这是一个分析字符串的方法,分别看是否有“,”,“、”返回一个数组,里面调用的另一个方法列在后面,里面bb[]是在方法外定义的。
public static String[] parseIp(String ipStr) {
StringTokenizer stoken = new StringTokenizer(ipStr ,",");
if(stoken.hasMoreTokens()){
StringTokenizer stoken1 = new StringTokenizer(ipStr,"/");
if(stoken1.hasMoreTokens()){
String[] aa = parserip(ipStr);
for(int i=0;i<=aa.length;i++){
//String []bb = null;
bb[i] = parserpath(aa[i],0);//总在这一行出问题,通不过,不知道什么原因?
return bb;
}
}else{
bb = parserip(ipStr);
return bb;
}
}else{
StringTokenizer stoken1 = new StringTokenizer(ipStr,"/");
if(stoken1.hasMoreTokens()){
String[] aa = parserip(ipStr);
for(int i=0;i<=aa.length;i++){
//String []bb =null;
bb[i] = parserpath(aa[i],0);
return bb;
}
}else{
bb = parserip(ipStr);
return bb;
}
}
return bb ;
}
public static String[] parserip(String filename) {
String fileName = filename;
String[] tokens = fileName.split(",");
return tokens;
}
public static String[] parseIp(String ipStr) {
StringTokenizer stoken = new StringTokenizer(ipStr ,",");
if(stoken.hasMoreTokens()){
StringTokenizer stoken1 = new StringTokenizer(ipStr,"/");
if(stoken1.hasMoreTokens()){
String[] aa = parserip(ipStr);
for(int i=0;i<=aa.length;i++){
//String []bb = null;
bb[i] = parserpath(aa[i],0);//总在这一行出问题,通不过,不知道什么原因?
return bb;
}
}else{
bb = parserip(ipStr);
return bb;
}
}else{
StringTokenizer stoken1 = new StringTokenizer(ipStr,"/");
if(stoken1.hasMoreTokens()){
String[] aa = parserip(ipStr);
for(int i=0;i<=aa.length;i++){
//String []bb =null;
bb[i] = parserpath(aa[i],0);
return bb;
}
}else{
bb = parserip(ipStr);
return bb;
}
}
return bb ;
}
public static String[] parserip(String filename) {
String fileName = filename;
String[] tokens = fileName.split(",");
return tokens;
}
bb是否已经初始化
String filePath = filepath;
String[] tokens = filePath.split("/");
return tokens[i];
}
String[] aa = parserip(ipStr);这句可以把aa初始化吗