Laravel - 从数据库事务关闭中获取变量
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Laravel - Get variable from a DB Transaction Closure
提问by Captain Hypertext
I am working with a Laravel 5 LAMP stack, and I am trying to process a CSV import with a database transaction. Code looks like this:
我正在使用 Laravel 5 LAMP 堆栈，我正在尝试使用数据库事务处理 CSV 导入。代码如下所示：
// Should contain any messages to pass back to the user
$results = [];
// Contains the total records inserted
$total = 0;
DB::transaction(function() use($csv_file, $results, $total) {
// Some Code ...
$total++;
$results[] = 'Row 10 has some weird data...';
});
return view('plan.import')
->with('results', $results)
->with('total', $total);
At the end of that, my records are imported, but my $total and $results are still empty, since they are outside the scope of the closure. I know they are being altered inside the function, because I've stepped through it, and seen them change. I just can't figure how to get them out of that transaction and return them to the user. Can anyone please help with this?
最后，我的记录被导入，但我的 $total 和 $results 仍然是空的，因为它们超出了闭包的范围。我知道它们在函数内部被改变，因为我已经通过它，看到它们发生了变化。我只是不知道如何让他们退出交易并将它们返回给用户。任何人都可以帮忙吗？
回答by The Alpha
You may replace the following line:
您可以替换以下行：
DB::transaction(function() use($csv_file, $results, $total)
with this:
有了这个：
DB::transaction(function() use($csv_file, &$results, &$total)
So the changes made inside the function will reflect in the variables because &creates a reference of the variable (Passes the variable reference) instead of passing them by value. Check Passing by ReferenceManual.
因此，函数内部所做的更改将反映在变量中，因为&创建了变量的引用（传递变量引用）而不是按值传递它们。检查通过参考手册。
Alternatively, you can return the variables from inside the closure like:
或者，您可以从闭包内部返回变量，例如：
$array = DB::transaction(function() use($csv_file, $results, $total) {
// Some Code ...
$total++;
$results[] = 'Row 10 has some weird data...';
return compact('total', 'results');
});
Then use it like:
然后像这样使用它：
return view('plan.import')
->with('results', $array['results'])
->with('total', $array['total']);
