Question

我已经给出了一些关于声明一个名为undefined的变量并用它来评估一个真实的if-else表达式的想法。我发现它在全球范围内是不可能的。但在MDN中，我发现了一些有关var person = { method : function(){ var undefined = 2 if(undefined){ console.log("undefined works as a variable") } } } person.method()数据类型的有趣见解。我直接引用MDN

undefined是全局对象的属性;即，它是一个变量在全球范围内。

它没有说明关于本地范围的任何事情。这意味着我可以在本地范围内创建一个。所以，我继续将这种见解进行测试。我创建了一个对象并为其指定了一个方法

console.log()

猜猜是什么！ if语句通过了测试，undefined内的字符串被打印在控制台上。这可能是一件危险的事情，当然也是一种不好的做法。有没有办法阻止在javascript的本地范围内声明以exception.txt关键字命名的变量？谢谢！

Answer 1

要解决undefined的意外修改，你不应该在你的代码中写下这个危险的词。

由于您只需要undefined的读取权限，因此建议总是使用void 0，而返回undefined。

var person = {
   method: function() {

       var undefined = 2

       if(void 0) {
          console.log("this should never be printed")
       } else {
          console.log("void 0 is always undefined, no matter what you do")
       }

   }
}

person.method()

如何使用void 0并完全摆脱“未定义”这个词？

// Turn
if(myVariable === undefined) { ... }
if(typeof myVariable === "undefined") { ... }
// into
if(myVariable === void 0) { ... }

// Turn
if((myVariable === undefined) || (myVariable === null)) { ... }
// into
if(myVariable == null) { ... }

// Turn
myObject.someKey = undefined
// into
myObject.someKey = void 0

Answer 2

欢迎来到精彩的JavaScript世界！

没有办法阻止某人这样做，但是如果你按照以下方式设置你的功能，有一种方法可以确保undefined意味着undefined（不是你真的应该这样做）这样做是因为任何人实际设置名为undefined的变量都是非常糟糕的做法。基本上较小的范围函数可以隐藏更高范围的undefined变量。

＆＃13;

// This is just a way to get a variable called "undefined"
function wrapper(){
  var undefined = 10;
  console.log("wrapper says undefined is: " + undefined);

  // This function declared a parameter called "undefined",
  // but if the function gets called with no argument, then
  // the value of this, more local, argument will truly be
  // undefined. If arguments are needed, just add "undefined" 
  // to the end of the list.
  function foo(undefined){
    // Formal test to ensure that undefined is what we believe it to be:
    if(typeof undefined === "undefined"){
      console.log("foo says undefined is: " + undefined);
    }
  }
  
  // When foo is called with no arguments passed in, the parameter
  // "undefined" will take on a value of undefined within the scope
  // of that function.
  foo();
}

wrapper();

＆＃13;

现在，由于你不打算使用＆＃34;假的＆＃34;设置你的所有功能，这有点做作。参数，但您可以测试undefined是否已被更改：

＆＃13;

function wrapper(){
  var undefined = 10;
  console.log(undefined);
  
  function foo(){
  
    if(typeof undefined === "undefined"){
       // undefined has not been altered
    } else {
       // undefined has been altered, so reset it for this scope:
       let undefined;
       console.log(undefined);
    }
 
  }
  
  foo();
}

wrapper();

＆＃13;

最后，您可以通过在函数中添加var undefined来阻止此功能影响您的功能。无论你在哪里声明它，吊装都会确保它在你的功能顶部开始。

Answer 3

我无法看到任何阻止它的方法，但是，你可以限制它通过yes inside如果你可以使用ES6使范围本地化。

你会发现现在如何改变范围，这不是一回事。

class DerexlUniversity(scrapy.Spider):
    name = 'drexel_university'
    allowed_domains = ['drexel.edu']
    start_urls = ['http://drexel.edu/search?t=people&q=']

    def parse(self, response):
        # with open('kw.txt') as file_object:
        #     last_names = file_object.readlines()

        for ln in ['Chong', 'Zhao']:
            last_name = ln.strip()
            print('-----------------------------------------------------')
            print("scraping last name: ", last_name)
            query = response.url + last_name
            yield scrapy.Request(query, callback=self.parse_item, meta=dict(last_name=last_name))

    def parse_item(self, response):
        last_name = response.meta['last_name']
        self.logger.info('This is item page %s', last_name)
        result_rows = response.xpath('//table//tr[@class="result-row"]')
        result_details = response.xpath('//table//tr[@class="result-details"]')

        for row, detail in zip(result_rows, result_details):
            full_name = row.xpath('.//span[@class="fullname"]/text()').extract_first()
            if full_name:
                full_name = clean_full_name(full_name)
                if last_name in full_name.split():
                    item = {}
                    item['fullname'] = full_name
                    item['university'] = 'Drexel University'
                    try:
                        item['email'] = row.xpath('.//span[@class="email-address"]/a/@href').extract_first()[7:]
                        item['phone'] = row.xpath('.//span[@class="phone-numbers"]/a/@href').extract_first()[4:]

                        person_detail = detail.xpath('.//span[@class="person-detail"]/text()').extract()
                    except ValueError:
                        pass
                    else:
                        person_detail_clean = ', '.join([pd.strip() for pd in person_detail[0].split(',')][1:])
                        item['person_detail'] = person_detail_clean

                    yield item

undefined为本地范围内的有效变量名

3 个答案: