如何調用另一個方法中具有參數的方法？ public class StackOverflow { static String blackListInput = "*[]#$"; public static boolean isBlackList(String val) { if (val == null) { return false; } for (int i = 0; i < val.length(); i++) { if (blackListInput.indexOf(val.charAt(i)) < 0) { return true; } } return false; } public static boolean cleanseInput() { if (isBlackList(val)) return true; return false; } }
2 回答
慕森王
TA貢獻1777條經驗 獲得超3個贊
嘗試這個
public class StackOverflow {
static String blackListInput = "*[]#$";
static char c[] = {'*','[',']','#','$'};
public static boolean isBlackList(String val) {
if("".equals(val) || val == null) {
return false;
}
if(val.contains(Character.toString(c[0]))) {
return true;
}
if(val.contains(Character.toString(c[1]))) {
return true;
}
if(val.contains(Character.toString(c[2]))) {
return true;
}
if(val.contains(Character.toString(c[3]))) {
return true;
}
if(val.contains(Character.toString(c[4]))) {
return true;
}
return false;
}
public static boolean cleanseInput(String val) {
if(isBlackList(val))
return true;
return false;
}
public static void main(String[] args) {
boolean b = cleanseInput("*");
System.out.println(b);
}
}
qq_笑_17
TA貢獻1818條經驗 獲得超7個贊
如果你想使用正則表達式模式
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexPattern {
public static boolean isBlackList(String value) {
String regex = "[a-zA-Z0-9\\-#\\*\\[\\]\\$\\s]{0,30}";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(value);
boolean b = m.matches();
return b;
}
public static boolean cleanseInput(String val) {
if(isBlackList(val))
return true;
return false;
}
public static void main(String[] args) {
boolean b = cleanseInput("fsa$assdhsahdfjfjkf");
System.out.println(b);
}
}
添加回答
舉報
0/150
提交
取消