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我的代码存在问题。我有一个名为“用户”与“ID”字段的表。我想将ID值复制到另一个名为“aircondition”的表。这是将值插入AIRCONDITION表。问题是代码,当我使用这个代码,我在新的id字段得到0代替user.id将值复制到另一个表
<?php
$con=mysqli_connect("localhost","george","george123","my_db");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// escape variables for security
$acname = mysqli_real_escape_string($con, $_POST['ACName']);
$btu = mysqli_real_escape_string($con, $_POST['BTU']);
$space = mysqli_real_escape_string($con, $_POST['Space']);
$energyclass = mysqli_real_escape_string($con, $_POST['EnergyClass']);
$sql="INSERT INTO aircondition (id, ACName, BTU, Space, EnergyClass)
VALUES ('SELECT id
FROM users', '$acname', '$btu', '$space', '$energyclass')";
if (!mysqli_query($con,$sql)) {
die('Error: ' . mysqli_error($con));
}
header('location:aircondition.php');
mysqli_close($con);
?>