执行下面的代码时，它返回错误“警告无法修改标头信息 - 标头已由......发送”。我已经编写了以 2 种不同方式提取数据的数组，但仍然收到问题。<?php //export.php $connect = mysqli_connect("localhost", "username", "password", "dbname"); $output = ''; if(isset($_POST["export"])){ $query = " select trim(ifnull(application, 'Grand Total' )) as Application, ifnull(sum(Totalcapacity) ,0.00) as 'Capacity (Gig)' from storagedata group by application with rollup "; $result1 = mysqli_query($connect, $query); $result2 = mysqli_query($connect, $query); $dataRow = ""; while($row2 = mysqli_fetch_array($result2)){ $dataRow = $dataRow."<tr><td>$row2[0]</td><td>$row2[1]</td></tr>";}?><html><head> <title>Returning Total Capacity Per DC</title></head><body><!-- Building table --> <table align = 'center' border='2' width='300' height='100'> <tr> <td colspan='2' align='center' bgcolor='lightblue'><b>Total Capacity by DC</b> </td> </tr> <tr> <th>Datacenter</th> <th>Total Capacity (Gig)</th> </tr> <?php while($row1 = mysqli_fetch_array($result1)):;?> <tr> <td><?php echo $row1[0];?></td> <td><?php echo $row1[1];?></td> </tr> <?php endwhile;?></body> <br><br><?php $output .= '</table>'; header('Content-Type: application/xlsx;'); header('Content-Disposition: attachment; filename=download.xlsx'); echo $output; }?>当单击我的导出按钮时，它应该导出数据。同样的代码也适用于其他“报告”。唯一的区别是顶部的选择查询。
1 回答
慕标琳琳
TA贡献1830条经验 获得超9个赞
地方：
header('Content-Type: application/xlsx;');
header('Content-Disposition: attachment; filename=download.xlsx');
在脚本的开头。
<?php
header('Content-Type: application/xlsx;');
header('Content-Disposition: attachment; filename=download.xlsx');
//...
并确保
<?php
位于文件的最开头，之前没有行或空格。
在您的脚本发送了一些输出之后，您将无法更改标头设置。
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