如何修复 PHP 中的“警告：
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。
慕田峪91588501回答
-
慕标琳琳
地方：header('Content-Type: application/xlsx;');header('Content-Disposition: attachment; filename=download.xlsx');在脚本的开头。<?phpheader('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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