dynamic,jqgrid,options , How to make jqGrid dynamically populate options list based on row data


How to make jqGrid dynamically populate options list based on row data

Question:

Tag: dynamic,jqgrid,options

I am using jQrid version 3.8.1 with inline editing and each row in the grid has several dropdown lists to populate. When the user edits the row, I need to do an AJAX query to get the values for each of these lists. I've seen this post regarding how to do that. It appears that the dataUrl and buildSelect features are the standard answer here. There are a few issues I can't figure out though:

  1. The row the user is editing has a value that must be passed into the dataUrl value. For example, say each row contains a field called "SpecialValue" and that for row 1, SpecialValue = 100. The dataUrl field for row 1 would be "http://my.services.com?SpecialValue=100". How do I do that?

  2. Each row in the grid has about 10 select boxes that need to be populated. For efficiency reasons, I don't want to make 10 separate AJAX calls. It would be much better to make one call to get all the data, split it up, and fill each select box accordingly. Is that possible? I tried doing this inside onSelectRow but the grid ended up ignoring the values I put in there (I'm guessing do the ordering of the events that fire when you edit a row).

Edit:

After reading Oleg's answers and working on it more, it's clear to me that using dataUrl and buildSelect are not going to work well for me. The version of jqGrid I'm using doesn't support using dataUrl the way I would need. And even if it did I don't want to send multiple separate requests for each dropdown list.

I've decided to do one request when gridComplete fires to pull all the data needed for all dropdown lists into a single JSON structure. Then when the user selects a row to do inline editing, I will populate each list in the row from that JSON structure (the code below uses the getSelectValuesFromJSON() function for that--I don't give its definition but you can imaging it looks through the structure and gets an appropriate list of values to but in the list box).

I have a few candidate solutions but I'm not 100% happy with either one.

Solution 1:

Inside onSelectRow, I call editRow overriding the on oneditfunc to get the data out of the grid that I need. Assume that the value in Field1 is required to get the values to be put into the list in Field2.

onSelectRow: function (index, status, e) {
    jQuery('#my_grid').jqGrid('editRow', index, true, function(rowId) {
        var f1Val = $('#my_grid').jqGrid('getCell', index, 'Field1');
        var selectVals = getSelectValuesFromJSON(f1Val); //gets data out of previously loaded JSON structure
        var select = $('#my_grid').find('tr[id="' + index + '"] td[aria-describedby="my_grid_Field2"] select');
        _.each(selectVals, function(selectVal) {
            $(select).append($("<option></option>").attr("value", selectVal).text(selectVal));
        });
    });
}

This works but I'm hesitant about the line

var select = $('#my_grid').find('tr[id="' + index + '"] td[aria-describedby="my_grid_Field2"] select');

which relies on this aria-describedby attribute that I don't know much about. Seems hacky and brittle.

Solution 2:

Make use of beforeSelectRow to dynamically change the model of the Field2 column when the user selects a row.

beforeSelectRow: function(index, e) {
    var f1Val = getGridCellValue('#my_grid', index, 'Field1');
    var values = getSelectValuesFromJSON(f1Val); //gets data out of previously loaded JSON structure
    var valStr = "";
    _.each(values, function(value) {
        valStr += value + ":" + value + ";"
    });
    jQuery('#grid_pipes').setColProp('Field2', { editoptions: { value: valStr } });
    return true;
}

This also works but I'm not sure about whether or not this is really a good idea. Is it valid to dynamically change the model of a column like that? What if the user has several rows selected at the same time? Isn't there only one model for a column? What would that mean?

To answer some of Oleg's questions, the dataType has been set to a function that uses $.ajax to post data to the server. I think I read that's not the recommended approach anymore. I inherited this code so I'm not sure why it was done that way but it probably won't change unless there is a really compelling reason.

The loadonce boolean is not specified so I guess that means it defaults to false.

Here is an abbreviated version of the column model (nothing terribly out of the ordinary):

colModel: [
    { name: 'PK', index: 'PK', hidden: true, editable: true, sortable: true, search: true },
    { name: 'Field1', index: 'Field1', hidden: true, editable: true, sortable: true, search: true },
    { name: 'Field2', index: 'Field2', sortable: false, editable: true, search: false, edittype: "select", editoptions: {} },
    { name: 'Field3', index: 'Field3', sortable: false, editable: true, search: false, edittype: "select", editoptions: {} },
    ...
]

Answer:

You don't wrote which version of jqGrid you use currently, but dataUrl can be defined as callback function with (rowid, value, name) parameters, which have to return the URL which you can build dynamically based on the information. The feature exist starting with v4.5.3 (see the line). You can use getCell, getRowData or getLocalRow inside of the callback to get the data from another columns of the row. Thus you can solve your first problem relatively easy.

