如何组合Select语句？

Question

有没有办法在2个Select语句之间应用逻辑运算符->或？

我对满足上述或条件的其他解决方案持开放态度。

如果第一个选择中没有数据，则转到下一个选择

示例

SELECT column1, column2
  FROM table-name1 JOIN table-name2 
    ON column-name1 = column-name2
 WHERE //all condition

或

SELECT column3, column4
  FROM table-name3 JOIN table-name4 
    ON column-name3 = column-name4
 WHERE //all condition

条件1/查询1

async getData() {
        try {
            const data = await this._conn.query(`

select first_name.value_name,quiz_table.answer, windows,player,first_name.value_id,country_place,current_name, pet_name, marker, relations
from schema_name.plugin,schema_name.quiz_table,schema_name.first_name, schema_name.value_version, schema_name.relationss
where plugin.answer= quiz_table.answer
and quiz_table.windows=first_name.value_id
and marker is not null
and schema_name.value_version.value_id= schema_name.first_name.value_id
and schema_name.value_version.caste= schema_name.first_name.caste
and schema_name.value_version.value_name= schema_name.first_name.value_name
and schema_name.value_version.version_number= schema_name.first_name.version_number
and schema_name.relationss.value_id= schema_name.first_name.value_id
and schema_name.relationss.caste= schema_name.first_name.caste
and schema_name.relationss.value_name= schema_name.first_name.value_name
and schema_name.relationss.version_number= schema_name.first_name.version_number
and  schema_name.quiz_table.windows= schema_name.first_name.value_id
and in_process='N'
}

或

条件2/查询2-显示数据

select schema_name.relationss."relations", schema_name.quiz_table."answer", schema_name.quiz_table."windows", schema_name.quiz_table."in_process", schema_name.quiz_table."object_name", schema_name.quiz_table."processed_date", schema_name.quiz_table."player", schema_name.quiz_table."country_place", schema_name.tools."mesh_scope_note", schema_name.plugin."current_name", schema_name.plugin."pet_name"
            from schema_name.quiz_table, schema_name.tools, schema_name.plugin, schema_name.relationss, schema_name.value_version 
            where (in_process = 'N' 
              and schema_name.quiz_table."windows" = schema_name.tools."value_id"
              and schema_name.quiz_table."player" = schema_name.tools."language"
              and schema_name.quiz_table."answer" = schema_name.plugin."answer"
              and schema_name.relationss."language" = schema_name.quiz_table."player"
              and schema_name.relationss."language" = schema_name.tools."language"
              and schema_name.relationss."caste" = schema_name.tools."caste"
              and schema_name.relationss."value_name" = schema_name.tools."value_name"
              and schema_name.relationss."version_number" = schema_name.tools."version_number"
              and schema_name.relationss."value_id" = schema_name.tools."value_id"
              and schema_name.value_version."value_id" = schema_name.tools."value_id"
              and schema_name.value_version."version_number" = schema_name.tools."version_number"
              and schema_name.value_version."caste" = schema_name.tools."caste"
)

注意-在where子句中，两个Select具有不同的条件

Answer 1

一个简单的方法是union all和exists--假设两个选择返回相同的列：

with cte as (
      select . . .
     )
select *
from cte
union all
select *
from t2
where not exists (select 1 from cte);