如何删除或取消设置Javascript变量

我在Javascript中有一个全局变量，它已经由前面的函数填充

我知道我可以给这个变量无意义的值进行测试，但我认为还有另一种方法。例如，我把

some\u var=undefined

放进去，它是为了测试some\u var==“undefined”的

typeofsome\u var==“undefined”

---

有没有更专业的方法来解决这个问题？

像这样定义变量：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

。。。或者像这样：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

然后，您可以按如下方式删除它们：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

。。。或者像这样：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

。。。或者像这样：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

---

注: 如果您这样定义变量，

delete

不起作用：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

---

如下定义变量：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

。。。或者像这样：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

然后，您可以按如下方式删除它们：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

。。。或者像这样：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

。。。或者像这样：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

---

注: 如果您这样定义变量，

delete

不起作用：

window.some_var = 1;

window["some_var"] = 1;

delete some_var;

delete window.some_var;

delete window["some_var"];

var some_var = 1;

我在Javascript中有一个全局变量，它已经由 前功能

如果是全球性的话

delete window["variable-name"]; //should do the trick

window.x = 1; //equivalent to var x = 1; if x is global
delete window.x;

如果您只有全局变量的名称，那么

window["x"] = 1; //equivalent to var x = 1; if x is global
delete window.x

我在Javascript中有一个全局变量，它已经由 前功能

如果是全球性的话

delete window["variable-name"]; //should do the trick

window.x = 1; //equivalent to var x = 1; if x is global
delete window.x;

如果您只有全局变量的名称，那么

window["x"] = 1; //equivalent to var x = 1; if x is global
delete window.x

---

delete window.some\u var
应该可以做到这一点。
删除窗口
应该这样做。
您可以将任何不是来自函数的值作为已删除值的指示符，但
未定义的
或
空的
可能是正确的选择。您不应该真正删除变量。变量应该继续存在，只要它在范围内，只有它的值应该影响任何东西。也许您应该重新构造您的作用域，以便在处理完变量后将其删除…？您可以将任何不是来自函数的值作为已删除值的指示符，但
未定义的
或
空的
可能是一种方法。您不应该真正删除变量。变量应该继续存在，只要它在范围内，只有它的值应该影响任何东西。也许你应该重新构造你的作用域，在处理完变量后去掉它…？如何把这个。。。犯错误不，那实际上不起作用。怎么把这个。。。犯错误不。这实际上不起作用。如果先用let或var声明某个_var，则不起作用。如果一个变量曾经在脚本中首先用let或var声明过，那么您就必须使用它。顺便说一句，这是一件好事。如果先用let或var声明某个_var，情况就不一样了。如果一个变量曾经在脚本中首先用let或var声明过，那么您就必须使用它。顺便说一句，这是件好事。