php - 您可以将资源传递给 PHP 中的查询参数吗？

Question

所以我有一个声明：

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

然后我可以将构造的资源应用于另一个查询吗？

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

我正在尝试做类似的事情

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

Answer 1

不，你不能这样做。

但是你可以结合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，您可以使用联接来代替子查询。

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