How to pivot columns in free jqgrid 4.15.4

How to pivot columns in free jqgrid 4.15.4

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How to pivot columns in free jqgrid 4.15.4
Tag : jquery , By : quicky
Date : November 25 2020, 07:06 PM

it helps some times What you are describing cannot easely be achieved using the jqPivot feature, as the data structure you are showing doesn't seem to follow a specific rule (some records contain value for one cell, while other records have values for two cells).
It seems that you only need to display one record with its values split on two lines, without much need of the features ofered by jqGrid, in which case I would recommend not using a grid plugin, but simply creating the DOM manually.
var dataConverted = convertData(data); //Convert your custom data to the structure of colModel

    datatype: "local",
    data: dataConverted,
    colModel: [
        {name: 'deadlineType', label: 'Deadline'},
        {name: 'deadlineValue', label: ''},
        {name: 'daysLeftType', label: 'Days left'},
        {name: 'daysLeftValue', label: ''},
        {name: 'participantsType', label: 'Participants'},
        {name: 'participantsValue', label: 'Total'},
        {name: 'qaPassed', label: 'QA passed'}

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How to show all columns in column context menu in free jqgrid

Tag : jquery , By : beebob
Date : March 29 2020, 07:55 AM
like below fixes the issue I posted updated version of jquery.jqgrid.showhidecolumnmenu.js to the github. It should solve the described problems.

How to fix Cannot read property 'cellIndex' of undefined dragging from one jqgrid column to other in free jqgrid

Tag : javascript , By : Jaya
Date : March 29 2020, 07:55 AM
I hope this helps you . It seems to me that you have to change your code a little. You should always test that you call $.jgrid.getCellIndex with DOM of
element and not with some other children of it. The code fragment of typical usage of $.jgrid.getCellIndex should be the following:
var $td = $(e.target).closest("tr.jqgrow>td");
if ($td.length > 0) {
    var iCol = $.jgrid.getCellIndex($td);

How to do natural sort/alpha numeric sort over columns for y-dimension in jqGrid pivot table

Tag : development , By : Habikki
Date : March 29 2020, 07:55 AM
Hope that helps Free jqGrid 4.9 contains full rewritten version of jqPivot. I tried to hold compatibility with the previous version, but it contains many advanced features. I tried to describe there in wiki.
Not so many people uses jqPivot. So I remind what it do. It gets an input data as source and generate new data, which will be input data for jqGrid. Additionally jqPivot generates colModel based on input data and yDimension parameter. During analyzing of input data jqPivot sorts input data by xDimension and by yDimension. The order or sorting of xDimension defines the order of rows of resulting grid. The order or sorting of yDimension defines the order of columns of resulting grid and the total number of resulting columns. The options compareVectorsByX and compareVectorsByY of allows to specify callback function which will be used for custom sorting by the whole x or y vector. It's important to understand that sorting function not only specify the the order of columns, but it informs jqPivot which vectors should be interpreted as the same. For example it can interpret the values 12, 12.0 and 012.00 as the same and specify that 12.0 is larger as 6.
yDimension: [
    { dataName: "sellyear", sorttype: "integer" },
    { dataName: "sell month",
        compare: function (a, b) {
            if (a === "Germany") { return b !== "Germany" ? -1 : 0; }
            if (b === "Germany") { return 1; }
            if (a > b) { return 1; }
            if (a < b) { return -1; }
            return 0;
compareVectorsByY: function (vector1, vector2) {
    var fieldLength = this.fieldLength, iField, compareResult;

    if (fieldLength === 2) {
        if (vector1[0] === "2011" && vector1[1] === "Germany") {
            if (vector2[0] === "2011" && vector2[1] === "Germany") {
                return {
                    index: -1,
                    result: 0
            return {
                index: vector2[0] === "2011" ? 1 : 0,
                result: -1
        // any vector1 is larger as vector2 ("2011", "Germany")
        if (vector2[0] === "2011" && vector2[1] === "Germany") {
            return {
                index: vector2[0] === "2011" ? 1 : 0,
                result: 1

    for (iField = 0; iField < fieldLength; iField++) {
        compareResult = this.fieldCompare[iField](vector1[iField], vector2[iField]);
        if (compareResult !== 0) {
            return {
                index: iField,
                result: compareResult
    return {
        index: -1,
        result: 0

Free-JQGrid: On Inline edit, the height of frozen columns is not matching with other columns

Tag : development , By : Tamizhvendan
Date : March 29 2020, 07:55 AM
hop of those help? Thank you for the bug report. I committed the fix of the problem to GitHub. Now the demo works correctly.

Free jqGrid - Showing/Hiding columns if empty

Tag : jquery , By : Jet Thompson
Date : March 29 2020, 07:55 AM
hope this fix your issue It's better to use beforeProcessing to analyse the data returned from the server. The code of beforeProcessing will heavy depend on the format of data returned from the server. Below is an example of beforeProcessing:
beforeProcessing: function (data) {
    var i, foundPrescription = false;
    if (data.rows != null) {
        for (i = 0; i < data.rows.length) {
            if (data.rows[i].prescription) { // if not empty string
                foundPrescription = true;
        $(this).jqGrid(foundPrescription ? "showCol" : "hideCol", "prescription");
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