Mysqli 更好的方法来更改使用的列？

Question

所以我的代码需要有一个重量跟踪器，并且需要在一个数据点中保持至少 20 个重量。因为我不能用“？”动态改变它。该位置我有 20 个 if () else if 语句能够做同样的事情。只是想知道是否有更好的方法来做到这一点，或者这是唯一的方法吗？

代码开头：

function addWeight($conn, $location, $weight,$dt){
//select the two variables changing 
if($location == "Weight1"){
    $sql ="UPDATE userweight SET Weight1=? WHERE userId = ?;";
    $sql2 ="UPDATE userweight SET timee1=? WHERE userId = ?;";
}elseif($location == "Weight2"){
    $sql ="UPDATE userweight SET Weight2=? WHERE userId = ?;";
    $sql2 ="UPDATE userweight SET timee2=? WHERE userId = ?;";
}elseif($location == "Weight3"){
    $sql ="UPDATE userweight SET Weight3=? WHERE userId = ?;";
    $sql2 ="UPDATE userweight SET timee3=? WHERE userId = ?;";
}elseif($location == "Weight4"){
    $sql ="UPDATE userweight SET Weight4=? WHERE userId = ?;";
    $sql2 ="UPDATE userweight SET timee4=? WHERE userId = ?;";
}
.....

Answer 1

不要这样存储数据。这只是自找麻烦。而是规范化您的模式并以关系方式存储数据。

如果您受困于当前的设计，可以通过动态构建 SQL 来解决这个问题。请务必使用硬编码值列表验证列名是否正确。

function addWeight(mysqli $conn, $location, $weight, $dt)
{
    $columns = [
        "Weight1" => 'timee1',
        "Weight2" => 'timee2',
        "Weight3" => 'timee3',
        "Weight4" => 'timee4',
    ];
    if (!isset($columns[$location])) {
        throw new \RuntimeException("Invalid location");
    }
    $sql = "UPDATE userweight SET {$location}=? WHERE userId = ?;";
    $sql2 = "UPDATE userweight SET {$columns[$location]}=? WHERE userId = ?;";
}

Answer 2

考虑使用两个表，并将位置更改为仅一个数字：

Table 1: userweight
userId | location_nr | weight |
   1    |      1      |  60    |
   1    |      8      |  93    |

Table 2: usertime
userId | location_nr | time           |
   1    |      1      |  2020-05-03    |
   1    |      8      |  2020-09-18    |

现在，根据 user_id 和位置，您可以使用如下查询：

UPDATE userweight SET weight = ? WHERE userId = ? and location_nr= ?
UPDATE usertime SET timee= ? WHERE userId = ? and location_nr= ?

Mysqli 更好的方法来更改使用的列？

Mysqli better way to change what column working with?

php

mysql

mysqli