Question

我正在编写一个脚本来在错误跟踪系统中添加错误报告。单击提交按钮后，弹出了SQL语法错误对话框。

这是我的编码

public partial class AddBugForm : Form
{
    public AddBugForm()
    {
        InitializeComponent();
       Fillcombo();
       Fillcombo1();
       Fillcombo2();
    }

    void Fillcombo()
    {
        string constring = "datasource = localhost; username = root; password = ";
        string Query = "select * from bug.type";
        MySqlConnection conDataBase = new MySqlConnection(constring);
        MySqlCommand cmdDataBase = new MySqlCommand(Query, conDataBase);
       MySqlDataReader myReader;
        try
        {
            conDataBase.Open();
            myReader = cmdDataBase.ExecuteReader();

            while (myReader.Read())
            {
                string type = myReader.GetString("Type_of_bug");
                comboBox1.Items.Add(type);

            }
        }
        catch (Exception ex)
        {
            MessageBox.Show(ex.Message);
        }

     }

      void Fillcombo1()
       {
           string constring1 = "datasource = localhost; username = root; password = ";
           string Query1 = "select * from bug.severity";
           MySqlConnection conDataBase1 = new MySqlConnection(constring1);
           MySqlCommand cmdDataBase1 = new MySqlCommand(Query1, conDataBase1);
           MySqlDataReader myReader;
           try
           {
               conDataBase1.Open();
               myReader = cmdDataBase1.ExecuteReader();

               while (myReader.Read())
               {

                   string severity = myReader.GetString("severity");
                   severity_combo.Items.Add(severity);

               }
           }
           catch (Exception ex)
           {
               MessageBox.Show(ex.Message);
           }

       }

    void Fillcombo2()
    {
        string constring2 = "datasource = localhost; username = root; password = ";
        string Query2 = "select * from bug.priority";
        MySqlConnection conDataBase2 = new MySqlConnection(constring2);
        MySqlCommand cmdDataBase2 = new MySqlCommand(Query2, conDataBase2);
        MySqlDataReader myReader;
        try
        {
            conDataBase2.Open();
            myReader = cmdDataBase2.ExecuteReader();

            while (myReader.Read())
            {

                string priority = myReader.GetString("priority");
                priority_combo.Items.Add(priority);

            }
        }
        catch (Exception ex)
        {
            MessageBox.Show(ex.Message);
        }

    }

    private void submit_button_Click(object sender, EventArgs e)
    {
        string constring = "datasource=localhost;username=root;password=";
        string Query = "INSERT INTO 'bug.bug' (Bug_ID, title, Type_of_bug, software, software_version, description, step_to_reproduction, severity, priority, symptom) values('" + this.bugid_txt.Text+"', '" + this.title_txt.Text + "','" + this.comboBox1.Text + "','" + this.software_txt.Text + "','" + this.software_version_txt.Text + "','" + this.description_txt.Text + "','" + this.step_to_reproduction_txt.Text + "','" + this.severity_combo.Text + "','" + this.priority_combo.Text + "','" + this.symptom_txt.Text + "');";

        MySqlConnection conDataBase = new MySqlConnection(constring);
        MySqlCommand cmdDataBase = new MySqlCommand(Query, conDataBase);
        MySqlDataReader myReader;
        try
        {
            conDataBase.Open();
            myReader = cmdDataBase.ExecuteReader();
            MessageBox.Show("Saved");
            while(myReader.Read())
            {

            }
        }catch(Exception ex)
        {
            MessageBox.Show(ex.Message);
        }
    }
}

}

请帮帮我:(（（

Answer 1

我在INSERT查询

中看到了两个语法错误上下文问题

首先，INSERT INTO 'bug.bug';删除那些单引号，否则它是一个文字值而不是表名。它应该是INSERT INTO bug.bug

其次，从查询语句的最后一个中删除分号

.... + this.symptom_txt.Text + "');";
                                  ^.... this semicolon

Answer 2

将此INSERT INTO 'bug.bug'替换为

INSERT INTO `bug.bug`

你的表名是字符串，而mysql引擎没有看到表。

Answer 3

您获得的语法错误是什么？

关于Insert语句的几点意见。

您不应该通过组合值字符串来构建SQL命令字符串，这可能会产生SQL注入问题并容易导致语法错误。相反，你应该使用参数。参数也使语法更简单。
您应该使用ExecuteNonQuery命令而不是Reader，因为Insert语句不读取任何数据

更新的语句（只有两个值用于使其更小）：

string Query = "INSERT INTO bug.bug (Bug_ID, title) values (@id, @title)"

    MySqlConnection conDataBase = new MySqlConnection (constring);
    MySqlCommand cmdDataBase = new MySqlCommand (Query, conDataBase);
    cmdDataBase.Parameters.AddWithValue ("@id", bugid_txt.Text)
    cmdDataBase.Parameters.AddWithValue ("@title", title_txt.Text)
    conDataBase.Open();
    cmdDataBase.ExecuteNonQuerty ();
    MessageBox.Show("Saved");

使用参数可能会解决您的语法错误。

在c＃中出现SQL语法错误

3 个答案: