如何解决这个问题C'/ mysql [英] How do is fix this problem C' / mysql

问题描述

嘿 - 如果Value = Fahrzeug，我想只选择一行。



问题：什么都没发生

Hey - i want to select only the row if Value = Fahrzeug.

Problem: Nothing happend

MySqlConnection connection = new MySqlConnection("datasource=xxx.xxx.xxx;port=3306;username=xxx;password=xxx");
MySqlCommand command;
MySqlDataReader mdr;


connection.Open();
string selectQuery = "SELECT * FROM mysqlcshap.Alamierung WHERE Fahrzeug= '" + label2.Text + "'";
command = new MySqlCommand(selectQuery, connection);
mdr = command.ExecuteReader();
if (mdr.Read())
{

    label1.Text = "";
    label2.Text = mdr.GetString("Transportziel");
    label3.Text = mdr.GetString("Transportziel");
    label4.Text = mdr.GetString("Transportziel");
    label4.Text = mdr.GetString("Transportziel").ToString();

}
else
{
    label1.Text = label2.Text;
    label2.Text = "";
    label3.Text = "";
    label4.Text = "";
    label4.Text = "";




}

我尝试过：



试图改变一切。



我想这个是问题，但我无法处理它。

What I have tried:

Tried to Change everything.

I think this is the Problem but i can't handle it.

string selectQuery = "SELECT * FROM mysqlcshap.Alamierung WHERE Fahrzeug= '" + label2.Text + "'";

推荐答案

首先，不要这样做。永远不要连接字符串来构建SQL命令。它让您对意外或故意的SQL注入攻击持开放态度，这可能会破坏您的整个数据库。总是使用参数化查询。



连接字符串时会导致问题，因为SQL会收到如下命令：

First off, don't do it like that. Never concatenate strings to build a SQL command. It leaves you wide open to accidental or deliberate SQL Injection attack which can destroy your entire database. Always use Parameterized queries instead.

When you concatenate strings, you cause problems because SQL receives commands like:

SELECT * FROM MyTable WHERE StreetAddress = 'Baker's Wood'

就SQL而言，用户添加的引号会终止字符串，并且您会遇到问题。但情况可能更糟。如果我来并改为输入：x'; DROP TABLE MyTable; - 然后SQL收到一个非常不同的命令：

The quote the user added terminates the string as far as SQL is concerned and you get problems. But it could be worse. If I come along and type this instead: "x';DROP TABLE MyTable;--" Then SQL receives a very different command:

SELECT * FROM MyTable WHERE StreetAddress = 'x';DROP TABLE MyTable;--'

哪个SQL看作三个单独的命令：

Which SQL sees as three separate commands:

SELECT * FROM MyTable WHERE StreetAddress = 'x';

完全有效的SELECT

A perfectly valid SELECT

DROP TABLE MyTable;

完全有效的删除表格通讯和

A perfectly valid "delete the table" command

--'

其他一切都是评论。

所以它确实：选择任何匹配的行，从数据库中删除表，并忽略其他任何内容。



所以总是使用参数化查询！或者准备好经常从备份中恢复数据库。你定期进行备份，不是吗？



其次，equals运算符要求双方完全匹配：所以除非你的Fahrzeug列只包含你的字母寻找，你将得不到任何结果。

所以首先直接查看你的数据库，找出你在Fahrzeug专栏中有什么价值。然后检查你的标签，确保它完全符合你的想法。然后查看标签中的文字是否确实存在。



我们不能为您做到这一点 - 我们无权访问您的数据库，或者你的标签！

And everything else is a comment.
So it does: selects any matching rows, deletes the table from the DB, and ignores anything else.

So ALWAYS use parameterized queries! Or be prepared to restore your DB from backup frequently. You do take backups regularly, don't you?

Second, the equals operator requires an exact match on both sides: so unless your Fahrzeug column contains just the letters you looked for, you will get no results.
So start by looking at your database directly, and find out what values you do have in the Fahrzeug column. Then check your label and make sure it holds exactly what you think it does. Then see if the text in your label is actually there.

We can't do that for you - we don't have any access to your DB, or to your label!

Quote：

问题：什么都没发生

如果 label2 的值与任何值都不匹配，这是正常的。

试试：

It is normal if label2 have a value that do not match anything.
Try with:

string selectQuery = "SELECT * FROM mysqlcshap.Alamierung WHERE Fahrzeug= 'known value'";

或

or

string selectQuery = "SELECT * FROM mysqlcshap.Alamierung WHERE Fahrzeug= 'known beginning%'";

string selectQuery = "SELECT * FROM mysqlcshap.Alamierung WHERE Fahrzeug= '" + label2.Text + "'";

不是你问题的解决方案，而是你遇到的另一个问题。

永远不要建立一个通过连接字符串进行SQL查询。迟早，您将使用用户输入来执行此操作，这会打开一个名为SQL注入的漏洞，这对您的数据库很容易并且容易出错。

名称中的单引号你的程序崩溃。如果用户输入像Brian O'Conner这样的名称可能会使您的应用程序崩溃，那么这是一个SQL注入漏洞，崩溃是最少的问题，恶意用户输入，并且它被提升为具有所有凭据的SQL命令。

SQL注入 - 维基百科 [ ^ ]

SQL注入 [ ^ ]

按示例进行SQL注入攻击 [ ^ ]

PHP：SQL注入 - 手册 [ ^ ]

SQL注入预防备忘单 - OWASP [ ^ ]

我该怎么办？解释没有技术术语的SQL注入？ - 信息安全堆栈交换 [ ^ ]

Not a solution to your question, but another problem you have.
Never build an SQL query by concatenating strings. Sooner or later, you will do it with user inputs, and this opens door to a vulnerability named "SQL injection", it is dangerous for your database and error prone.
A single quote in a name and your program crash. If a user input a name like "Brian O'Conner" can crash your app, it is an SQL injection vulnerability, and the crash is the least of the problems, a malicious user input and it is promoted to SQL commands with all credentials.
SQL injection - Wikipedia[^]
SQL Injection[^]
SQL Injection Attacks by Example[^]
PHP: SQL Injection - Manual[^]
SQL Injection Prevention Cheat Sheet - OWASP[^]
How can I explain SQL injection without technical jargon? - Information Security Stack Exchange[^]

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