Question

生成以下两个结果集：

1）。查询OCCUPATIONS中所有姓名的按字母顺序排列的列表，紧接着是每个职业的第一个字母作为括号（即：括在括号中）。例如：AnActorName（A），ADoctorName（D），AProfessorName（P）和ASingerName（S）。

2）。查询OCCUPATIONS中每个职业的发生次数。按升序对事件进行排序，并按以下格式输出：

There are total [occupation_count] [occupation]s.

表名：职业

总列数：两个=＆＃39;名称＆＃39;和＆＃39;职业＆＃39;，演示表如下所示：

Table: Occupations

示例输出：

Ashely(P)
Christeen(P)
Jane(A)
Jenny(D)
Julia(A)
Ketty(P)
Maria(A)
Meera(S)
Priya(S)
Samantha(D)
There are total 2 doctors.
There are total 2 singers.
There are total 3 actors.
There are total 3 professors.

我的方法：

(SELECT NAME, '(', SUBSTRING(OCCUPATION, 1, 1), ')' 
    FROM OCCUPATIONS ORDER BY NAME) 
UNION ALL 
(SELECT COUNT(*) FROM OCCUPATIONS GROUP BY OCCUPATION ORDER BY ASEC);

错误：

ERROR 1222 (21000) at line 1: 
    The used SELECT statements have a different number of columns

谢谢！

Answer 1

您忘记使用CONCAT功能粘贴所选数据

尝试这样的事情（也见sqlfiddle上的这个例子）：

(
  SELECT CONCAT(NAME, '(', SUBSTRING(OCCUPATION, 1, 1), ')') as THETEXT, '1' as SELECTNUMBER
    FROM OCCUPATIONS
)
UNION ALL 
(
  SELECT CONCAT('There are total ', COUNT(*),' ', OCCUPATION, (IF (COUNT(*) > 1, 's',''))) as THETEXT, '2' as SELECTNUMBER
   FROM OCCUPATIONS GROUP BY OCCUPATION
)
ORDER BY SELECTNUMBER ASC, THETEXT ASC;

Answer 2

这将起作用！

SELECT CONCAT(Name, 
              CONCAT('(', LEFT(Occupation,1),')'))  
FROM OCCUPATIONS
ORDER BY Name ASC;

SELECT 
    CONCAT("There are a total of", 
           " ",
           COUNT(occupation), 
           " ",
           LCASE(occupation),
           "s",
           ".")
                     AS stat
FROM OCCUPATIONS
GROUP By occupation
ORDER BY COUNT(occupation) ASC, occupation;

Answer 3

SELECT concat(NAME,concat("(",LEFT(occupation,1),")")) 
FROM OCCUPATIONS 
ORDER BY NAME ASC;

select CONCAT("There are a total of", " ",COUNT(occupation), " ",LCASE(occupation),"s",".")AS stat
from OCCUPATIONS
group by occupation
order by COUNT(occupation) ASC,occupation

Answer 4

致力于hackerrank

SELECT Name || '(' || SUBSTR(Occupation,1,1) || ')'-- AS THETEXT, '1' AS SELECTNUMBER
    FROM OCCUPATIONS
    order by Name;

    SELECT 'There are a total of' || ' ' || COUNT(*) || ' ' || lower(Occupation) || 's.'
    FROM OCCUPATIONS  GROUP BY OCCUPATION                            
    ORDER BY COUNT(Occupation), lower(Occupation);

Answer 5

    SELECT CONCAT(NAME ,'('||SUBSTR(occupation,1,1)||')')
      FROM occupations 
  ORDER BY NAME ASC;

    SELECT CONCAT('There are a total of ',COUNT (occupation)||' '||LOWER(occupation)||'s.')
      FROM occupations 
  GROUP BY occupation 
  ORDER BY COUNT(occupation),occupation ASC;

Answer 6

对于 Oracle SQL 特别试试这个：

SELECT NAME || '(' || SUBSTR(OCCUPATION, 1, 1) || ')' FROM OCCUPATIONS ORDER BY NAME; 

SELECT 'There are a total of ' || count(OCCUPATION) || ' ' ||  lower(OCCUPATION) || 's.'  FROM OCCUPATIONS  GROUP BY OCCUPATION ORDER BY count(OCCUPATION), OCCUPATION ASC;

Answer 7

试试这个代码..

