我正在尝试为地址解析器创建switch语句,通过正则表达式解析String,但我遇到了困难。
我的挑战是这部分:
if(m.matches() && i==0){
b.street(m.group(1));
break;
如何使这成为
switch(SOMETHINGHERE) {
case SOMETHINGHERE: SOMECODE HERE
}
这里是正则表达式和patternList,它是一个Arraylist(或Linkedlist,我还没有决定:
static String streetReg = "([a-zæøåäöëüéèA-ZÆØAÄÖËÜÉÈ -./]*)";
static String symbolsReg = "[ ,.-]*";
public static void addPatterns() {
patternList.add(Pattern.compile(streetReg + "" + symbolsReg));
}
如何将以下内容转换为switch语句?
public static Address parse(String s) {
addPatterns();
Builder b = new Builder();
boolean noMatch = false;
for(int i = 0; i<patternList.size(); i++){
Matcher m = patternList.get(i).matcher(s);
if(m.matches() && i==0){
b.street(m.group(1));
break;
}
else if(m.matches() && i==1){
b.street(m.group(1));
b.city(m.group(2));
break;
}else if(m.matches() && i==2) {
b.postcode(m.group(1));
b.city(m.group(2));
break;
}else if(m.matches() && i== 3){
b.street(m.group(1));
b.house(m.group(2));
b.city(m.group(3));
break;
我试过这样做:
switch (s){
case "st":
b.street(m.group(1));
break;
case "street and house":
b.street(m.group(1));
b.street(m.group(2));
break;
case "noMatch":
noMatch =true;
break;
}
}
if(noMatch)return null;
else return b.build();
}
但我得到的错误是为开关和外壳选择的类型不匹配。例如boolean vs int或string vs int ..
if ( m.matches()) {
switch(i) {
case 0: b.street(m.group(1));
checkNewVariable = true;
break; // WARNING!! this break is a break for the switch, not for the FOR loop
// you'll need to add a variable (checkNewVariable) so you can break after the switch if needed
case 1: b.street(m.group(1));
b.city(m.group(2));
checkNewVariable = true;
break;
// ... rest of your cases
}
if ( checkNewVariable ) { break; } // to break out of the for loop
}
是一种选择。