在jQuery中将一列的内容复制到另一列

Question

以下jQuery非常慢(约7秒).我显然做错了!

我正在尝试将列的内容复制col到0HTML表中的列,因此如果col为2,那么我需要将第2列复制到第0列.

for (var i=0;i<31;i++)
  $('.grid tr:nth-child(' + i + ') td:first-child').text(
    $('.grid tr:nth-child(' + i + ') td:nth-child(' + col + ')').text()
   );

HTML:

<table>
  <tr><td>A</td><td>D</td><td>G</td></tr>
  <tr><td>B</td><td>E</td><td>H</td></tr>
  <tr><td>C</td><td>F</td><td>I</td></tr>
  <!-- etc. -->
</table>

Answer 1

您不需要单独选择每个表格单元格。您可以选择源列和目标列并迭代它们：

// Get the target column table cells.  This will select the first cell from
// each row in the table.
var target = $('.grid tr td:first-child');

// Iterate over each cell in the source column and copy its text to the
// corresponding cell in the target column.
$('.grid tr td:nth-child(' + (col + 1) + ')').each(function (rowIndex) {
    target.slice(rowIndex, rowIndex + 1).text($(this).text());
});