Question

我有5张桌子。我需要从一个表中选择4个最近的帖子，并在四个单独的变量中获取这些帖子的ID，以用于其他表中的查询：

SELECT user FROM table1 WHERE subcategory = "$input" ORDER BY id DESC LIMIT 4;

所以我有我需要的东西。现在，我如何将这四个用户标识结果分配给新变量，以便在同一页面的未来查询中使用。

我需要像这样存储它：

$user1 = "Sue";
$user2 = "Dave";
$user3 = "Alicia";
$user4 = "Tim";

所以我可以这样使用它：

SELECT * FROM table2 WHERE user = $user1;
SELECT * FROM table2 WHERE user = $user2;
SELECT * FROM table2 WHERE user = $user3;
SELECT * FROM table2 WHERE user = $user4;

不要混淆问题，但是：最终目标是拥有多行但只有四列，其中行中的所有数据都由USER排在第一行。重要的是要注意，因为每组行由对其下面的数据中包含的内容的描述分隔。这些只是直列，很容易。像这样：

Sue            Dave             Alicia             Tim
Sue Profile    Dave Profile     Alicia Profile     Tim Profile
>>Answers to questions about work:
Sue Income     Dave Income      Alicia Income      Tim Income
Sue Years      Dave Years       Alicia Years       Tim Years
>>Answers to questions about school:
Sue Degree     Dave Degree      Alicia Degree      Tim Degree
Sue Schools    Dave Schools     Alicia Schools     Tim Schools
>>Answers to questions about life:
Sue Born       Dave Born        Alicia Born        Tim Born
Sue Parents    Dave Parents     Alicia Parents     Tim Parents

如果我可以将id结果读入单独的变量，那么对于每个部分（问题）我可以做类似的事情：

SELECT * FROM income WHERE user = "Sue";
SELECT * FROM income WHERE user = "Dave";
SELECT * FROM income WHERE user = "Alicia";
SELECT * FROM income WHERE user = "Tim";

SELECT * FROM degree WHERE user = "Sue";
SELECT * FROM degree WHERE user = "Dave";
SELECT * FROM degree WHERE user = "Alicia";
SELECT * FROM degree WHERE user = "Tim";

Answer 1

您最好使用JOINS然后运行多个查询。以下是代码......

SELECT t1.user, t2.* FROM table1 as t1 LEFT JOIN table2 as t2 ON t1.id = t2.user_id WHERE t1.subcategory = "$input" ORDER BY t1.id DESC LIMIT 4

我希望您可以按照 t1.id = t2.user_id

的步调更改两个表中的相同键

Answer 2

虽然你要求的东西可以这样实现：

$id=1;
${"user".$id}="Sue";

这样$ user1将具有“Sue”，这样就可以很容易地使用数组：

$id=1;
$user[$id]="Sue";

或者更好的是，尝试在查询中使用Join一次获取所有数据，然后根据需要重新格式化。

Answer 3

如果您不想使用JOINS，那么在执行第一个查询后，您应该使用foreach循环......

$user=array(); // i would suggest you for making an array
foreach($result_array as $key->$value)
{
    //here you will get the ids in $value['user']
    // and then you can use them in your queries
    // so no need to store them in different variables and remembering them.
    Edited:
    $user[$key] = $value['user'];
}

查询结果到变量

3 个答案: