Such as the title
I need to add some details after load from json data format, however After I swap the two columns ( one is first column which used as rowid ), 'checkId' is no longer the rowid. In setRowData method, if each row of the swapped column' values are same, it can not identify each row.
I just a beginner of jqgrid, did I miss something?
jqgrid like this:
colModel:[
{name:'checkId',index:'checkId', width:80, key:true},
{name:'mobjName',index:'mobjName'},
{name:'sabilityName',index:'sabilityName'},
...
gridComplete: function(){
var _ids = jQuery("#my_grid").jqGrid('getDataIDs');
$(_ids).each(function(i, n) {
var row = jQuery("#my_grid").jqGrid('getRowData', n);
var _optcolstr = ...
...do something.
jQuery("#my_grid").jqGrid('setRowData', n, {optcol: _optcolstr});
});
Correspond to the documentation the syntax of the usage of setRowData should be following:
jQuery("#my_grid").jqGrid('setRowData', n, row , {optcol: _optcolstr});
or
jQuery("#my_grid").jqGrid('setRowData', n, false, {optcol: _optcolstr});
if you don't want modify the row data and only change some css property with the name optcol.
Related
In a sheet I have two rows that I'd like to sort on:
ColA: Archived - True/False checkbox
ColB: SortOrder - number, a way I have at the moment to group things
Cols C-G: various details
I have a function that does this from a macro, as I'll add a button to the sheet later to run this:
function SortServices() {
var ss = SpreadsheetApp.getActive();
ss.getRange('A1:B').activate();
ss.getActiveRange().offset(1, 0, spreadsheet.getActiveRange().getNumRows() - 1).sort([{column: 1, ascending: true}, {column: 2, ascending: true}]);
};
I'm having a problem, as there are 1000 rows but only ~300 rows currently have data. Of these, about 200 are unchecked and 100 checked (as Archived). So this sorts the entire colA, then colB, which doesn't correspond to the data cols.
Struggling to explain this, imgs might be better. Here's an example, originally:
After applying the Sort:
The issue must be the rows without data, but ColA has checkboxes, which by default are unchecked. Is there a simple way to get around this? (I don't know much about Sheets)
[Edit] Sorry, should have added what I would like... the rows with no data at the end, same as the first img. When a row is checked as Archived, using Sort would then move it down into the lower section together with other Archived items:
Solution
Divide the data in two sets and sort in the desired order those that meet the requirements (have data in column B).
Code
function myFunction() {
var ss = SpreadsheetApp.getActiveSpreadsheet().getActiveSheet()
var lr = ss.getLastRow()
// original values
var range = ss.getRange(1, 1, lr, 3).getValues()
// cells without data in colB
var filtered = range.filter(function (value, index, arr) {
return value[1] == "";
});
// cells with data in colB
var original = range.filter(x => !filtered.includes(x));
// write
ss.getRange(1, 1, original.length, 3).setValues(original).sort([{ column: 1, ascending: true }, { column: 2, ascending: true }]);
ss.getRange(1 + original.length, 1, filtered.length, 3).setValues(filtered)
}
Thought there would be an easier answer, but this works OK:
Inserted a new colA with formula to create a single sortable column:
=if( isblank(D2), 20000, if(B2=true, 10000+C2, C2 ) )
Then used this script:
function SortServices() {
var ss = SpreadsheetApp.getActive();
var sheet = SpreadsheetApp.getActive().getActiveSheet();
sheet.getRange(1, 1, sheet.getMaxRows(), sheet.getMaxColumns()).activate();
ss.getActiveRange().offset(1, 0, ss.getActiveRange().getNumRows() - 1).sort({column: 1, ascending: true});
};
Probably add this to a custom menu.
JQ grid having column name mobile
I have jq grid which has column 'Mobile Login',It contains values 'Yes' and 'No' only.
My requirement is to get count of 'Yes' value and show in footer.
Currently i am using below code in load complete event of JQ Grid
grid.jqGrid("getCol", colName, false, "count")
Is their any way to add condition where value = 'Yes' in jqgrid.
There are many ways to do this, but it is needed to have more information on your jqGrid configuration - like datatype, loadonce and etc.
To solve the problemin based on your information, you can just use the getCol, but to return array from object values and then looping this array can give you the needed result.
var colvals = grid.jqGrid("getCol", colName, true),
len = colvals.length,
i=0,
yescounter=0;
while(i<len) {
if(colvals[i].value === 'Yes') {
yescounter++;
}
i++;
}
Hope this helps
I'm using
getGridParam('selarrrow');
to get the rows that are selected,where the method returns me the selected row ids according to their selection,but I want the ids according to their index.Do I have to write a method to sort the ids or is there a inbuilt method which returns me the selected row ids in order of their indexes.
thanks in advance
If you mean the index of the row in the grid then you have to resort the array or id returned by $("#gridId").jqGrid("getGridParam", "selarrrow"). You can use sort() method of Array with your custom sort function. You can just use the fact that ids are the ids of <tr> elements. So the DOM elements of <tr> has native implemented rowIndex property which you can get by $("#"+rowid)[0].rowIndex.