You second question seems to me absolutely independent from the first one. It's better to separate such questions in different posts to allow the searching engine better to index the information and so to help other people to find it.

There are no simple way how to solve the second problem, but one can sure suggest a solution, but one have to know much more details what you do and how you do. How you start inline editing (do you use inlineNav, formatter: "actions" or you call editRow directly)? Which version of jqGrid (till version 4.7), free jqGrid or Guriddo jqGrid JS you use? How the columns with selects are defined in colModel? Which datatype you use and whether loadonce: true you use? I recommend you to post separate question with the information.

UPDATE: If you have to use old version of jqGrid then you can't generate dataUrl full dynamically, but because you need to add only SpecialValue=100" part to the URL you can follow the trick which I described in many my old answers (the first one was probably here, but the choice on property names which asked the user could be misunderstood). You can use ajaxSelectOptions.data which will define the data parameters of jQuery.ajax request. The problem only that you can define only one ajaxSelectOptions.data property. So you can add the following jqGrid option:

ajaxSelectOptions: {
    data: {
        SpecialValue: function () {
            var rowid = $myGrid.jqGrid("getGridParam", "selrow");
            return $myGrid.jqGrid("getCell", rowid, "SpecialValue");
        }
    }
}

($myGrid is something like $("#grid"))

UPDATED: You used unknown functions getSelectValuesFromJSON, getLookupValuesFromJSON in the updated part of your question. Both of there seems to use synchronous Ajax request which is not good. Moreover you set editoptions.value for only one Field2 instead of setting all selects.

onSelectRow: function (rowid) {
    var $myGrid = $(this);
    $.ajax({
        url: "someUrl",
        dataType: "json";
        data: {
            specialValue: $myGrid.jqGrid("getCell", rowid, "Field1")
        },
        success: function (data) {
            // for example response data have format:
            // { "Field2": ["v1", "v2", ...], "Field3": ["v3", "v4", ...] }
            var filed, str;
            for (filed in data) {
                if (data.hasOwnProperty(filed)) {
                    str = $.map(data[filed], function (item) {
                                return item + ":" + item
                            }).join(";");
                    $myGrid.jqGrid("setColProp", filed, {
                        editoptions: {
                            value: str
                        }
                    });
                }
            }
            $myGrid.jqGrid("editRow", rowid, true);
        }
    });
}

Nevertheless the "Solution 2" is more close to what I would recommend you. It's not really important whether to use onSelectRow or beforeSelectRow. You can make asynchronous Ajax request to the server which returns information for all select which you need. After you get the response from the server (inside of success callback) you can set editoptions.value for all selects and only then you can start editRow. In the way you will be sure that editing of the line will use row specific options in all select.

Some additional remarks. I recommend you to verify gridview: true option in the grid. Additionally I suspect that you fill the grid in not full correct way because you have hidden PK column and you use index instead of rowid as the first parameter of beforeSelectRow and onSelectRow. It's very important to understand that the current implementation of jqGrid always assign id attribute on every row (<tr> element) of the grid. So you have to provide id information in every item of input data. If you want to display the id information to the user (and so to have column in colModel with the primary key) then you should just include key: true property in the column definition. For example you can add key: true to the definition of PK column and so you will have rowid (or index in your case) with the same value like PK. It simplify many parts of code. For example jqGrid send id parameter in the editing request to the server. It's practical to have PK in the request. Moreover if you use repeatitems: false format of jsonReader the you can include id: "PK" in the jsonReader instead of having hidden PK column. It informs jqGrid to get rowid from PK. jqGrid will save PK in id attribute of <tr> and you will don't need to have additional <td style="display:none"> with the same information in the grid.

The last remark. I would strictly recommend you to update the retro version jqGrid 3.8.1 to some more recent version, for example to free jqGrid. Even if you would use no features (like Font Awesome for example) you will have performance advantages, and the look of modern web browsers will looks much better. You should understand the jqGrid 3.8.1 was tested with old (and slow jQuery 1.4.2). The version used with Internet Explorer 8 as the latest IE version (IE9 was published later in March 2011) and it's more oriented on IE6/IE7. The look in modern Chrome/Firefox/Safari can be bad. Is it what you want?


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