Question

我有两个表table1和table2。

table1 - tID,title,description

table2 - sID,tID,title,description

table1（t1）始终包含值。在table2（t2）中，如果存在一行，则tID（外键）将始终与t1（主键）中的tID匹配。

我想创建一个select语句，我在其中选择t1中的所有列，并创建一个带有true或false值的附加列“ex”，具体取决于t2中是否有记录，其中tID匹配... < / p>

Select t1.*, (some code) as ex from table1 t1,table2 t2;

明白了吗？例如，如果Select t1.*, t2.* from table1 t1, table2 t2 where t1.tID = t2.tID返回任何行，ex同样应该是真的...我该怎么做？

编辑：没有列可以为空

编辑：忘记提及，我希望select语句只匹配我指定的sID = someIdValue ....

Answer 1

使用<html> <head><title>Add info...</title></head> <body> <form action="tfs.php" method="post"> <table> <tr> <td>First Name :</td> <td><input type="text" name="fsname" /></td> </tr> <td>Last Name :</td> <td><input type="text" name="lsname" /></td> </tr> <tr> <tr> <td>rent on :</td> <td><input type="text" name="rent_on" /></td> </tr> <tr> <td>room status :</td> <td><input type="text" name="status" /></td> </tr> <td>city :</td> <td><input type="text" name="city" /></td> </tr> <tr> <td>area :</td> <td><input type="text" name="area" /></td> </tr> <tr> <td>college :</td> <td><input type="text" name="college" /></td> </tr> <tr> <td>room for :</td> <td><input type="text" name="rent_for" /></td> </tr> <tr> <td>quantity :</td> <td><input type="text" name="quantity" /></td> </tr> <tr> <td>bed status :</td> <td><input type="text" name="bed" /></td> </tr> <tr> <td>study table :</td> <td><input type="text" name="s_table" /></td> </tr> <tr> <td>wifi :</td> <td><input type="text" name="wifi" /></td> </tr> <tr> <td>gizer :</td> <td><input type="text" name="gizer" /></td> </tr> <tr> <td>Contact :</td> <td><input type="text" name="contact" /></td> </tr> <tr> <td>password :</td> <td><input type="text" name="password" /></td> </tr> <tr> <td>mail id :</td> <td><input type="text" name="mail_id" /></td> </tr> <tr> <td>Address :</td> <td><input type="text" name="address" /></td> </tr> <tr> <td>postal code :</td> <td><input type="text" name="pcode" /></td> </tr> <tr> </table> <input type="submit" value="Submit Info" /> </form> </body> </html> the connection between database is successful but some values like pcode, geo, rent shows doesn't have default values. The error is as shown in image.（外部联接），然后使用LEFT JOIN表达式检查tID中是否存在table2

IS NULL

demo

编辑：＆＃34;如果我只想要sID匹配某些值的值，该怎么办？＆＃34;

Select t1.*, t2.tid is not null as ex 
from table1 t1
left join table2 t2 on t1.tid = t2.tid

Answer 2

试试这个

select *, coalesce(tID = true, false) = true as is_exists from table1 t1
left join table2 t2 using (tid)

Answer 3

如果这对你有用，请尝试下面的一个。

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="appName_list">

  <div class="appName_main" data-type="header" data-id="1">
    Header Data
  </div>
  <div class="appName_main" data-type="header" data-id="2">
    Header Data Two
  </div>

  <div class="appName_main" data-type="footer" data-id="3">
    Footer Data
  </div>
  <div class="appName_main" data-type="header" data-id="4">
    Footer Data Two
  </div>

  <div class="appName_main" data-type="offers" data-id="5">
    Offers Data
  </div>
  <div class="appName_main" data-type="offers" data-id="6">
    Offers Data Two
  </div>
</div>

<div class="Menu_list">

  <div class="appNameSelect" data-type="header" data-id="1">
    Header One
  </div>
  <div class="appNameSelect" data-type="header" data-id="2">
    Header Two
  </div>
  <div class="appNameSelect" data-type="footer" data-id="3">
    Footer One
  </div>
  <div class="appNameSelect" data-type="footer" data-id="4">
    Footer Two
  </div>
  <div class="appNameSelect" data-type="offers" data-id="5">
    Offers One
  </div>
  <div class="appNameSelect" data-type="offers" data-id="6">
    Offers Two
  </div>

</div>

<div id="appendTo">


</div>

输出

 declare @T1 as Table 
(
tId int,
title varchar(25),
description varchar(50)
)

declare @T2 as Table 
(
sid int,
tId int,
title varchar(25),
description varchar(50)
)
insert into @t1 (tId,title,description) values(1,'abc','test')
insert into @t1 (tId,title,description) values(2,'abc','test')
insert into @t1 (tId,title,description) values(3,'abc','test')
insert into @t1 (tId,title,description) values(4,'abc','test')
insert into @t1 (tId,title,description) values(5,'abc','test')

insert into @t2 (tId,title,description) values(10,'abc','test')
insert into @t2 (tId,title,description) values(2,'abc','test')
insert into @t2 (tId,title,description) values(3,'abc','test')
insert into @t2 (tId,title,description) values(14,'abc','test')
insert into @t2 (tId,title,description) values(5,'abc','test')

Select distinct t1.*, 
   case when t2.tid is null then 'False' else 'True' end as ex from @T1 t1    
   left join @T2 t2 
   on t1.tId = t2.tId

Answer 4

EXISTS(subquery)产生一个可以直接使用的布尔值：

SELECT t1.*
        , EXISTS(SELECT13 FROM table2 t2 WHERE t2.tID = t1.tID) AS does_exist
FROM table1 t1
        ;

这将产生所有来自t1的行，加上一个标志，指示t2中存在一个或多个相应的行。
如果仅想要t2中匹配行的t1记录，请将exists()移至where子句：

SELECT t1.*
FROM table1 t1
WHERE EXISTS(SELECT13 FROM table2 t2 WHERE t2.tID = t1.tID) 
        ;

如果存在行，如何将结果作为列值返回？

4 个答案: