refresh: false
});
Thanks for the assistance.
While I didn't see any way to have jqGrid respond to select (doesn't even seem that that footer is selectable) or a click. The footer row is specified by a ui-jqgrid-sdiv class. You could attach a click event handler as below.
$('.ui-jqgrid-sdiv').click(function() {alert('Bong')});
Edit: In response to Gill Bates question to add a footer event but only on a single cell the selector would be:
$('.ui-jqgrid-sdiv').find('td[aria-describedby="GridName_ColumnName"]').click(function() { alert("Bong");});
GridName_ColumnName is the format for all the footer td aria-describedby, and you can see the exact name via firebug element inspector (or any of it's equivalents).
jqGrid registers click event on main <table> of the grid, but it calls onCellSelect not always. First of all (see here) it tests some additional conditions and then returns (ignore click event) if the conditions failed. For example if one clicks on grouping headers of the grid the callback onCellSelect will not be processed.
The problem with footer row because it exists outside of the grid. The main <table> element are placed inside of div.ui-jqgrid-bdiv, but the footer is inside of another table which is inside of div.ui-jqgrid-sdiv. One can examine the HTML structure of jqGrid using Developer Tools of Internet Explorer, Google Chrome, Firebug or other. One will see about the following
The main <table> element (<table id="list"> in the picture above and which get the class "ui-jqgrid-btable") and another table element with the footer (which get the class "ui-jqgrid-ftable") are separate.
So the fist answer of Mark on your question was correct. If one has multiple grids on the page one could specify footer of specific grid using
var $grid = $('#jqgSummaryResults'); // one specific grid
.... // here the grid will be created
$grid.closest(".ui-jqgrid-view").find(".ui-jqgrid-sdiv").click(function() {
// do in case of the footer is clicked.
var $td = $(e.target).closest("td"),
iCol = $.jgrid.getCellIndex($td); // or just $td[0].cellIndex,
colModel = $grid.jqGrid("getGridParam", "colModel");
// $td - represents the clicked cell
// iCol - index of column in footer of the clicked cell
// colModel[iCol].name - is the name of column of the clicked cell
});
P.S. In the old answer are described many other elements of the grid. The descriptions are not full, but it could be probably helpful.
Here little implementation of this problem, i'm new in jquery and jqgrid, but i had same problem and thanks two posts above of #Oleg and #Mark, Im implemented something like that:
//Raport1Grid - name of my jqgrid
//endusers, adminusers,decretusers - name of my rows in colModel
//Raport1Grid_endusers - GridName_ColumnName
var endUsers = $("[aria-describedby='Raport1Grid_endusers']").click(function(){
//remove previous style of selection
$('.ui-jqgrid-ftable').find('.selecteClass').removeClass('selecteClass');
//set selection style to cell
$(endUsers).addClass('selecteClass');
});
//Also can get value of selectedCell
var qwer = $("[aria-describedby='Raport1Grid_endusers']").text();
alert(qwer);
Demo here
http://jsfiddle.net/Tobmai/5ju3py83/
I have a JQGrid with 4 columns as below
<script type="text/javascript">
$(function(){
jQuery("#toolbar").jqGrid({
url:'<%=request.getContextPath()%>/TestServlet?q=1&action=fetchData',
datatype: "xml",
mtype: 'POST',
height: '100%',
width: '100%',
colNames:['LocationImage','City','State','LocationID'],
colModel:[
{name:'LocationImage',index:'LocationImage',align: 'center', formatter: imageFormatter},
{name:'City',index:'City', width:200,fixed:true,sortable:true,align:'center',editable:true, editoptions:{readonly:false,size:500}},
{name:'State',index:'State', width:200,fixed:true,sortable:true,align:'center',editable:true, editoptions:{readonly:false,size:50}},
{name:'LocationID',index:'LocationID',width:60,align:'center',hidden:true,editable:false, editoptions:{readonly:true,size:30}
],
paging: true,
rowNum:16,
rowTotal:2000,
rownumbers: true,
rownumWidth: 25,
rowList:[16,32,48],
viewrecords: true,
loadonce: true,
gridview: true,
pager: $("#ptoolbar"),
sortname: 'Name',
sortorder: "asc",
caption: "Test Grid"
})
jQuery("#toolbar").jqGrid('navGrid','#ptoolbar',{del:false,add:false,edit:false});
jQuery("#toolbar").jqGrid('filterToolbar',{stringResult: true,searchOnEnter : false});
});
function imageFormatter(el, cval, opts) {
var num = '2';
return ("<center><img src='../gd/images/" + num+".png' height='180px' width='180px' /></center>")
}
Here I have hardcoded num as '2' and so displaying 2.png in the first column i.e. LocationImage, what i need is that the value for num should be from the 4th column i.e. LocationID (note its a hidden column).
Please let me know how to do this.
Thanks,
Deepna
You should change your imageFormatter function to be like this:
function imageFormatter(cellvalue, options, rowObject){
var num = $("#" + options.rowId).find("td[aria-describedby='toolbar_LocationID']").text();
return "<center><img src='../gd/images/" + num + ".png' height='180px' width='180px' /></center>";
}
okay, I'm not rally sure about it, I can't test now. but here's what you can do
From the options parameter you can get the row id, check here and then you can get Location Id with that row Id using getCell or getGridParam. and you can use this value for setting the image.
If this doesn't work for custom formatter, you can use setCol method in gridComplete property of JqGrid.
Now this is going to work for only one id. If you want to set image for all rows, get row ids using getDataIDs method in a variable and t hen go with for loop and use setCol.
I Hope this helps
I was able to fix the issue in the below way
Changed the first column in the Grid to Location ID
Now in the formatter - when I say "cellvalue", it actually
returns the location id
This location id is then set into the variable num (i.e. num =
cellvalue)
Now the formatter code replaces the id value with an image based
on id.png
Thanks,
Deepna