In the simplified form the code could be about the following
var selRowIds = $("#gridId").jqGrid("getGridParam", "selarrrow");
selRowIds.sort(function (id1, id2) {
// one can use document.getElementById alternatively
return $("#" + id1)[0].rowIndex - $("#" + id2)[0].rowIndex;
});
or you can use namedItem method instead
var $grid = $("#gridId"),
selRowIds = $grid.jqGrid("getGridParam", "selarrrow"),
rows = $grid[0].rows;
selRowIds.sort(function (id1, id2) {
return rows.namedItem(id1).rowIndex - rows.namedItem(id2).rowIndex;
});
Probably you should include more verification in the code to be sure that the item with id will be found and it has the rowIndex property.
I have local data in a grid. How can I get all of the rows or IDs that are not removed after a user uses the filter toolbar? I need to get all filtered rows, regardless of pagination.
For example, say I begin with 50 rows in the grid. The user uses the filter toolbar and the set of rows decreases to 10 rows. How can I get those ten rows?
There are no direct way to get the information which you need. Internally jqGrid uses $.jgrid.from to filter local data. The main code which uses $.jgrid.from in inside of addLocalData. To get results which you need without studying all the code I suggest to use the fact that all filtered data will be returned by select method of $.jgrid.from (see the line of code). My suggestion is to catch the data before the data will be cut to the page size.
To do this I suggest to use sub-classing: overwriting of the method select method of $.jgrid.from. I demonstrate the technique in the examples created for the answer and this one.
In your case the code will be
var oldFrom = $.jgrid.from,
lastSelected;
$.jgrid.from = function (source, initalQuery) {
var result = oldFrom.call(this, source, initalQuery),
old_select = result.select;
result.select = function (f) {
lastSelected = old_select.call(this, f);
return lastSelected;
};
return result;
};
Now the variable lastSelected will save the array of elements which are results of the last sorting or filtering operation. Because $.jgrid.from is global the data are not connected to the grid. If you have more as one grid on the page it will be uncomfortable. One can fix the small disadvantage with the following line in the code of loadComplate of every grid:
loadComplete: function () {
this.p.lastSelected = lastSelected; // set this.p.lastSelected
}
In the way we introduce new jqGrid parameter lastSelected which will have close structure as data parameter, but will hold only last filtered data.
The following code will display the ids of filtered data in alert message
$("#getIds").click(function () {
var filteredData = $grid.jqGrid('getGridParam', 'lastSelected'), i, n, ids = [],
idName = $grid.jqGrid('getGridParam', 'localReader').id;
if (filteredData) {
for (i = 0, n = filteredData.length; i < n; i++) {
ids.push(filteredData[i][idName]);
}
alert("tolal number of filtered data: " + n + "\n" +
"ids of filtered data:\n" + ids.join(', '));
}
});
I used localReader.id parameter because property name used for local data are typically id or _id_. The _id_ will be used in case of data loaded from the server if one uses loadonce: true option.
The demo demonstrate the approach. If one filter for example only the data from FedEx and then clicks on "Show Ids" button one will see information about all filtered and not only about the data displayed on the current page:
UPDATED: free jqGrid provides new lastSelectedData option. See the demo in the list of demos.
You colud use afterSearch option of the search toolbar:
var filteredIDs = new Array(); //Global variable
$("#"+gridId).jqGrid("filterToolbar", { stringResult:true, searchOnEnter:false,
afterSearch:function(){
filteredIDs = $("#"+gridId).getDataIDs();
}
});
If you want to get the filtered rows instead the filtered IDs, use getRowData() instead of getDataIDs().
All, I found another answer which is far easier to include
loadComplete: function (gridData) {
var isSearchPerformed = $grid.getGridParam("postData")._search;
if (isSearchPerformed) {
$("#spanFilterTotal").text(gridData.records);
}
All you want is below:
$.each($grid.getRowData(), function( index, value ) {
a.push(value["COLUMN_NAME"]); //Get the selected data you want
});
I tried to access the rowObject in custom formatter function by column name but it didn't give any values. I have tried this with both JSON and XML data type .
Is there any way to get the column number by column name in jqgrid.
function Draw_Link ( cellvalue , options , rowObject )
{
return "<a href='someurl.php?col_name="+rowobject.col_name+"'>"+cellvalue+"</a>";
}
The column index for the column is the same as the index of the column in the colModel array before the jqGrid initialization (it is the same as in the input parameter colModel). If you use rownumbers:true, multiselect:true or subGrid:true additional columns will be addid to the grid as the first rows, so the column index which one has in the colModel array as the jqGrid parameter can be other as one have after the grid initialization. You can use for example this simple function to get the index
var getColumnSrcIndexByName = function(grid,columnName) {
var cm = grid.jqGrid('getGridParam','colModel'),
i=0, index=0, l=cm.length, cmName;
while (i<l) {
cmName = cm[i].name;
i++;
if (cmName===columnName) {
return index;
} else if (cmName!=='rn' && cmName!=='cb' && cmName!=='subgrid') {
index++;
}
}
return -1;
};
var index = getColumnSrcIndexByName($("#list"),'MyColumn');
UPDATED: Free jqGrid fork simplifies getting column index from column name because it holds internally the parameter iColByName, which is the map by the column name. One can just get iColByName via
var iColByName = $("#list").jqGrid("getGridParam", "iColByName");
and iColByName["MyColumn"] will be the required column index (iCol). I remind that one can use getGridParam without any parameter to get the reference to all parameters of jqGrid:
var p = $("#list").jqGrid("getGridParam");
The value
var iCol = p.iColByName["MyColumn"];
will be the column index and p.colModel[iCol].name will be "MyColumn".