-- for Query 01
SELECT CONCAT(Name, '(', LEFT(Occupation,1),')') 
FROM OCCUPATIONS 
ORDER BY Name;

-- for Query 02
SELECT CONCAT('There are a total of ', COUNT(Occupation), ' ', LOWER(Occupation), 's.') 
FROM OCCUPATIONS 
GROUP BY Occupation 
ORDER BY COUNT(Occupation), Occupation;

Answer 8

请按照此代码进行 Mysql。这对我最有效。我已经尝试并得到了正确的结果。

select concat(name,'(',left(occupation,1),')') from occupations
order by name;

select concat('There are a total of ',count,' ',lower(occupation),'s.') from (select occupation, count(*) as count from occupations group by occupation)a

order by count, occupation;

对于第一次查询，我在列占用中用括号和值的第一个字符连接了列“名称”。 later order by 是按升序排列它们（默认是升序）。
第二个查询，按计数排序（职业）对我不起作用，所以为了解决这个问题，我使用了嵌套查询。括号中的查询是提供我的职业列，分组时的职业计数。然后我用计数和职业名称连接字符串。稍后按升序按计数排序。此外，如果 2 个职业的数量相同，则也使用按职业的顺序将它们按升序排列。

谢谢

Answer 9

select concat(Name,'(',substring(Occupation,1,1),')') from OCCUPATIONS order by Name;

select 
concat('There are a total of ',count(*),' ',LCASE(Occupation),
       If(count(*)>1,'s.','')) 
       from OCCUPATIONS group by Occupation order by count(*) ASC
       ;

Answer 10

一个人可以利用MySQL中的UNION运算符来合并问题的每个部分。

注意：：要将列名组合为一列，可以使用CONCAT()函数。

SELECT DISTINCT CONCAT(name, '(', left(occupation, 1), ')') AS n
from OCCUPATIONS
union
select concat('There are a total of ', (select count(*) from OCCUPATIONS group by occupation having occupation = 'doctor'), ' doctors.')
union
select concat('There are a total of ', (select count(*) from OCCUPATIONS group by occupation having occupation = 'singer'), ' singers.')
union
select concat('There are a total of ', (select count(*) from OCCUPATIONS group by occupation having occupation = 'actor'), ' actors.')
union
select concat('There are a total of ', (select count(*) from OCCUPATIONS group by occupation having occupation = 'professor'), ' professors.')

order by n asc;

Answer 11

select Name+'('+substring(Occupation,1,1)+')' from occupations order by name asc;

select concat('There are a total of',' ', count(occupation),' ',Lower(occupation),'s','.') from Occupations group by Occupation order by count(occupation), occupation asc

Answer 12

  /* sort answer is here */ 
     ( select concat(name,'(',left(occupation,1),')') as ocps from 
      Occupations) union 
     (select distinct concat('There are a total of 
     ',count(occupation),' ',lower(occupation),'s.') from Occupations 
     group by occupation)order by ocps;

Answer 13

检查以下代码。希望这有助于其100％正常工作

select concat(Name,"(",left(Occupation,1),")")
  from occupations 
 order by Name;

select concat("There are a total of ",count(Occupation),' ',lower(occupation),"s.") 
  from Occupations
 group by occupation 
 order by count(Occupation)asc, Occupation;

Answer 14

select concat(name, '(',substring(occupation,1,1),')') from OCCUPATIONS order  by name;
select concat('There are a total of ',count(occupation),' ',lower(occupation),'s.') as 'sentence' from OCCUPATIONS group by occupation order by count(occupation), occupation,field(occupation,'doctor','singer','actor','professor');

Answer 15

select CONCAT(NAME, '(', left(OCCUPATION,1),')') from OCCUPATIONS ORDER BY NAME;

select CONCAT("There are a total of ", count(OCCUPATION), " ", lower(OCCUPATION),(if(count(OCCUPATION) >1,"s","")), ".") from OCCUPATIONS group by occupation order by count(OCCUPATION), OCCUPATION;

Answer 16

SELECT CONCAT(NAME,'(',LEFT(OCCUPATION,1),')')
FROM OCCUPATIONS ORDER BY NAME;

SELECT CONCAT("There are total of ",COUNT_OCCUPATION," ",OCCUPATION,".")
FROM
(SELECT COUNT(OCCUPATION) AS COUNT_OCCUPATION,OCCUPATION
FROM OCCUPATIONS
GROUP BY OCCUPATION
ORDER BY COUNT_OCCUPATION ASC,OCCUPATION ASC) AS OCP

Answer 17

这对我有用：

select CONCAT(name,'(',left(occupation,1),')') 
from OCCUPATIONS 
order by name;

select CONCAT('There are a total of',' ',count(occupation),' ',lower(occupation),'s.') 
from OCCUPATIONS 
group by occupation 
order by count(occupation), occupation;

