PHP和MySQL的新手。
我在我的php脚本中创建了一个insert语句,用于将某行数据从一个表传输到下一个表中。唯一的问题是,它似乎没有起作用?
有人可以看到问题所在吗?
<?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>";
?>
答案 0 :(得分:0)
首先你应该修改作业:
$Reference=$_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.",".$Forename.","...
以及其他值。
另外
while($rows=mysql_fetch_array($result)){
不起作用,因为结果只会在成功时包含true。
也许有更多的错误我不确定。但你也应该检查这一点,以了解如何避免注射: What's the best method for sanitizing user input with PHP?
答案 1 :(得分:0)
如果要将数据从一个表传输到另一个表,则应在某处选择此表。您的代码中没有任何地方,您只是指定了列,您的脚本应该如何知道它们来自哪里?
INSERT INTO table1 (col1, col2, col3) SELECT correspondingColumn1, correspondingColumn2, correspondingColumn3 FROM table2
P.S。:你不使用$ Reference,但你仍在覆盖它
答案 2 :(得分:0)
尝试这个
1)你提到所有var名称为$ Reference,已更改
2)查询不正确plz研究如何写查询..
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>";
?>