插入语句不起作用。不转移到其他表

Question

对PHP和MySQL来说很新颖。

我已经在我的php脚本中创建了一条insert语句，将一行数据从一个表转移到另一个表的某些字段。唯一的事情是，它似乎没有工作？

任何人都可以看到问题出在哪里吗？

<?php 

require_once('auth.php'); 

$host=""; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name=""; // Database name 
$tbl_name="Instruction"; // Table name 

// Connect to server and select database. 
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB"); 

$sql="INSERT INTO Triage (Reference,Forename,surname,D.O.B,Mobile Number,Home Number,Address,Postcode1,Email,Accident,Details); 
VALUES (Reference,Forename,surname,DOB,Mobile,Home,Address,Postcode1,Email,Accident,Details)"; 
$result=mysql_query($sql); 

// 
while($rows=mysql_fetch_array($result)){ 
echo '<a href="update.php?Reference='.$rows['Reference'].' ">update test</a>'; 
} 
// 
// end of while loop 

echo "Successful"; 
echo "<BR>"; 
echo "<a href='list_records.php'>View result</a>"; 

?> 

更新

// Connect to server and select database. 
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB"); 

$Reference=$_REQUEST['Reference']; 
$Forename=$_REQUEST['Forename']; 
    $surname=$_REQUEST['surname']; 
$DOB=$_REQUEST['DOB']; 
$Mobile=$_REQUEST['Mobile']; 
$Home=$_REQUEST['Home']; 
$Address=$_REQUEST['Address']; 
$Postcode=$_REQUEST['Postcode1']; 
$Email=$_REQUEST['Email']; 
$Accident=$_REQUEST['Accident']; 
$Details=$_REQUEST['Details']; 


//semi colon removed 
$sql="INSERT INTO Triage (Reference,Forename,surname,D.O.B,Mobile Number,Home Number,Address,Postcode1,Email,Accident,Details) 
VALUES('.$Reference.','.$Forename.','.$surname.','.$DOB.','.$Mobile.','.$Home.','.$Address.','.$Postcode1.','.$Email.','.$Accident.','.$Details.')"; 
$result=mysql_query($sql); 


echo "Successful"; 
echo "<BR>"; 
echo "<a href='list_records.php'>View result</a>"; 


?> 

Answer 1

首先应该解决的任务：

$Reference=$_REQUEST['Reference']; 
$Reference=$_REQUEST['Forename']; 
... 

应该是这样的：

$Reference=$_REQUEST['Reference']; 
$Forename=$_REQUEST['Forename']; 
$surname=$_REQUEST['surname']; 

然后在更新查询：

$sql="INSERT INTO Triage (Reference,Forename,surname,D.O.B,Mobile Number,Home Number,Address,Postcode1,Email,Accident,Details) 
VALUES (".$Reference.",".$Forename.","... 

等与其余的值。

而且

while($rows=mysql_fetch_array($result)){ 

不会因为工作的结果将只包含成功返回真。

也许还有更多的错误我不确定。但你也应该检查这个以了解如何避免注入： What's the best method for sanitizing user input with PHP?

Answer 2

如果你想将数据从一个表转移到另一个表，你应该在某处选择这个表。你的代码中没有任何地方，你只是指定了列，你的脚本应该如何知道它们来自哪里？

INSERT INTO table1 (col1, col2, col3) SELECT correspondingColumn1, correspondingColumn2, correspondingColumn3 FROM table2 

PS：不要使用$参考，不过，你overwritting它

Answer 3

试试这个

1）你提到的所有变量名称为$引用其改变

2）查询不正确plz研究如何写查询..

3）REFER：http://www.w3schools.com/php/php_mysql_intro.asp

<?php 

require_once('auth.php'); 

$host=""; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name=""; // Database name 
$tbl_name="Instruction"; // Table name 

// Connect to server and select database. 
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB"); 

$Reference=$_REQUEST['Reference']; 
$Forename=$_REQUEST['Forename']; 
$surname=$_REQUEST['surname']; 
$DOB=$_REQUEST['DOB']; 
$Mobile=$_REQUEST['Mobile']; 
$Home=$_REQUEST['Home']; 
$Address=$_REQUEST['Address']; 
$Postcode=$_REQUEST['Postcode1']; 
$Email=$_REQUEST['Email']; 
$Accident=$_REQUEST['Accident']; 
$Details=$_REQUEST['Details']; 


//semi colon removed 
$sql="INSERT INTO Triage (Reference,Forename,surname,D.O.B,Mobile Number,Home Number,Address,Postcode1,Email,Accident,Details) 
VALUES ('$Reference','$Forename','$surname','$DOB','$Mobile','$Home','$Address','$Postcode1','$Email','$Accident','$Details')"; 
$result=mysql_query($sql); 



// 
while($rows=mysql_fetch_array($result)){ 
echo '<a href="update.php?Reference='.$rows['Reference'].' ">update test</a>'; 
} 
// 


// end of while loop 

echo "Successful"; 
echo "<BR>"; 
echo "<a href='list_records.php'>View result</a>"; 


?>