检查对象是否是 Javascript 中的字符串
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check if an object is string in Javascript
提问by stackunderflow
I was following a tutorial that suggested to check if an object is string and not empty as the following:
我正在学习一个教程，该教程建议检查对象是否为字符串而不是空的，如下所示：
var s = "text here";
if ( s && s.charAt && s.charAt(0))
it is said that if s is string then it has a method charAt and then the last component will check if the string is empty or not.
据说如果 s 是字符串，那么它有一个方法 charAt 然后最后一个组件将检查字符串是否为空。
I tried to test it with the other available methods like ( typeofand instanceof) using some of the SO questions and here and here too !!
我尝试使用其他可用方法（例如（typeof和instanceof））使用一些SO 问题以及此处和此处对其进行测试！
so I decided to test it in Js Bin : jsbin code hereas follow:
所以我决定在 Js Bin 中测试它：这里的 jsbin 代码如下：
var string1 = "text here";
var string2 = "";
alert("string1 is " + typeof string1);
alert("string2 is " + typeof string2);
//part1- this will succeed and show it is string
if(string1 && string1.charAt){
alert( "part1- string1 is string");
}else{
alert("part1- string1 is not string ");
}
//part2- this will show that it is not string
if(string2 && string2.charAt ){
alert( "part2- string2 is string");
}else{
alert("part2- string2 is not string ");
}
//part3 a - this also fails !!
if(string2 instanceof String){
alert("part3a- string2 is really a string");
}else{
alert("part3a- failed instanceof check !!");
}
//part3 b- this also fails !!
//i tested to write the String with small 's' => string
// but then no alert will excute !!
if(string2 instanceof string){
alert("part3b- string2 is really a string");
}else{
alert("part3b- failed instanceof check !!");
}
Now my questions are :
现在我的问题是：
1- why does the check for string fails when the string is empty using the string2.charAt???
1- 为什么使用string2.charAt???在字符串为空时检查字符串失败
2- why does the instanceofcheck failed??
2-为什么instanceof检查失败？
回答by user2864740
string valuesare not String objects(which is why the instanceof fails)2.
字符串值不是String 对象（这就是 instanceof 失败的原因）2。
To cover both cases using "type-checking" it would be typeof x === "string" || x instanceof String; the first only matches stringsand the latter matches Strings.
要使用“类型检查”覆盖这两种情况，它将是typeof x === "string" || x instanceof String；第一个只匹配字符串，后者匹配Strings。
The tutorial assumes that [only] String objects - or string values which are promoted1- have a charAtmethod and so uses "duck-typing". If the method does exist, then it is called. If charAtis used out-of-bounds then an empty string "", which is a false-y value, is returned.
本教程假定 [only] String 对象 - 或提升为1 的字符串值- 具有charAt方法，因此使用"duck-typing"。如果该方法确实存在，则调用它。如果charAt越界使用，则返回一个空字符串 ""，它是一个假 y 值。
The tutorial code would also accept a string of "\0", while s && s.lengthwould not - but it would also "work" on arrays (or jQuery objects, etc). Personally, I trustthe caller to provide the allowed values/types and use as little "type-checking" or special-casing as possible.
教程代码也会接受一个字符串“\0”，而s && s.length不会 - 但它也可以在数组（或 jQuery 对象等）上“工作”。就个人而言，我相信调用者会提供允许的值/类型并尽可能少地使用“类型检查”或特殊大小写。
1For primitive values of string, number, and boolean there is a corresponding object type of String, Number, and Boolean, respectively. When x.propertyis used on one of these primitive values the effect is ToObject(x).property- hence the "promotion". This is discussed in ES5: 9.9 - ToObject.
1对于字符串、数字和布尔值的原始值，分别有对应的字符串、数字和布尔对象类型。当x.property用于这些原始值之一时，效果是ToObject(x).property- 因此是“提升”。这在ES5: 9.9 - ToObject 中进行了讨论。
Neither the null or undefined values have corresponding objects (or methods). Functions are already objects but have a historically different, and useful, typeofresult.
null 或 undefined 值都没有相应的对象（或方法）。函数已经是对象，但具有历史上不同且有用的typeof结果。
2See ES5: 8 - Typesfor the different types of values. The String Type, eg., represents a stringvalue.
2请参阅ES5：8 -不同类型值的类型。字符串类型，例如，代表一个字符串值。
回答by David Sherret
1- Why does the check for string fails when the string is empty using the string2.charAt?
1- 为什么在字符串为空时使用 string2.charAt 检查字符串会失败？
The following expression evaluates to false because the first condition fails:
以下表达式的计算结果为 false，因为第一个条件失败：
var string2 = "";
if (string2 && string2.charAt) { console.log("doesn't output"); }
That second line is basically equivalent to:
第二行基本上相当于：
if (false && true) { console.log("doesn't output"); }
So for example:
例如：
if (string2) { console.log("this doesn't output since string2 == false"); }
if (string2.charAt) { console.log('this outputs'); }
2- Why does the instanceof check fail?
2- 为什么 instanceof 检查失败？
This fails because, in javascript, string can be literals or objects. For example:
这失败了，因为在 javascript 中，字符串可以是文字或对象。例如：
var myString = new String("asdf");
myString instanceof String; // true
However:
然而：
var myLiteralString = "asdf";
myLiteralString instanceof String; // false
You can reliably tell if it's a string by checking both the type and the instanceof:
您可以通过检查类型和 来可靠地判断它是否是字符串instanceof：
str instanceof String || typeof str === "string";
