为什么PHP将MySQL INT列视为字符串？ [英] Why PHP considers MySQL INT columns as strings?
问题描述
我正在使用Laravel查询构建器来搜索类别。
这里是：
I am using a Laravel query builder to search for categories. Here it is:
$array[] = $categories->where('pk_i_id', $category_id)->first();
我必须手动将类别ID转换为字符串，即使在数据库中它是Integer类型。为什么要这样做？
I have to manually convert the category id to string, even though in database it is Integer type. Why do I have to do this?
其实 - 我必须在安装了灯泡堆叠的Linux机器上执行此操作。在具有Xampp的Windows机器上，它将该列视为整数。
Actually - I have to do this on Linux machine with Lamp stack installed. On Windows machine with Xampp it considers the column as integer as it should.
推荐答案
该问题听起来更像是 $ categories 已经是 Collection ，而不是 Builder 。 Collection 和 Builder 有一个，其中（）方法，但它们的逻辑不一样。
The issue sounds more like that $categories is already a Collection, not a Builder. Both the Collection and the Builder have a where() method, but their logic is not the same.
where（）方法 Builder 将在数据库的查询运行中添加一个参数化的where子句。在这种情况下，变量的类型并不重要。
The where() method on the Builder will add a parameterized where clause to the query run against the database. In this case, the type of variable doesn't matter.
但是，在$（$）code 集合将循环遍历集合，并返回第一个参数中的字段严格相等的结果（ === ）到第二个参数中传递的值。要改变这个，你可以通过 false 作为第三个参数，它将使用一个松散的比较（ == ）而不是严格。此外，您可以使用 whereLoose（），这是其中（）的快捷方式，第三个参数为 false 。
However, the where() method on the Collection will loop through the collection and return those results where the field in the first parameter is strictly equal (===) to the value passed in the second parameter. To change this, you can pass false as the third parameter, and it will use a loose comparison (==) instead of strict. Additionally, you could use whereLoose(), which is a shortcut for where() with the third parameter as false.
$array[] = $categories->where('pk_i_id', $category_id, false)->first();
// or
$array[] = $categories->whereLoose('pk_i_id', $category_id)->first();
如果不正确的字段类型导致更多的问题你所描述的，那么你可能想要解决基本问题。在链接的帖子中已经指出，在LAMP堆栈中，您需要用 mysqlnd 替换 mysqld 司机。
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