Answer 18

SELECT * FROM (
    select distinct Name + '('+left(Occupation,1)+')'as a
    from OCCUPATIONS 
    union
    select  ('There are total ' +cast( count( Occupation) as nvarchar) + ' '+ lower(Occupation) +'s.') as b 
    from OCCUPATIONS
    GROUP BY OCCUPATION
    ) AS N
ORDER BY a

这适用于MS SQL Server。

Answer 19

union要求两个查询具有相同的列数，您可以在第二个查询中添加任何一些列，使它们像第一个像SELECT COUNT（*），＆＃39; col-1＆＃39;，＆＃39; col-2＆＃39;，＆＃39; col-3＆＃39;从

Answer 20

SELECT CONCAT(name, '(', SUBSTRING(occupation, 1, 1), ')') as occ from occupations ORDER BY name;

SELECT concat('There are a total of ', COUNT(*), ' ', LCASE(occupation),  's.') from occupations GROUP BY occupation ORDER BY COUNT(*), occupation;

这很有魅力。

Answer 21

(Select name||'('||substr(Occupation,0,1)||')'  
from OCCUPATIONS)
Union All
(Select 'There are a total of '||count(*)||' '||lower(Occupation)||'s.'  
from OCCUPATIONS group by Occupation) 
order by 1;

Answer 22

我刚试过hackerrank并且它有效，你不需要使用Union。

select concat(name,'(',upper(substring(occupation,1,1)),')') from occupations
order by name;

select concat("There are a total of",' ',count(occupation),' ',lower(occupation),'s',".")  from occupations
group by occupation
order by count(occupation) asc;

Answer 23

select concat(name,'(',left(occupation,1),')') from occupations order by name;


select concat('There are total of',' ',count(occupation),' ',lower(occupation),"s.") from occupations group by occupation order by count(occupation),occupation;

Answer 24

(
  SELECT CONCAT(NAME, '(', SUBSTRING(OCCUPATION, 1, 1), ')') as THETEXT
    FROM OCCUPATIONS
)
UNION ALL 
(
  SELECT CONCAT('There are total ', COUNT(*),' ', LOWER(OCCUPATION) ,(IF (COUNT(*) > 1, 's','')), '.') as THETEXT
   FROM OCCUPATIONS  group by  OCCUPATION
)
ORDER by THETEXT ASC;

Answer 25

//Just be careful for the format!

(select concat(name, '(',left(occupation,1),')') as text
from occupations
)
union all
(
   select concat(
       'There are total ', 
       count(*), ' ', lower(occupation), 
       case
        when COUNT(*) > 1 
        then 's'
       end
       ,
       '.'
    ) as text
    from occupations
    group by occupation
)
order by text asc;

Answer 26

GetParent()

Answer 27

select (name||'('||UPPER(substr(occupation,1,1))||')') from occupations order by name;

select ('There are a total of '||count(name)||' '||lower(occupation)||'s.') from occupations group by occupation order by count(name), occupation;

Answer 28

select Name + '('+ substring(OCCUPATION, 1,1) +')' from OCCUPATIONS 
union  
select 'There are total ' + cast(count(OCCUPATION) as varchar) +' '+ 
lower(OCCUPATION)+'s.' from OCCUPATIONS   
group by OCCUPATION

Answer 29

value before = 23
value in while loop = 23

Answer 30

SELECT * FROM (
    select distinct Name + '('+left(Occupation,1)+')'as a
    from OCCUPATIONS 
    union
    select  ('There are total ' +cast( count( Occupation) as nvarchar) + ' '+ lower(Occupation) +'s.') as b 
    from OCCUPATIONS
    GROUP BY OCCUPATION
    ) AS N
ORDER BY a

Answer 31

SELECT CONCAT (NAME, CONCAT ('(',LEFT(OCCUPATION,1),')'))FROM OCCUPATIONS ORDER BY NAME ASC;

SELECT CONCAT ('There are a total of ',count(OCCUPATION),' doctors.' )FROM OCCUPATIONS WHERE OCCUPATION='Doctor';

SELECT CONCAT ('There are a total of ',count(OCCUPATION),' singers.' )FROM OCCUPATIONS WHERE OCCUPATION='Singer';

SELECT CONCAT ('There are a total of ',count(OCCUPATION),' actors.' )FROM OCCUPATIONS WHERE OCCUPATION='Actor';

SELECT CONCAT ('There are a total of ',count(OCCUPATION),' professors.' )FROM OCCUPATIONS WHERE OCCUPATION='Professor';

使用UNION / UNION ALL和Group By的MySQL Query错误

31 个答案: