laravel 查询生成器：将参数传递给匿名函数
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Query builder: passing argument to anonymous function
提问by petwho
I got a problem in passing variable to query builder closure, here is my code:
我在将变量传递给查询构建器闭包时遇到问题，这是我的代码：
function get_usersbyname($name){
dd($name);
$resultset = DB::table('users')->where(function($query){
$query->where('username', 'LIKE', $name);
});
....
}
if I run it, it returns an error "undefined name variable", but I already passed $namevariable and checked its existence.
Also I cann't find any resouce explains how to pass variable to query builder anonymous function.
Could you help me with this problem?
如果我运行它，它会返回一个错误“ undefined name variable”，但我已经传递了$name变量并检查了它的存在。此外，我找不到任何解释如何将变量传递给查询构建器匿名函数的资源。你能帮我解决这个问题吗？
回答by Aran
You need to the tell the anonymous function to use that variable like...
您需要告诉匿名函数使用该变量，例如...
Because that variable is outside the scope of the annonymous function it needs to be passed in using the usekeyword as shown in the example below.
因为该变量在匿名函数的范围之外，所以需要使用use关键字传入，如下例所示。
function get_usersbyname($name){
dd($name);
$resultset = DB::table('users')->where(function($query) use ($name) {
$query->where('username', 'LIKE', $name);
});
....
}
