I want to set viewrecords property of jqgrid dynamically. By default this property is set as false. I want to set this to true or false (sometimes to show and at times not to show the recordText at the table footer) depending upon the data that I am populating in the grid dynamicaly. I tried with the following but with no avail-
jQuery("#gridID").jqGrid({viewrecords : true});
jQuery("#gridID").setGridParam({viewrecords : true});
I recommend you to use viewrecords: true and just hide the div.ui-paging-info inside of loadComplete depend from the current number of records. For example
loadComplete: function (data) {
if (parseInt(data.records, 10) > 10) {
$("#pager div.ui-paging-info").show();
} else {
$("#pager div.ui-paging-info").hide();
}
}
The demo demonstrate the approach. If you open on the demo the searching dialog and filter for the client data equal to test you will see only one record and the viewrecords field will be not visible:
Clicking on the "Reload Grid" navigator button will follow to show the viewrecords field back.
Related
I am trying to implement filtering in my grid.Panel component. The filtering works fine (date, string and numbers) but I can not uncheck the "Filters" checkbox by clicking on it. As far as I have seen in the Ext js documentation there is nothing that I should add extra to be able to uncheck the Filters checkbox (see image below)
When I click on the Filters checkbox, nothing happens, no error also. The only way to disable this particular "Wholesaler" filter is if I go to the text box and remove the entry. This of course would be just fine, but at the date filters, I also can not click the checkbox and thus can not remove the filter once applied (see image below) The only way to remove it is to refresh the page.
I am using Ext JS version 5.1.2, and here is the sample code:
{
dataIndex: 'Wholesaler',
text: 'Wholesaler',
filter: {
type: 'string'
},
}
And the Panel componenet itself:
Ext.apply(this, {
region: 'center',
api: new Dextop.api('sample'),
border: false,
plugins: 'gridfilters',
columns: sample.getController().getGridColumns(),
storeOptions: {
autoLoad: false
}
});
Any help is appreciated!
I have a checkbox selection model that I am using on a gridpanel. I have configured the CheckboxModel like so:
Ext.create('Ext.selection.CheckboxModel', {
mode: 'Single',
checkOnly: true,
showHeaderCheckbox: false
});
This configuration allows the user to select only one record from the gridpanel. I have removed the checkbox in the header since I want the user to have to explicitly check the box in the record row. Now I have just a blank header in that column....is there any way to add text to that header column? From what I can tell it seems like my only options are to either leave the checkbox in the header or remove it and have a blank column header...is that correct?
You can override the getHeaderConfig method of the CheckboxModel, like:
Ext.define('Fiddle.override.selection.CheckboxModel', {
override: 'Ext.selection.CheckboxModel',
getHeaderConfig: function() {
return Ext.apply(this.callParent(arguments), {
width: 100,
header: 'Select'
});
},
});
Working example: https://fiddle.sencha.com/#fiddle/1b3g
I am trying to create a grid that allows user to add, edit and delete the record. I have done working for populating grid now I am going to display dialog for add record. But, I don't why it doesn't work. Dialog does not getting displayed.
This is the snapshot of my grid.
and this is my js function
InfoDesk.GridManager.postsGrid = function (gridName, pagerName) {
//Create the grid
$(gridName).jqGrid({
//server url and other ajax stuff
url: '/Admin/Posts',
datatype: 'json',
mtype: 'GET',
height: 'auto',
//Columns
colNames: colNames,
colModel: columns,
//pagination options
toppager: true,
pager: pagerName,
rowNum: 10,
rowList: [10, 20, 30],
//row number columns
rownumbers: true,
rownumWidth: 40,
//default sorting
sortname: 'PostedOn',
sortorder: 'desc',
//display the no. of records message
viewrecords: true,
jsonReader: { repeatitems: false },
afterInsertRow: function (rowid, rowdata, rowelem) {
var tags = rowdata["Tags"];
var tagStr = "";
$.each(tags, function (i, t) {
if (tagStr) tagStr += ", "
tagStr += t.Name;
});
$(gridName).setRowData(rowid, { "Tags": tagStr });
}
});
$(gridName).navGrid(pagerName,
{
cloneToTop: true,
search: false,
add:true
}, editOptions, addOptions, deleteOptions);
};
When I click on add button. Nothing is happening. This is my first project with jqGrid. So, I am completely blank about it.
