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Home/ Questions/Q 8027387
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T23:47:47+00:00 2026-06-04T23:47:47+00:00

I have code which generate table. You can see below the code: function createDynamicTable(tbody,

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I have code which generate table. You can see below the code:

function createDynamicTable(tbody, rows, cols) {
         if (tbody == null || tbody.length < 1) return;
         for (var r = 1; r <= rows; r++) {
             var trow = $("<tr>");
             for (var c = 1; c <= cols; c++) {
                 var cellText = "Cell " + r + "." + c
                 $("<td>")
                         .addClass("tableCell")
                         .text(cellText)
                         .data("col", c)
                         .appendTo(trow);
             }
             trow.appendTo(tbody);
         }
     }      

the results are:

<table id="tbl" border="1">
<tbody>
<tr>
<td class="tableCell">Cell 1.1</td>
<td class="tableCell">Cell 1.2</td>
<td class="tableCell">Cell 1.3</td>
<td class="tableCell">Cell 1.4</td>
</tr>
</tbody>
</table>

I wish to add each <td></td> => <td><input><input/></td> Tagas. How I can do that?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-04T23:47:48+00:00Added an answer on June 4, 2026 at 11:47 pm

    Modify your function a bit, and it will add input fields to each cell:

    function createDynamicTable(tbody, rows, cols) {
        if (tbody == null || tbody.length < 1) return;
        for (var r = 1; r <= rows; r++) {
            var trow = $("<tr>");
            for (var c = 1; c <= cols; c++) {
                var input = $("<input />");
                $("<td>").addClass("tableCell")
                    .append(input)
                    .data("col", c)
                    .appendTo(trow);
            }
            trow.appendTo(tbody);
        }
    }
    

    DEMO: http://jsfiddle.net/Xt33h/

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