Laravel - 从控制器传递数组到视图
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Laravel - Passing array from Controller to view
提问by Peter Jewicz
I'm having an issue passing a variable from a database query to my view. The variable itself passes fine, and I can var_dump it to see the contents of the array. The problem is when I try to call a part of the array, I get an undefined index error.
我在将变量从数据库查询传递到我的视图时遇到问题。变量本身通过得很好，我可以通过 var_dump 来查看数组的内容。问题是当我尝试调用数组的一部分时，出现未定义的索引错误。
Controller:
控制器：
public function displayCustomer()
{
$id = $_GET['id'];
$user = DB::table('customers')->where('id', '=', $id)->get();
return View::make('single')->withuser($user);
}
View
看法
<body>
<?php
if (Auth::check() != TRUE)
{
return Redirect::guest('login')->with('failed', 'Login Failed');
}
?>
<?php
//var_dump($user);
echo($user['name'])
?>
</body>
So as of right now, I'm able to var_dump the $user array and it works fine, However, when I try to echo out a part of the array I get undefined index, even though I know for a fact 'name' is one of my indexes. I've also tried $user[1] and get got the same error.
所以到目前为止，我可以对 $user 数组进行 var_dump 并且它工作正常，但是，当我尝试回显数组的一部分时，我得到了未定义的索引，即使我知道一个事实 'name' 是我的索引之一。我也试过 $user[1] 并得到同样的错误。
var_dump output
var_dump 输出
array(1) {
[0]=> object(stdClass)#148 (11) {
["id"] => int(1)
["name"]=> string(9) "paul rudd"
["email"]=> string(18) "[email protected]"
["phone"]=> string(10) "6305555555"
["address"]=> string(15) "123 fake street"
["item"]=> string(13) "6 pcs"
["total_price"]=> float(500)
["paid_amount"]=> float(400)
["balance"]=> float(100)
["created_at"]=> string(19) "0000-00-00 00:00:00"
["updated_at"]=> string(19) "0000-00-00 00:00:00"
}
}
回答by Gary Green
Based on your var_dump output try;
根据您的 var_dump 输出尝试；
<?php echo $user[0]->name; ?>
Your user is an array of objects. In your original query as your specifically picking out one id it might be better to get the firstresult instead of using a get, like so;
您的用户是一组对象。在您的原始查询中，因为您专门挑选了一个 id，所以获得first结果而不是使用 a可能会更好get，就像这样；
$user = DB::table('customers')->where('id', '=', $id)->first();
Notice the ->first()then you can simply do;
注意->first()then 你可以简单地做；
<?php echo $user->name; ?>
