Question

SqlConnection sqlConnection = new SqlConnection(sqlConnectionString);
SqlCommand command = new SqlCommand();

command.CommandText = 
   "INSERT INTO [Users] 
    VALUES (Username=@username, Firstname=@firstname
          , Lastname=@lastname, ProfilePic=NULL, Bio=@bio, Email=@email
          , EmailIsVerified=@emailverified, Hash=@hash, CompanyID=@companyid)";
command.Parameters.AddWithValue("@username", m_username);
command.Parameters.AddWithValue("@firstname", m_firstname);
command.Parameters.AddWithValue("@lastname", m_lastname);
command.Parameters.AddWithValue("@bio", m_bio);
command.Parameters.AddWithValue("@email", m_email);
command.Parameters.AddWithValue("@emailverified", (m_emailIsVerified ? "yes" : "no"));
command.Parameters.AddWithValue("@hash", m_hash);
command.Parameters.AddWithValue("@companyid", m_companyID);
command.CommandType = CommandType.Text;
command.Connection = sqlConnection;

sqlConnection.Open();

command.ExecuteNonQuery();

sqlConnection.Close();

根据上述代码,我发现“Syntax误差接近=”错误。我做了哪些错误?

Answer 1

INSERT在发言中没有使用用户名称=用户名称等,因为这一名称无效。相反,它希望:

command.CommandText = "INSERT INTO [Users] VALUES (@username, @firstname, @lastname, NULL, @bio, @email, @emailverified, @hash, @companyid)";

假设这些数值与数据库各栏的顺序相同,否则,你也需要具体说明各栏。

Answer 2

您需要修改INSERT的声明。

INSERT INTO [Users] (UserName, FirstName, ...)
SELECT @UserName, @firstname, ...

http://msdn.microsoft.com/en-us/library/ms174335.aspx” rel=“nofollow”

明确列出您的栏目<代码>被认为是较好的做法。 INSERT INTO。如果修改了你的表格的栏目,则防止出现任何(可能难以解决的)问题。这可能导致以下因素:(将价值归入无效类别)或将数值列入意外栏目(重排栏,但具有相同类型,因此,插入将取得成功)。

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