Question

我发现下面的代码几乎就像id一样。但是，当我去复制数据透视数据工作表并专门将其粘贴到资源成本核算... xls 中名为“数据”的标签中时我似乎无法使它工作。一旦它复制了如何关闭它复制的excel文件？任何帮助，建议都非常感谢。

<div class="panel panel-default">
  <div class="panel-heading"><?php echo $heading_title; ?></div>
  <div class="list-group">
    <?php foreach ($filter_groups as $filter_group) { ?>
    <a class="list-group-item"><?php echo $filter_group['name']; ?></a>
    <div class="list-group-item">
      <div id="filter-group<?php echo $filter_group['filter_group_id']; ?>">
        <?php foreach ($filter_group['filter'] as $filter) { ?>
        <div class="checkbox">
          <label>
            <?php if (in_array($filter['filter_id'], $filter_category)) { ?>
            <input type="checkbox" name="filter[]" value="<?php echo $filter['filter_id']; ?>" checked="checked" />
            <?php echo $filter['name']; ?>
            <?php } else { ?>
            <input type="checkbox" name="filter[]" value="<?php echo $filter['filter_id']; ?>" />
            <?php echo $filter['name']; ?>
            <?php } ?>
          </label>
        </div>
        <?php } ?>
      </div>
    </div>
    <?php } ?>
  </div>
  <div class="panel-footer text-right">
    <button type="button" id="button-filter" class="btn btn-primary"><?php echo $button_filter; ?></button>
  </div>
</div>
<script type="text/javascript"><!--
$('#button-filter').on('click', function() {
    filter = [];

    $('input[name^=\'filter\']:checked').each(function(element) {
        filter.push(this.value);
    });

    location = '<?php echo $action; ?>&filter=' + filter.join(',');
});
//--></script>

Answer 1

在下面的代码中，我将了解如何关闭工作簿的示例。使用您要打开的工作簿的变量。之后，您可以关闭它而不保存它。

Dim wbLatestFile As Workbook

'Open the latest file and copy Pivot Data Tab
Set wbLatestFile = Workbooks.Open MyPath & LatestFile
wbLatestFile .Worksheets("Pivot Data").Copy After:=CopiedData

wbLatestFile.Close SaveChanges:=False

将工作表复制到另一个工作簿并关闭源工作表

1 个答案: