2014-08-31 109 views
0

我的代碼存在問題。我有一個名爲「用戶」與「ID」字段的表。我想將ID值複製到另一個名爲「aircondition」的表。這是將值插入AIRCONDITION表。問題是代碼,當我使用這個代碼,我在新的id字段得到0代替user.id將值複製到另一個表

<?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

使用該查詢

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