使用perl cgi更新db行

Question

我正在尝试使用输入到用户的一些新值来更新我的sql表。由于某种原因，sql命令没有更新我的数据库。我得到了我验证的正确值。这是我的代码

#!/usr/bin/perl 
#This is going to be the user login check and will set a cookie

use DBI;
use CGI qw(:standard);

use strict;

#Connection error 
sub showErrorMsgAndExit {
    print header(), start_html(-title=>shift);
    print (shift);
    print end_html();
    exit;
}

#Connecting to the database
my $dbUsername = "root";
my $dbPassword = "password";

my $dsn = "DBI:mysql:f18final:localhost";
my $dbh = DBI->connect($dsn, $dbUsername, $dbPassword, {PrintError => 0});

#error checking
if(!$dbh) {
    print header(), start_html(-title=>"Error connecting to DB");
    print ("Unable to connec to the database");
    print end_html();
    exit;
}

print header;
print start_html(-title=>'Add Classes');

#Get the information the user entered
my $id = param('classid');
my $className = param('classname');
my $department = param('department');
my $classnum = param('classnum');
my $grade = param('grade');
my $credits = param('credit');
print "$id $className, $department, $classnum, $grade, $credits";
#first sql check to see if username is already taken
my $check = "UPDATE tblclasses(classname, department, classnum, grade, credits) VALUES (?, ?, ?, ?, ?) WHERE classID = $id";
my $sth = $dbh->prepare($check);
$sth->execute($className, $department, $classnum, $grade,$credits);
print "<h1>Success</h1>";
print "<form action=http://localhost/cgi-bin/edit.pl method = 'post'>";
print "<input type = 'submit' name = 'submit' value = 'Update Another'>";
print "</form>";
print "<form action=http://localhost/cgi-bin/actions.pl method = 'post'>";
print "<input type = 'submit' name = 'submit' value = 'Back to actions'>";
print "</form>";


print end_html();
exit;

当我尝试在mysql workbench中运行sql命令时，它成功更新了该行。我的问题是什么？

Answer 1

SQL语句的语法有错误：

UPDATE tblclasses(classname, department, classnum, grade, credits) 
VALUES (?, ?, ?, ?, ?)
WHERE classID = $id

应写成：

UPDATE tblclasses
SET classname = ?, 
       department = ?,
       classnum = ?,
       grade = ?,
       credits = ?
WHERE classID = ?

见the mysql docs。

附注（由@Grinnz评论）：

• 你应该总是«使用严格»
• 你应该在你的数据库或语句句柄上将DBI属性«RaiseError»设置为1;因此所有DBI错误都会致命;禁用«RaiseError»和«PrintErrror»导致DBI既不会死于错误也不会报告错误，因此您必须手动检查每个DBI调用的返回码以确保它有效 - 请参阅the DBI docs
• 你应该将SQL语句中的所有变量绑定到void SQL注入（你没有绑定$ id，我在上面的查询中改了）

Answer 2

在不知道DBMS的情况下，我不能100％确定，但看起来好像你混合了insert和update命令的语法。更新的正确语法应该是：

UPDATE tblclasses
set
  classname = ?,
  department = ?,
  classum = ?,
  grade = ?,
  credits = ?
WHERE classID = $id

此外，对于它的价值，您还应该能够将$id变量作为参数传递，而不是对其进行插值。从理论上讲，这将对数据库更友好，因为它将编译一次并反复执行相同的SQL语句，只有不同的绑定变量值：

my $check = qq{
  UPDATE tblclasses
  set
    classname = ?,
    department = ?,
    classum = ?,
    grade = ?,
    credits = ?
  WHERE classID = ?
};

my $sth = $dbh->prepare($check);
$sth->execute($className, $department, $classnum, $grade,$credits, $id);