2014-08-31 108 views
0

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); 
?> 

回答

2

使用该查询

INSERT INTO aircondition (id, ACName, BTU, Space, EnergyClass) 
SELECT id, '$acname', '$btu', '$space', '$energyclass' 
FROM users 
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