php - 在html表中传递$ _POST [“variable”]

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我想在html表中传递$ _POST [“ variable”]，但我不能，应该如何获取？
我显示代码。

$number = $_POST["limit"];
$variable = $_POST["variable"];
//echo "numero " . $numero;
echo "variable " . $variable;


$result = mysqli_query($con,"SELECT City, Data, $variable FROM Net2 where Data BETWEEN '" . split($_POST["date6"]) ."' AND '" . split($_POST["date7"]) ."' order by TMax desc limit $number");

在这里，我创建了表，我想传递$ variable，但是我不能...如何？

echo "<div id= 'capatabla'>";
echo "<table  border='1'>
<thead>
<tr bgcolor= '#3399FF'>
<th>City</th>
<th>$variable</th>
<th>Data</th>
</tr>
</thead>";


echo "<tbody>";
while ($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td >" . $row['City'] . "</td>";
echo "<td>" . $row['$variable'] . "</td>";
echo "<td>" . $row['Data'] . "</td>";
echo "</tr>";
}

最佳答案

我不确定您要完成什么，可以尝试在mysql查询中使用别名。它可能会解决它：

$result = mysqli_query($con,"SELECT City, Data, $variable as variable_test FROM Net2 where Data BETWEEN '" . split($_POST["date6"]) ."' AND '" . split($_POST["date7"]) ."' order by TMax desc limit $number");

一会儿应该是这样的：

echo "<tbody>";
while ($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td >" . $row['City'] . "</td>";
echo "<td>" . $row['variable_test'] . "</td>";
echo "<td>" . $row['Data'] . "</td>";
echo "</tr>";
}