如何从查询Laravel获得相反的结果 [英] how to get opposite results from a query Laravel

问题描述

如何从 SELECT 查询中获得相反的结果

How to get the opposite results from a SELECT query

下面是我的查询

  $fod = FodMap::select('*')
        ->where('fructose_level', $fructose == 1 ? '=' : '>=', $fructose)
        ->where('lactose_level', $lactose == 1 ? '=' : '>=', $lactose)
        ->where('polyols_level', $polyols == 1 ? '=' : '>=', $polyols)
        ->where('fructan_level', $fructan == 1 ? '=' : '>=', $fructan)
        ->get();

我想获取不属于上述查询范围的所有内容.请咨询

I want to get everything that doesn't fall under the above query. pls advice

推荐答案

好吧，我想了一段时间，我认为解决方案并不像您首先提出的那样复杂(当我先阅读时，我以为您想要反转结果.

Ok, I thought about your question for some time and I think the solution is not as complicated as you asked first (when I've read it first, I thought you want inverted result).

1. 在您的评论之一中，您写道:

以上结果给出了用户无法食用的食物列表. 希望它得到上面不匹配的结果

above results gives the list of foods which user can't eat.. what i want it to get the results which not matching above

1. 我已经读过果糖含量等信息.总的来说，高于1的果糖是不好的，低于1的果糖是好的.

那么，解决方案太简单了:

So, solution is way too simple then:

$fod = FodMap::select('*')
        ->where('fructose_level', '<', $fructose)
        ->where('lactose_level', '<', $lactose)
        ->where('polyols_level', '<', $polyols)
        ->where('fructan_level', '<', $fructan)
        ->get();

另外，原始查询中的三元运算符也没有意义，因为子句如下:

Also, ternary operators in your original query do not make sense, because clause like:

->where('fructose_level', $fructose == 1 ? '=' : '>=', $fructose)

将始终用作:

->where('fructose_level', '>=', $fructose)

考虑一下.

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