繁体   English   中英

PHP MySQL从同一结果行中将列提取为不同的数组

[英]PHP MySQL fetching columns as different arrays from the same results row

假设我有一个事件表,并且想获取某天的所有事件(例如2018年1月2日)

|   e1    |     d1     |   e2    |     d2     |   e3    |     d3     |
----------------------------------------------------------------------
| event A | 2018-01-01 | event B | 2018-01-02 | event C | 2018-01-02 |
| event D | 2018-01-01 | NULL    | NULL       | NULL    | NULL       |
| NULL    | NULL       | event E | 2018-01-02 | event F | 2018-01-03 |

问题是当我尝试获取这样的结果时:

"SELECT * FROM table WHERE d1='2018-01-02' OR d2='2018-01-02' OR d3='2018-01-02' "

我得到了整行结果(在本例中是所有三行)。 但是我想获得带有事件和日期的单独结果数组。 像这样:

Array
(
    [0] => Array
        (
            event => event B
            date => 2018-01-02
        )
    [1] => Array
        (
            event => event C
            date => 2018-01-02
        )
)

到目前为止,我以这种丑陋的方式进行了操作,执行了3个查询并将它们的结果合并在一起:

$Sql=$conn->prepare("SELECT e1 AS event, d1 AS date FROM table WHERE d1='2018-01-02'");
$Sql->execute();
$Arr1 = $Sql->fetchAll(PDO::FETCH_ASSOC);

$Sql=$conn->prepare("SELECT e2 AS event, d2 AS date FROM table WHERE d2='2018-01-02'");
$Sql->execute();
$Arr2 = $Sql->fetchAll(PDO::FETCH_ASSOC);

$Sql=$conn->prepare("SELECT e3 AS event, d3 AS date FROM table WHERE d3='2018-01-02'");
$Sql->execute();
$Arr3 = $Sql->fetchAll(PDO::FETCH_ASSOC);

$Arr=array();
foreach($Arr1 AS $a){
 $Arr[]=array($a['event'],$a['date']);
}
foreach($Arr2 AS $a){
 $Arr[]=array($a['event'],$a['date']);
}
foreach($Arr3 AS $a){
 $Arr[]=array($a['event'],$a['date']);
}

print_r($Arr);

但我确信应该有一些更适当的方法来做到这一点。 通过修改SQL或PHP。 提前致谢。

看看https://dev.mysql.com/doc/refman/5.7/en/union.html尝试这样的事情

$Sql=$conn->prepare("(SELECT e1 AS event, d1 AS date FROM table WHERE d1='2018-01-02') UNION (SELECT e2 AS event, d2 AS date FROM table WHERE d2='2018-01-02') UNION (SELECT e3 AS event, d3 AS date FROM table WHERE d3='2018-01-02') ");
$Sql->execute();
$Arr = $Sql->fetchAll(PDO::FETCH_ASSOC);

还请注意,日期是保留字,因此您可能需要用回勾号写出来……不过不确定,因为我只是不使用它们;

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM