Question

想要在查询中生成多个结果。

    $now = Carbon::now();
    $currentStartDay = $now->startOfDay();
    $currentEndDay = $now->endOfDay();
    $currentStartWeek = $now->startOfWeek();
    $currentEndWeek = $now->endOfWeek();
    $currentStartMonth = $now->startOfMonth();
    $currentEndMonth = $now->endOfMonth();
    $currentStartYear = $now->startOfYear();
    $currentEndYear = $now->endOfYear();

    $query = DB::table('transactions AS t');

    $queryExpenseToday = $query;
    $queryExpenseWeek = $query;
    $queryExpenseMonth = $query;
    $queryExpenseYear = $query;

    dump($queryExpenseToday->whereBetween('t.created_at', [$currentStartDay, $currentEndDay])->toSql());
    dump($queryExpenseWeek->whereBetween('t.created_at', [$currentStartWeek, $currentEndWeek])->toSql());
    dump($queryExpenseMonth->whereBetween('t.created_at', [$currentStartMonth, $currentEndMonth])->toSql());
    dump($queryExpenseYear->whereBetween('t.created_at', [$currentStartYear, $currentEndYear])->toSql());

问题和输出如下：

第一次转储

"select * from `transactions` as `t` where `t`.`created_at` between ? and ?"

第二次转储

"select * from `transactions` as `t` where `t`.`created_at` between ? and ? and `t`.`created_at` between ? and ?"

第三次转储

"select * from `transactions` as `t` where `t`.`created_at` between ? and ? and `t`.`created_at` between ? and ? and `t`.`created_at` between ? and ?"

第四个转储

"select * from `transactions` as `t` where `t`.`created_at` between ? and ? and `t`.`created_at` between ? and ? and `t`.`created_at` between ? and ? and `t`.`created_at` between ? and ?"

正在附加查询，如何停止此查询？或如何取消绑定最后一个绑定参数？

我正在使用Laravel 5.6

寻找最佳答案。

Answer 1

关于重新执行查询该怎么办？代替：

    ...
    $query = DB::table('transactions AS t');
    $queryExpenseToday = $query;
    $queryExpenseWeek = $query;
    ...

您可以这样做：

    ...
    $query = DB::table('transactions AS t');
    $queryExpenseToday = DB::table('transactions AS t');
    $queryExpenseWeek = DB::table('transactions AS t');
    ...

或者您也可以克隆它：

    ...
    $query = DB::table('transactions AS t');
    $queryExpenseToday = clone($query);
    $queryExpenseWeek = clone($query);
    ...

解释很简单。使用DB时，它会在内部创建一个Object的新实例，因此，如果使用$anyQuery=$query，则将引用引用到同一个第一个Object。因此，您必须创建一个新的（第一个选项）或克隆它，然后在内部创建一个新的（第二个选项）

您可以使用dd($anyQuery)进行检查，并在概念上稍作改动。好看！

Answer 2

认为您正在此处处理对象，因此实际上您正在处理相同的实例。

因此，通常情况下，您可以设置通用查询代码，就您而言：

// https://mvnrepository.com/artifact/com.cloudinary/cloudinary-http43
compile group: 'com.cloudinary', name: 'cloudinary-http43', version: '1.2.2'

但是如果您想为其添加不同的条件并有多个查询，则需要克隆，而不是：

$query = DB::table('transactions AS t');

您应该使用：

$queryExpenseToday = $query;
$queryExpenseWeek = $query;
$queryExpenseMonth = $query;
$queryExpenseYear = $query;

当然，您可以在一行中完成此操作，因此也可以像这样转储它（并在确保其确实起作用后将其分配给变量）：

$queryExpenseToday = clone $query;
$queryExpenseWeek = clone $query;
$queryExpenseMonth = clone $query;
$queryExpenseYear = clone $query;

Laravel查询绑定和取消绑定

2 个答案: