您可以將資源傳遞給PHP中查詢的參數嗎？

Question

所以我有一個說法：

$r = mysql_query("Select type from boats where type like '%speed%'"); 

可我那麼構建資源應用到另一個查詢？

$r2 = mysql_query("select * from assets where type in ".$r); 

我試圖做同樣的事情到

select * from assets where type in (select type from boats where type like '%speed%') 

Answer 1

不，你不能做到這一點。

但你可以結合sql。

$sql = "Select type from boats where type like '%speed%'"; 
$r = mysql_query($sql); 
$r2 = mysql_query("select * from assets where type in ($sql)"); 

或者，您可以使用聯接而不是子查詢。

$r2 = mysql_query("select a.* from assets a 
        join boats b on a.type = b.type and b.type like '%speed%'");