如何在 typeorm querybuilder 中显示原始 SQL [英] How to show raw SQL in typeorm querybuilder

问题描述

我开发了 typeorm querybuilder .为了调试的目的，我想展示一下sql.

I developed typeorm querybuilder . For the aiming of debugging,I'd like to show sql.

我测试了 printSql() 方法.但它没有显示任何 SQL.

I tested printSql() method. but it didn't show any SQL.

      const Result = await this.attendanceRepository
      .createQueryBuilder("attendance")
      .innerJoin("attendance.child", "child")
      .select(["attendance.childId","child.class","CONCAT(child.firstName, child.lastName)"])
      .where("attendance.id= :id", { id: id})
      .printSql()
      .getOne()
      console.log(Result);

它返回了以下内容.

Attendance { childId: 4, child: Child { class: 'S' } }

我想要的结果是得到SQL

有什么不对的地方吗?有没有什么好办法获取sql?

Are there any wrong point ? Are there any good way to get sql ?

如果有人有意见，请告诉我.

If someone has opinion, please let me know.

谢谢

推荐答案

.getQuery() 或 .getSql()

const sql1 = await this.attendanceRepository
      .createQueryBuilder("attendance")
      .innerJoin("attendance.child", "child")
      .select(["attendance.childId","child.class","CONCAT(child.firstName, child.lastName)"])
      .where("attendance.id= :id", { id: id})
      .getQuery();
      console.log(sql1);

---

const sql2 = await this.attendanceRepository
      .createQueryBuilder("attendance")
      .innerJoin("attendance.child", "child")
      .select(["attendance.childId","child.class","CONCAT(child.firstName, child.lastName)"])
      .where("attendance.id= :id", { id: id})
      .getSql();
      console.log(sql2);

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