插入语句不起作用。 不转移到其他表

Question

PHP和MySQL的新手。

我已经在php脚本中创建了一条insert语句，用于将某些表中的一行数据从一个表传输到下一个表。 唯一的事情是，它似乎不起作用？

有人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>";

 ?> 

更新资料

// 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，您应该修正作业：

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

应该是这样的：

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

然后在以下位置更新查询：

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

以此类推。

也

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

将不起作用，因为结果仅在成功时包含true。

也许还有更多我不确定的错误。 但是，您还应该检查一下以了解如何避免注入： 用PHP清理用户输入的最佳方法是什么？

Answer 2

如果要将数据从一个表传输到另一个表，则应在某处选择该表。 您的代码中没有任何位置，只是指定了列，脚本应该如何知道它们来自何处？

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

PS：您没有使用$ Reference，但是仍然覆盖了它

Answer 3

试试这个

1）您将所有var名称都称为$ Reference更改了

2）查询不正确的plz学习怎么写查询..

3）参考： 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>";


 ?>