我想创建一个运算符来分割problem1 :(“+ 6x”,“2”),get(“+ 3x”); problem2 :(“12”,“2”),get(“6”)。
我想在答案后加上x,我试过了,但是一种方法只能有一个返回。我想创建一个运算符来分割problem1 :(“+ 6x”,“2”),get(“+ 3x”); problem2 :(“12”,“2”),get(“6”)。
第一个如果是从字符串中去掉x,除以后我想把它加回去。
public String apply(Vector args) {
//define two string variables
String expString1 = (String)args.get(0);
String expString2 = (String)args.get(1);
String s = "";
//move the x if the last char is x.
if (expString1.charAt(expString1.length()-1) == 'x'){s = expString1.substring(1, expString1.length()-1);
}else if (expString1.charAt(expString1.length()-1) != 'x') {s = expString1;}
//convert string to int, and do the divide operator.
int n2 = Integer.parseInt(expString2);
int n1 = Integer.parseInt(s);
int result = n1/n2;
//get result, but not the one I want, especially for x string.
return String.valueOf(result);
}
这就是我想要的。
public void testApplyVector() { Vector arg = new Vector();
arg.add("+6x");
arg.add("2");
Divide add = new Divide();
assertEquals("+3x", add.apply(arg));
Vector arg2 = new Vector();
arg2.add("12");
arg2.add("2");
assertEquals("6", add.apply(arg2));
}
但是这是JUnit得到的,第一个条件不能得到答案。答案后我不怎么加“x”。
public void testApplyVector() { Vector arg = new Vector();
arg.add("+6x");
arg.add("2");
Divide add = new Divide();
assertEquals("3", add.apply(arg));
Vector arg2 = new Vector();
arg2.add("12");
arg2.add("2");
assertEquals("6", add.apply(arg2));
}
回答:
我通过为结果添加另一个if语句来解决问题。代码是这样附加的。
public String apply(Vector args) {
//define two string variables
String expString1 = (String)args.get(0);
String expString2 = (String)args.get(1);
String s = "";
//move the x if the last char is x.
if (expString1.charAt(expString1.length()-1) == 'x'){
s = expString1.substring(0, expString1.length()-1);
}else if (expString1.charAt(expString1.length()-1) != 'x') {
s = expString1;}
//convert string to int, and do the divide operator.
int n2 = Integer.parseInt(expString2);
int n1 = Integer.parseInt(s);
int div = n1/n2;
//this is what i add, to check is the expString1 has a x.
String result = String.valueOf(div);
if (expString1.contains("x")) { return result = result + "x"; }
else return result;
}
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