ajax.dataSrc server-side processing parameter

ajax.dataSrc server-side processing parameter

DystDyst Posts: 8Questions: 3Answers: 0

Good day,

I'm using DataTable 2.0.8 and I'm having a problem in displaying data. I've followed the documentation for ajax.dataSrc and Server-side processing.processing on DataTables.net, but the table remains empty.

Here is the parsedServerData after calling the function in visual basic:

Here is my code for the DataTable:

    var empTable = $('#employeeTable').DataTable({
        stateSave: true,
        processing: true,
        serverSide: true,
        ajax: {
            url: "Index.aspx/GetServerSideData",
            type: "POST",
            contentType: "application/json; charset=utf-8",
            data: function (d) {
                return JSON.stringify({ parameters: d });
            },
            dataSrc: function (json) {
                let parsedServerData = JSON.parse(json.d)
                console.log(parsedServerData)
                return parsedServerData;
            }
        },
        columns: [

            { data: 'IDNumber' },
            { data: 'FullName' },

        ],
        columnDefs: [
            { targets: 0, className: 'dt-body-center' },
            { targets: 1, className: 'dt-body-left' }
        ],
        order: [],
    });

I also tried using this format:

        dataSrc: function (json) {
            var parsedData = JSON.parse(json.d);
            return {
                draw: parsedData.draw,
                recordsTotal: parsedData.recordsTotal,
                recordsFiltered: parsedData.recordsFiltered,
                data: parsedData.data
            };
        },

After trying all of that, I still can't make it work. Is there something I'm missing? Any help would be greatly appreciated. Thank you.

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Answers

  • DystDyst Posts: 8Questions: 3Answers: 0
    edited July 16

    Update.

    By using this dataSrc:

                dataSrc: function (json) {
                    var parsedData = JSON.parse(json.d);
                    return parsedData.data; 
                },
    

    I finally got a response from the DataTable debugger and the table now displays data, but the filter and pagination are still not working properly.

    Here is the table information:

    And here is the table:

  • allanallan Posts: 62,858Questions: 1Answers: 10,344 Site admin
    Answer ✓

    If you need to do JSON.parse(json.d); that is because the server-side is returning the JSON data as a string in the d parameter. I've never understood why Microsoft decided to do that in .NET - it just doesn't make sense to me. But, that is what it is!

    DataTables 2.1 has built in support for that and you could try the nightly version if you fancy it? You'll need to remove the ajax.dataSrc function that you are currently using to let it handle itself.

    2.1 is planned for the next few days, so if you are willing to give it a crack, I'd appreciate any feedback.

    If that isn't an option, you need to use ajax.dataSrc as an object with each of the server-side processing properties specified and pulling the information out of the string. See the "object" part of the documentation. Each would need a function. The reason for this is that ajax.dataSrc on its own specifies where the data is, not the other parameters as well.

    Allan

  • DystDyst Posts: 8Questions: 3Answers: 0

    Thank you for your response.

    After experimenting with json.d, as you mentioned, I've learned that I can't adjust it due to the parameters of ajax.dataSrc and the need to parse the data first. Instead, I used HttpContext.Current.Response.Write(jsonResponse) to output the correct JSON data. Dealing with encapsulation has always been a pain.

    Ah yes, I'll be sure to try version 2.1.

    As always, thank you.

  • allanallan Posts: 62,858Questions: 1Answers: 10,344 Site admin

    HttpContext.Current.Response.Write(jsonResponse)

    Perfect! Thanks - I need to put that into a "useful information" file, as this does come up on a semi-regular basis (hence the addition in 2.1!).

    Allan

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