Question

我正在尝试创建一个查询，将一个现有表中的数据输入到另一个现有表中。两者之间没有共同的id字段。

我有以下现有表t1

class EasyFile{
    std::string fileContent;
    char* filePath;
public:
    EasyFile(char* filePath){
       this->filePath = filePath;
       std::fstream file(filePath);
       getline(file,fileContent);
    }
    char* getFilePath(){
        return filePath;
    }
    std::string getFileContent(){
        return fileContent;
    }
    void setContent(std::string content,char* filePath){
        std::ofstream file(filePath);
        file<<content;
    }
    void operator=(EasyFile f);
 };
void EasyFile::operator=(EasyFile f){
   this->setContent(f.getFileContent(),f.getFilePath());
}

int main(int argc,char** argv)
 {
   EasyFile efile1(argv[1]);
   EasyFile efile2(argv[2]);
   efile2 = efile1;
   std::string output="";
   std::ifstream file(argv[2]);
   std::getline(file,output);
   std::cout<<output;
   return 0;
}

有大量用户和（大约100个标准中）6个标准我有兴趣插入以下现有表格

    ----------+------+-------+
    |user | criteria | record|
    ----------+------+-------+   
    | 1    | 11      | K     |
    ----------+------+-------+
    | 1    | 12      | L     |
    ----------+------+-------+
    | 1    | 13      | M     |
    ----------+------+-------+
    | 1    | 16      | P     |
    ----------+------+-------+
    | 1    | 18      | R     |
    ----------+------+-------+
    | 1    | 20      | T     |
    ----------+------+-------+
    | 2    | 11      | K     |
    ----------+------+-------+
    | 2    | 12      | L     |
    ----------+------+-------+
    | 2    | 13      | M     |
    ----------+------+-------+
    | 2    | 16      | P     |
    ----------+------+-------+
    | 2    | 18      | R     |
    ----------+------+-------+
    | 2    | 20      | T     |
    ----------+------+-------+

t1用户与t2中用户的列相同。从t1为t1用户插入的数据必须匹配t2中已存在的同一用户的数据。

我的最终查询就是这样（这不起作用）

 table t2

 +----------+----------+----------+----------+----------+----------+
 | Label u  | Label v  | Label w  | Label x  | Label y  | Label z |
 +----------+----------+----------+----------+----------+----------+
 | record K | record L | record M | record P | record R  | record T|
 | record K | record L | record M | record P | record R  | record T|
 +----------+----------+----------+----------+----------+----------+

 where Criteria number 11 = Label u
       Criteria number 12 = Label v
       Criteria number 13 = Label w
       Criteria number 16 = Label x
       Criteria number 18 = Label y
       Criteria number 20 = Label z

 Note* Line 1 in t2 corresponds to user 1
       Line 2 in t2 corresponds to user 2

       There is no "user" column in t2 for the fields "user 1, user 2"

       t2 already contains data in other columns

Answer 1

这是您的数据透视表。如果您不想拥有用户ID，可以删除2.行

SELECT
  user,
  GROUP_CONCAT(IF (criteria = 11,record,NULL)) AS Label_u,
  GROUP_CONCAT(IF (criteria = 12,record,NULL)) AS Label_v,
  GROUP_CONCAT(IF (criteria = 13,record,NULL)) AS Label_w,
  GROUP_CONCAT(IF (criteria = 14,record,NULL)) AS Label_x,
  GROUP_CONCAT(IF (criteria = 15,record,NULL)) AS Label_y,
  GROUP_CONCAT(IF (criteria = 16,record,NULL)) AS Label_z
FROM mytable
WHERE user IN (1,2)
GROUP BY USER
ORDER BY USER;

结果

+------+---------+---------+---------+---------+---------+---------+
| user | Label_u | Label_v | Label_w | Label_x | Label_y | Label_z |
+------+---------+---------+---------+---------+---------+---------+
|    1 | P1      | P1      | P1      | P1      | R1      | T1      |
|    2 | P2      | P2      | P2      | P2      | R2      | T2      |
+------+---------+---------+---------+---------+---------+---------+
2 rows in set (0.00 sec)

MariaDB >

Answer 2

为了设计这个查询，可以从tc和tr同时更新其他字段 - 这个答案适用于tc.a = tr.cid和tr.cid = t1.user

的地方

SELECT ' Label_u'
  ,' Label_u'
  ,' Label_v'
  ,' Label_w'
  ,' Label_x'
  ,' Label_y'
  ,' Label_z'
  ,(SELECT `record` FROM `t1` WHERE `user` = b.id AND `criteria` = 11)
  ,(SELECT `record` FROM `t1` WHERE `user` = b.id AND `criteria` = 12)
  ,(SELECT `record` FROM `t1` WHERE `user` = b.id AND `criteria` = 13)
  ,(SELECT `record` FROM `t1` WHERE `user` = b.id AND `criteria` = 16)
  ,(SELECT `record` FROM `t1` WHERE `user` = b.id AND `criteria` = 18)
  ,(SELECT `record` FROM `t1` WHERE `user` = b.id AND `criteria` = 20)
  FROM tc a
     , tr b
  WHERE a .id = b .cid 
    AND b .print_id IS NOT NULL ;

MySQL Pivot查询到现有表

2 个答案: