如何获取类和方法的名称？

Question

我有这样的PHP代码：

if(DEBUG){      
        g_start = new Date().getTime();
        console.log("start graph render: " + g_start );
    }
    var data_ajax = [{
            "color" : "#A8B400",
            "label" : "R1 Graph",
            "lines" : {
                "show" : true,
                "lineWidth" : 1
            },
            "points" : {
                "show" : false
            },
            "yaxes" : 1,
            "data" : arr,
            animator: { start: 100, steps: 99, duration: 1000, direction: "left" }
        }
    ];
    $('#network-graph').empty();
    //plot = $.plotAnimator($("#network-graph"),  data_ajax, options);
    plot = $.plot("#network-graph",  data_ajax, options);

    bind();
    if(DEBUG){      
        g_end = new Date().getTime();
        console.log("finish graph render: " + g_end );
        console.log("process graph render duration : " +  ((g_end - g_start)/1000) + "s" );
    }

正如您在代码评论中看到的那样，我需要class class_name { public function function_name() { /** I need to "class_name" and "function_name" in here * How can I get them? */ } } 和class_name。我怎么能得到它们？我认为function_name可以获得课程的名称。但实际上我的主要问题是获得方法的名称......

Answer 1

根据评论：

class class_name
{
    public function function_name()
    {
        echo __CLASS__;     // output: class_name
        echo __FUNCTION__;  // output: function_name
    }
}

但是，为了获得这两者，你必须使用返回__METHOD__的{​​{1}}。