mysql语法错误？答案

【问题标题】：error in mysql syntax?mysql语法错误？
【发布时间】：2011-07-21 05:27:02
【问题描述】：

我的 sql 语法有问题我收到错误：

ERROR [HY000] [MySQL][ODBC 3.51 Driver][mysqld-5.5.9]Not unique table/alias: 'WallPosting'

代码：

{

    string theUserId = Session["UserID"].ToString();
    using (OdbcConnection cn = new OdbcConnection("Driver={MySQL ODBC 3.51 Driver}; Server=localhost; Database=gymwebsite2; User=x; Password=x;"))
    {
        cn.Open();
        using (OdbcCommand cmd = new OdbcCommand("SELECT WallPosting.Wallpostings FROM WallPosting LEFT JOIN WallPosting ON User.UserID = WallPosting.UserID WHERE User.UserID=" + theUserId + "", cn))
// problem in select statement

        using (OdbcDataReader reader = cmd.ExecuteReader())
        {
            var divHtml = new System.Text.StringBuilder("<div id=mysqlcontent>");
            while (reader.Read())
            {
                divHtml.Append(String.Format("{0}", reader.GetString(0)));
            }

            divHtml.Append("</div>");
            test1.InnerHtml = divHtml.ToString();

        }
    }
}

我的sql脚本和表结构：

SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0;
SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0;
SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='TRADITIONAL';
CREATE SCHEMA IF NOT EXISTS `gymwebsite2` DEFAULT CHARACTER SET latin1 COLLATE latin1_swedish_ci ;
USE `gymwebsite2` ;

-- -----------------------------------------------------
-- Table `gymwebsite2`.`User`
-- -----------------------------------------------------
CREATE  TABLE IF NOT EXISTS `gymwebsite2`.`User` (
  `UserID` INT NOT NULL AUTO_INCREMENT ,
  `Email` VARCHAR(245) NULL ,
  `FirstName` VARCHAR(45) NULL ,
  `SecondName` VARCHAR(45) NULL ,
  `DOB` VARCHAR(45) NULL ,
  `Location` VARCHAR(45) NULL ,
  `Aboutme` VARCHAR(1045) NULL ,
  `username` VARCHAR(45) NULL ,
  `password` VARCHAR(45) NULL ,
  PRIMARY KEY (`UserID`) )
ENGINE = InnoDB;

-- -----------------------------------------------------
-- Table `gymwebsite2`.`WallPosting`
-- -----------------------------------------------------
CREATE  TABLE IF NOT EXISTS `gymwebsite2`.`WallPosting` (
  `idWallPosting` INT NOT NULL AUTO_INCREMENT ,
  `UserID` INT NOT NULL ,
  `Wallpostings` VARCHAR(2045) NULL ,
  PRIMARY KEY (`idWallPosting`) ,
  INDEX `fk_WallPosting_User` (`UserID` ASC) ,
  CONSTRAINT `fk_WallPosting_User`
    FOREIGN KEY (`UserID` )
    REFERENCES `gymwebsite2`.`User` (`UserID` )
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;

-- -----------------------------------------------------
-- Table `gymwebsite2`.`Pictures`
-- -----------------------------------------------------
CREATE  TABLE IF NOT EXISTS `gymwebsite2`.`Pictures` (
  `idPictures` INT NOT NULL AUTO_INCREMENT ,
  `UserID` INT NOT NULL ,
  `picturepath` VARCHAR(1045) NULL ,
  PRIMARY KEY (`idPictures`) ,
  INDEX `fk_Pictures_User1` (`UserID` ASC) ,
  CONSTRAINT `fk_Pictures_User1`
    FOREIGN KEY (`UserID` )
    REFERENCES `gymwebsite2`.`User` (`UserID` )
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB;

SET SQL_MODE=@OLD_SQL_MODE;
SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS;
SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS;

【问题讨论】：

我认为问题是左连接不知道如何选择它或者是否需要它？

标签： c# asp.net mysql sql html

【解决方案1】：

当您执行自连接（将表连接到自身）时，您需要为至少一个表实例指定别名，以便 SQL 在任何给定时间知道您正在谈论的实例。 ... LEFT JOIN WallPosting AS a ON ... 之类的东西，当您指的是连接表时，请使用 a.WallPosting。

【讨论】：

【解决方案2】：

不应该

"SELECT WallPosting.Wallpostings FROM WallPosting LEFT JOIN User ON User.UserID = WallPosting.UserID WHERE User.UserID=" + theUserId + ""

您将再次加入WallPosting 表（这需要您分配别名），但如果您要使用User 表中的列，则应该加入User 表。

【讨论】：

【解决方案3】：

更正你的查询

      SELECT WallPosting.Wallpostings FROM WallPosting LEFT JOIN USER ON User.UserID = WallPosting.UserID WHERE User.UserID=" + theUserId + "", cn

你所做的是

              WallPosting LEFT JOIN WALLPOSTING ON User.UserID = WallPosting.UserID

【讨论】：