Question

我正在PHP中运行此SQL LEFT JOIN查询，但它显示了一个重要列，即 user_id 为null，即非null 。

MYSQL查询

SELECT
      a.*,
      b.*,
      c.*,
      d.*
  FROM ts_users_skills a
  LEFT JOIN ts_students_log b
      ON a.`user_id` = b.`user_id`
  LEFT JOIN ts_students_info c
      ON a.`user_id` = c.`user_id`
  LEFT JOIN `geo_records` d
      ON a.`user_id` = d.`user_id`
  WHERE (a.`skillsTeach` = '$skill_1' OR a.`skillsTeach` = '$skill_2'
      OR a.`skillsTeach` = '$skill_3')
  GROUP BY a.`user_id`;

JSON输出

    {
    "status": "success",
    "nearby_teachers": [
        {
            "user_id": null,
            "skillsTeach": "PHP",
            "email": "praveenkumarkp666@gmail.com",
            "country_code": "91",
            "username": "praveenkum",
            "name": "Procusa Founder",
            "gender": "M",
            "birthday": "1997-02-25",
            "location": "Bhubaneswar"
        },
        {
            "user_id": null,
            "skillsTeach": "marketing",
            "email": "ashok@procusa.in",
            "country_code": "91",
            "username": "ashok",
            "name": "Ashok the founder 2",
            "birthday": "1993-11-18",
            "location": "Bhubaneswar"
        },
        {
            "user_id": null,
            "skillsTeach": "html",
            "email": "talspo@gmail.com",
            "country_code": "91",
            "username": "talspo",
            "name": "Procusa Founder",
            "gender": "M",
            "birthday": "1997-01-01",
            "location": "Bhubaneswar"

        }
    ]
}

谢谢。

Answer 1

您的代码存在的问题是它返回了几列名为user_id的列。来自LEFT JOIN的某些值可能会以NULL的形式出现。

当序列化为键唯一的JSON时，很可能已选择«错误»字段，从而产生了NULL值。

要解决此问题，您想精确列出要从每个表中检索的字段，而不要使用*。当然，user_id应该来自未经LEFT JOIN编辑的表：

SELECT
  a.user_id, a.field1, a.field2
  b.field11, b.field12
  c.field21, c.field22, c.field23
  d.field31
...

详细信息

您正在使用*选择字段：

SELECT
  a.*,
  b.*,
  c.*,
  d.*
...

但是您的查询显示您有多个名为user_id的字段：

...
LEFT JOIN ts_students_log b
  ON a.`user_id` = b.`user_id`
LEFT JOIN ts_students_info c
  ON a.`user_id` = c.`user_id`
LEFT JOIN `geo_records` d
  ON a.`user_id` = d.`user_id`
 ...

Answer 2

您在不同的表中有几列user_id。在您的JSON中，只有一个user_id字段，不清楚从哪个表开始。

很明显，此时数据正在丢失，JSON包含来自数据集中的user_id字段，而没有填充user_id字段。

尝试使用与user_id不同的表中的ts_users_skills a。

MYSQL LEFT JOIN未显示所有数据

2 个答案: