javascript - Laravel 5.2 和 jQuery 验证 :Trying to get property of Non-Object

Question

我正在使用 jQuery 验证插件来验证用户名是否已经存在。

验证插件的文档说:

The serverside resource is called via jQuery.ajax (XMLHttpRequest) and gets a key/value pair corresponding to the name of the validated element and its value as a GET parameter. The serverside response must be a JSON string that must be "true" for valid elements, and can be "false", undefined, or null for invalid elements, using the default error message. If the serverside response is a string, eg. "That name is already taken, try peter123 instead", this string will be displayed as a custom error message in place of the default.

Laravel Controller :

public function verifyName(Request $request)
{
    $username = $request->input('name', '');
    $user = User::where('name', $username)->first();
    if ($username == $user->name) {
        return json_encode(false);
    }else{
        return json_encode(true);
    }
}

JavaScript:

<script>
    $(function () {
        $.ajaxSetup({
            headers: {
                'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content')
            }
        });
        $("#register").validate({
            rules: {
                name: {
                    required: true,
                    rangelength: [4, 30],
                    remote: {
                        url: "{{ url('services/validation/verify_name') }}",
                        type: "post"
                    }
                }
            },
            errorClass: "has-danger",
            validClass: "has-success",
            //success: "valid",

            highlight: function (element, errorClass) {
                $(element).fadeOut(function () {
                    $(element).fadeIn();
                });
                $(element).closest(".form-group").addClass(errorClass);
            },
            unhighlight: function (element, errorClass) {
                $(element).closest(".form-group").removeClass(errorClass);
            },
            errorPlacement: function (error, element) {
                error.insertAfter(element);
            }
        });
    });
</script>

问题:

当return false时，它可以工作，但是当return true时，出现错误:“尝试获取非对象的属性”，我不知道为什么.

Answer 1

如果您没有包含给定 $username 的用户记录，则您的 $user 变量将为 null。如果 $user 为 null，当您尝试在 if 语句中访问 $user->name 时，您将得到 “尝试获取非对象的属性” 错误。

你想做这样的事情:

public function verifyName(Request $request)
{
    $username = $request->input('name', '');
    $user = User::where('name', $username)->first();

    // if the user was found, this will be true;
    // if the user wasn't found, $user will be null and this will be false
    if ($user) {
        return json_encode(false);
    } else {
        return json_encode(true);
    }
}