Php 如何使用YEAR函数从数据库中读取数组
这是选择:Php 如何使用YEAR函数从数据库中读取数组,php,mysql,Php,Mysql,这是选择: select distinct YEAR(ovl_dat) from db; 我得到的数组如下所示: Array ( [0] => stdClass Object ( [YEAR(ovl_dat)] => 1995 ) [1] => stdClass Object ( [YEAR(ovl_dat)] => 1957 ) [2] => stdClass Object ( [YEAR(ovl_dat)] => 1994 ) [3] =>
select distinct YEAR(ovl_dat) from db;
我得到的数组如下所示:
Array (
[0] => stdClass Object ( [YEAR(ovl_dat)] => 1995 )
[1] => stdClass Object ( [YEAR(ovl_dat)] => 1957 )
[2] => stdClass Object ( [YEAR(ovl_dat)] => 1994 )
[3] => stdClass Object ( [YEAR(ovl_dat)] => 1982 )
[4] => stdClass Object ( [YEAR(ovl_dat)] => 1997 ) )
这是将数组读入选项(joomla)的代码
我在阅读$enkeljaar->YEAR(ovl_dat)时遇到问题。有人能告诉我怎么做吗
关于Jan您应该为列名称和
select distinct YEAR(ovl_dat) as my_year from db;
参考别名获取值,例如假设结果为$row中的returne
$row['my_year'];
您应该为列名称和名称使用别名
select distinct YEAR(ovl_dat) as my_year from db;
参考别名获取值,例如假设结果为$row中的returne
$row['my_year'];
获取/设置具有任意名称的对象属性的语法如下:
$foo = (object)null;
$foo->{'[YEAR(ovl_dat)]'} = 2018;
$foo->{'One
Two->Three'} = 'Hi';
var_dump($foo->{'[YEAR(ovl_dat)]'}, $foo->{'One
Two->Three'});
()
但是,除非您参加了代码混淆比赛,否则您可能需要指定一个正确的名称:
select distinct YEAR(ovl_dat) as distinct_year from db;
获取/设置具有任意名称的对象属性的语法如下:
$foo = (object)null;
$foo->{'[YEAR(ovl_dat)]'} = 2018;
$foo->{'One
Two->Three'} = 'Hi';
var_dump($foo->{'[YEAR(ovl_dat)]'}, $foo->{'One
Two->Three'});
()
但是,除非您参加了代码混淆比赛,否则您可能需要指定一个正确的名称:
select distinct YEAR(ovl_dat) as distinct_year from db;
你的代码不够。。因此,您使用查询结果的其余代码您的代码还不够。。接下来的代码将使用查询结果