EDITED:
I have found a fiddle example and modified for navigation bar. It is working fine but, still I could not recognize the issue where i am doing mistake in my code.
fiddle
It seems to me that your main goal to understand how you can correctly use navGrid with top and bottom pagers. So I'll explain exactly how all works.
It's not clear whether you need to add the navigator icons on the top pager or on the both top and bottom pager. It's unclear whether you use the bottom pager at all.
jqGrid have two main option which specify the page: pager and toppager options. To use pager option you need create <div> which have id attribute, place the div on the page somewhere and to use the id selector or the id name as the value of pager option. For example you can place <div id="mypager"></div> and to use pager: "#mypager" as jqGrid option. If you would use the value of pager option in another supported form: pager: "mypager" or pager: $("#mypager") then jqGrid will normalize the option to id selector. If you would use var thePager = $(gridName).jqGrid("getGridParam", "pager"); to get the value of "pager" option directly after the grid is created you will get thePager === "#mypager" independent from the way in which you used input pager parameter.
The option toppager works a little in another way. You should use toppager: true to create the top pager. In the case jqGrid create the corresponding <div> itself. The id of the div will be constructed from the id of the grid and the string _toppager. So in case of having gridName equal to #mygrid you would have the top pager with id="mygrid_toppager". If you would get the value of toppager option after the grid is created
var theTopPager = $(gridName).jqGrid("getGridParam", "pager");
you will get back the id selector of the top pager instead of true: theTopPager will be equal to "#mygrid_toppager" (gridName + "_toppager").
The value of the first parameter of navGrid should be the id selector of the pager on which you want to place the navigator buttons. So it should be $(gridName).navGrid(gridName + "_toppager", ...); to add the buttons on the top pager and $(gridName).navGrid(pagerName, ...); if you want to add buttons on the bottom pager. The usage $("#grid").navGrid('setGridParam', ... like you do in jsfiddle demo is wrong because 'setGridParam' is not the id selector of any pager.
In case if you have two pagers (at the bottom and at the top of the grid) you can use pager selector as the first parameter of navGrid and to use additional option cloneToTop: true. By the way the method navButtonAdd which can be used to add custom button don't have any cloneToTop: true option and you would have to specify the id selector of the pager directly.
So if you need create the grid with one top pager only you can remove unneeded jqGrid option pager: pagerName and to use the following code to create the navigator bar with Add, Edit, Delete and Refresh buttons:
$(gridName).navGrid($(gridName).getGridParam("toppager"), { search: false });
If you would like to create grid on both top and bottom pager then you have to use both options of jqGrid toppager: true and pager: pagerName and can use either
$(gridName).navGrid($(gridName).getGridParam("pager"), { search: false });
$(gridName).navGrid($(gridName).getGridParam("toppager"), { search: false });
or the short form
$(gridName).navGrid($(gridName).getGridParam("pager"),
{ search: false, cloneToTop: true });
By the way I use $(gridName).getGridParam("pager") instead of pagerName as the parameter of navGrid because I'm not sure whether you use id name (like "mypager") or id selector (like "#mypager") as the parameter of InfoDesk.GridManager.postsGrid. The method navGrid accept only the id selector.
If you would need to specify additional parameters of form editing methods used during Add, Edit or Delete operation you should specify additional optional parameters editOptions, addOptions, deleteOptions, searchOptions, viewOptions (see the documentation). You should of cause define the objects editOptions, addOptions, deleteOptions, searchOptions, viewOptions before the usage. The current code which you posted don't contains the definition of the option.
UPDATED: By the way I implemented some new features in the fork of jqGrid which I publish on GitHub here. 1) One can now use pager: true option like toppager: true. 2) one can use navGrid without the pager (like $(gridName).navGrid({search: false});). In the case jqGrid will add the buttons on all pagers which jqGrid have. 3) I added new option navGrid: iconsOverText:true which allows another look of navigator buttons with texts (see the demo). 4) the navigator buttons will be wrapped on the next row of pager automatically if too many icons is added and the grid have not so large width. You can see more demos of the features at the bottom of readme on the page.
