Warning: file_get_contents(/data/phpspider/zhask/data//catemap/8/mysql/65.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Mysql 内部联接没有匹配的列_Mysql_Sql - Fatal编程技术网

Mysql 内部联接没有匹配的列

Mysql 内部联接没有匹配的列,mysql,sql,Mysql,Sql,我有下表,需要以下输出- doctor professor <--column names tom mary harry layla 你不想加入组织。您想要全部联合: select doctor as name, 'doctor' as occupation from t union all select professor as name, 'professor' as occupation from t; 你不想加入组织。您想要全部联合: select doctor as

我有下表,需要以下输出-

doctor professor <--column names
tom    mary
harry  layla

你不想加入
组织
。您想要
全部联合

select doctor as name, 'doctor' as occupation
from t
union all
select professor as name, 'professor' as occupation
from t;

你不想加入
组织
。您想要
全部联合

select doctor as name, 'doctor' as occupation
from t
union all
select professor as name, 'professor' as occupation
from t;

“医生作为名字”——医生不是列名..这怎么行?我遗漏了什么?看看你的问题。将医生和教授显示为列名称。此查询将生成您要求的结果。@user1050619。你的问题明确地说“医生-教授'医生作为名字'——医生不是列名..这怎么行?我遗漏了什么?看看你的问题。你将医生和教授作为列名显示。这个查询将产生你要求的结果。@user1050619…你的问题明确地说“医生-教授”
select
  doctor as name,
  'doctor' as occupation
from
  tb1
union all
select
  professor as name,
  'professor' as occupation
from
  tb1;