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.
Related
The React Bootstrap Table documents provide an example of how to select a row.
However, I do not see any mention of how to wrap a row in an anchor tag (or a click handler), so that when a user clicks on a table row, they are then redirected to a new url.
How do I do this? Thanks!
Additionally, I see that in an onRowSelect, you could do a window.location = ______.
However, the mode: radio/checkbox is required, and that means the table will have that additional column with a radio/checkbox. Which, for this use case (just redirecting on click) feels superfluous.
Woohoo! Got it! You can add the attribute hideSelectColumn: true and you will be able to register the click callback, and won't have that extra column of radio elements!
var selectRowProp = {
mode: "radio",
hideSelectColumn: true,
clickToSelect: true,
bgColor: "rgb(238, 193, 213)",
onSelect: onRowSelect
};
I am trying to link the variables in the first column of my jqgrid to another page. Right now, the variables do not appear link-like OR exclusive (meaning no matter where I click on the row, I am taken to the destination page).
I need to use the custom formatter because the URL in my search bar should not change (I am running this on my local server). But with the custom formatter, I cannot seem to also use 'showlink' (which ultimately seems to make my data appear link-like when I'm not using the custom formatter). I want the finger cursor when hovering over the data in my first column, all I'm getting right now is the "I".
Is there someway I can use both
formatter:'showlink' from the predefined formatter AND formatter: returnHyperLink(name) from the custom formatter? I want to be able to ONLY click the first column's data to be taken to the page, and I want this data to appear link-like ( should not be able to click anywhere on the row).
My relevant jqgrid code is:
colNames:['Name','Status', 'Created On', 'Update By', 'Updated On', 'RetentionDays','ValidityTime','Edit'],
colModel:[
{
name: 'name', width:100,editable: true, edittype:'select',
formatter: returnHyperLink(name),
xmlmap: function (obj) {
return $(obj).attr('name');
}
},
And my function, returnHyperLink appears as:
function returnHyperLink(name){
$(this).click(function() {
$( "#contents" ).load("jsp/consumers.jsp");
console.log(this, "Hello world");
});
};
...okay obviously something is wrong if all of this code isn't even showing up in the code box. I was thinking I could call the javascript function from inside of an href, but I also do not know how to do this.
It seems to me that you misunderstand how the custom formatter works. jqGrid creates the whole body of the grid (<tbody>) as the string. Thus the custom formatter should return the string and no $(this).click inside of the formatter do not what you expect. The <td> element not yet exist during the custom formatter is working.
You don't wrote, which version of jqGrid and from which fork of jqGrid you use. If you would use free jqGrid, then you can use onClick callback option of formatter: "showlink":
formatter: "showlink", formatoptions: {
onClick: function (options) {
// this is DOM of the grid
// options.iCol is the column index
// options.iRow is the row index
// options.rowid is the rowid
// options.cm is the element from colModel
// options.cmName is the column name
// options.cellValue is the text from the <a>
// options.a - DOM element of the clicked <a>
// options.event - Event object of the click event
alert("it's clicked!");
}
}
If you have to use an old version of jqGrid and can't migrate to free jqGrid then you can download jQuery.jqGrid.dynamicLink.js from here and to use formatter: "dynamicLink", which I described in the old answer. See the answer too.
If you would do prefer to use your custom formatter, then you can use beforeSelectRow callback to implement onClick action. See the answer and this one for more details.
I use grid with DataTables component.
I want to switch between two grid.
the second one is hidden at startup.
when I show the hidden grid, columns headers are not aligned with columns values
like this
you can see here in live
you can change the showed grid with radio at the top
an idea ?
An idea which correct your display error but which is not really attractive :
The idea is that your display became correct after sorting of a column so you can bypass it by adding myTable.fnSort([[0, 'asc']]); (sort first column by ascending order) after your datatable is initialized.
See here
A little more information would be helpful, but your jquery is not formatting your headers. show some more code or format them manually or get jquery to quit being that way. possibly a way to trick jquery would be to have those headers visible the whole time, but have your text font equal to your background and change the font color when it comes time
jQuery DataTables does not properly align columns and headers when the table is hidden. Temporarily show the table while applying dataTables to it:
$(document).ready(function() {
Table1 = $('#Table1').dataTable({
"bSort": false,
"sScrollY": "400px",
"bJQueryUI": true,
"bPaginate": false
});
$('#Table1Container').hide();
$('#Table2Container').show();
Table2 = $('#Table2').dataTable({
"bSort": false,
"sScrollY": "400px",
"bJQueryUI": true,
"bPaginate": false
});
$('#Table2Container').hide();
$('#Table1Container').show();
$("#rdTable1").click(function() {
$('#Table2Container').hide();
$('#Table1Container').show();
});
$("#rdTable2").click(function() {
$('#Table1Container').hide();
$('#Table2Container').show();
});
});
Since DataTables 1.10, this issue can be resolved by calling the columns.adjust() method when you show your table.
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.
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.