如何将弹出模式的值传递给原始表

Question

Original form

<table>
     for($i=0;$i<5;$i++)
     {
      <tr>
      <td><input name="data[$i]"/> <button onclick="getElementById('modal').style.display('block')"> 
          </button></td>
      </tr>
     }
</table>

The pop up modal

<div id="modal">
         <table>
         <tr>
             <td>Value 1</td>
             <td>Value 2</td>
         </tr>
         </table>
</div>

How do I get the td value from popup modal to the original form Table tr td upon click on that particular row?

Answer 1

First of all, you need to add an event to the <td> tags.

<td onclick="return sendValue(this);">Value 1</td>

In the calling (parent) window add such JS code:

function HandlePopupEvent(result) {
    alert("the selected popup value: " + result);
}

In the child window(popup) code add this:

function sendValue(sender) {
    try {
        window.opener.HandlePopupResult(sender.innerHTML);
    }
    catch (err) {}
    return false;
}