I am new to ExtJS and I want to add something like a favorite button to each row of a data grid. I have gone through almost all sources after a lot of googling but I have not found anything. If anyone has a clear cut idea about how this can be done, please let me know.
First of all adding ExtJS Component inside grids is not supported by default, and the tutorials I've seen out there are kinda hacky. So this is what I would do.
I assume your grid is bound to a store, and each record in the store is a row in your grid.
I assume you have a field in each record that represents the "fav" status of that record, maybe a boolean value.
if the above assumptions are true, I've done something like this before:
add a column in your grid with id "fav-col" that has the dataIndex pointing to the fav field of your store.
{
id : 'fav-column',
dataIndex : 'fav',
sortable : true,
hideable : false,
menuDisabled : true,
fixed : true,
width : 20,
renderer : renderFav
}
add a renderer to that column that render different HTML depending if the row is fav'ed.
function renderFav(favAdded, metaData, record){
if (favAdded === true){
return 'fav added'; //something to represent already added to favourite ;
}else{
return 'fav not added'; //something to represent non-fav'ed row;
}
}
add a listener to 'cellclick' event on the grid, check if the cell being clicked on is a fav cell and toggle the fav value of the record, the grid will re-render automatically once the data in the store is changed
cellclick : function(grid, cellEl, cellIdx, record, rowEl, rowIdx, evtObj){
if (this.columns[cellIdx].getId() === 'fav-col'){
record.set('fav', !record.get('fav')); //toggle the fav state
grid.getStore().sync(); //if the store is a REST store, update backend
record.commit(); //commit the record so the red triangle doesn't show
this.doLayout(); //might not need this.
}
}
I want to remove the paging buttons on the grid, but I want to keep the add, edit, refresh, etc buttons on the bottom left. I don't want the pager there because I will be displaying all records in this particular grid implementation.
I want to keep what is in GREEN but remove what is in RED:
Currently, my solution is to empty out the center of the grid's navigation
$('#pager_center').empty();
But this means that the pager renders to the page, and then gets emptied, I'm wondering if I can just prevent it from even being rendered in the first place.
You can use my following JqGrid option to disable RED zone from JqGrid. It should be the best way to solve this question because you don't need to hack JqGrid rendering via CSS style sheet that be caused of problem if JqGrid change pattern for generating pager or you change pager id.
$('#grid').jqGrid
({
rowList: [], // disable page size dropdown
pgbuttons: false, // disable page control like next, back button
pgtext: null, // disable pager text like 'Page 0 of 10'
viewrecords: false // disable current view record text like 'View 1-10 of 100'
});
You could apply a CSS style to hide it...?
#pager1_center {
visibility: hidden;
}
There are also options like pgbuttons and recordtext that settings in the init might cause that part not to render any HTML.
jQuery("#grid_id").jqGrid({pgbuttons:false, recordtext: ''});
Using this will remove the paging/view records area with buttons and everything.
jQuery("#WebsitesGrid").jqGrid({
...
pginput: false,
pgbuttons: false,
viewrecords: false,
....
Or if you would like to have more space in the footer of your jqGrid, you can simply hide desired part of
navigation.
gridComplete: function()
{
$( '#' + gridId + 'Pager_center' ).hide();
$( '#' + gridId + 'Pager_left' ).hide();
},
where gridId is id of your jqGrid.
$('#grid').jqGrid({pgbuttons:false, recordtext:'', pgtext:'')}
If you're looking for a solution to avoid Pager in jqGrid then just add following code in loadcomplete callback or as a statement after your jqgrid call, with or without #Soul_Master's solution,
$("#divPager").css({ "height": "0px", "border": "0px" });
It worked for me.
even you can set align property of pager details like no of records dropdown, pager text, record text. to acheive this need to change pagerpos and recordpos to right or left or center as we required. Details has to be updated in jquery.jqGrid.min.js or just do search for these keywords in your js files and do the update.