如何将参数传递给 Laravel DB::transaction()

问题描述

限时删除！！

来自 laravel 文档:数据库事务.它说:

From the laravel documentation: Database Transaction.It says that:

DB::transaction(function() {
    DB::table('users')->update(array('votes' => 1));
    DB::table('posts')->delete();
});

此处明确输入 1 以更新用户...我用一个变量试过这个，

Here, 1 is explicitly entered to update the users...I tried this using a variable,

$id = 3;
DB::transaction(function() {
    DB::table('users')->where('id','=',$id)->get();
});

它抛出一个错误:

Undefined variable: id

我也尝试将 $id 作为这样的参数:

I also tried to place to $id as a parameter like this:

$id = 3;
DB::transaction(function($id) {
    DB::table('users')->where('id', '=', $id)->get();
});

仍然是一个错误:

类 IlluminateDatabaseMySqlConnection 的对象无法转换为字符串

我做错了什么吗?请指教.谢谢...

Have I done anything wrong? Please advise. Thanks...

推荐答案

use 关键字正是您所需要的:

The use keyword is what you need:

$id = 3;
DB::transaction(function($id) use ($id) {
    DB::table('users')->where('id', '=', $id)->get();
});

对于 PHP 7(未经测试，按照以下答案的要求进行编辑):

For PHP 7 (untested, edited as requested by the answer below):

$id = 3;
DB::transaction(function() use ($id) {
    DB::table('users')->where('id', '=', $id)->get();
});

这篇关于如何将参数传递给 Laravel DB::transaction()的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

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