结果1来自表1:
select types from table1 where id = 123456;
结果:
934046, 934049
然后,将结果1作为table2的参数:
select GROUP_CONCAT(name) from table2 where id in (
select types from table1 where id = 123456
);
但是结果1是一个字符串值, 这样的结果:
NULL
事实上我想得到这个:
select GROUP_CONCAT(name) from table2 where id in ('934046', '934049');
or
select GROUP_CONCAT(name) from table2 where id in (934046,934049);
但它可能仍然像这样:
select GROUP_CONCAT(name) from table2 where id in ('934046,934049');
1.
select REPLACE(types,',','\',\'') types from table1 where id=123456
1.1
result: 934046',' 934049
1.2
select GROUP_CONCAT(name) from table2 where ids in (
select REPLACE(types,',','\',\'') types from table1 where id=123456
);
2.
select CONCAT('\'',REPLACE(types ,',','\',\''),'\'') from table1 where
id=123456
2.1
reuslt: '934046',' 934049'
2.2
select GROUP_CONCAT(name) from table2 where ids in (
select CONCAT('\'',REPLACE(types ,',','\',\''),'\'') from table1 where
id=123456);
该字符串值不能作为参数,我该怎么做才能以正确的方式获取ID?
答案 0 :(得分:0)
$query = "select types from table1 where id = 123456";
$result = mysql_query($query) or die(mysql_error());
$array = array();
while($row = mysql_fetch_array($result))
{
$array[] = $row['types'];
}
$array = implode(",",$array);
$query2 = "select GROUP_CONCAT(name) from table2 where id in ($array)";
我用其他查询创建了代码。
答案 1 :(得分:0)
select GROUP_CONCAT(t.name) from table2 t
where exists(
select 1 from table1
where id = 123456
and LOCATE(concat(', ', t.id, ',') , concat(', ', types, ',') )>0
)