java - 进行 OOP 时，无法在 if else 中输出字符串

Question

当我在 get 方法上执行 if else 语句时我无法在 int length 中输出检查 String 并希望将 Gh 设置为 "AAAAA"代码是

public String getGh(){
        int length = Gh.length();
        if (length == 0){
            Gh="AAAAA";
        }else{
            Gh= null;
        }
    return this.Gh;
}
public String ShowGh() {
    return ("Title: "+this.Gh);

输出不正常，请帮忙

Answer 1

您应该记住，String 很可能为 null，因此

int length = Gh.length();

将抛出异常。这就是为什么在执行 Gh.length 之前必须手动检查 null null == Gh 的原因。我建议这样:

private String Gh = ""; // no null but empty string by default

...

public String getGh {
  // If Gh is null or empty, assign "AAAAA"; empty otherwise
  Gh = (null == Gh || 0 == Gh.length()) 
    ? "AAAAA"
    : "";

  return Gh;
}

//TODO: Turn ShowGh() into camel case: showGh()
public String ShowGh() {
    //DONE: you probably mean "getGh()" instead of Gh
    // return ("Title: " + Gh);
    // getGh() tries to get Gh, change its value to "AAAAA" and we have "Title: AAAAA"
    return "Title: " + getGh();
}

编辑:如果您想消除 getGh() 中的副作用并以典型方式实现逻辑:

public String getGh {
  // Just return Gh value
  return Gh;
}

public String setGh(String value) {
  // If Gh is null or empty, assign "AAAAA"; empty otherwise
  Gh = (null == value || 0 == value.length()) 
    ? "AAAAA
    : value;
}