Question

我有这个程序要编写，不知道如何控制键盘输入：

编写一种方法来检查输入的单词是否有效。   有效的词应该：


至少有10个字符

以字母开头

包含大写字母

至少包含3位数字

包含特殊字符（例如@，$。％...等）

包含空格

Answer 1

您不需要在每次击键时验证它，等到用户输入了该单词，然后进行验证。假设您正在使用控制台输入：

System.out.print("Enter something > ");
Scanner input = new Scanner(System.in);

String inputString = input.nextLine();

//perform validations on inputString, heres the first one:
//regex could be used instead of multiple if statements
if(inputString.length() < 10) {
    System.out.println("Validation failed, word was too short");
}
else if ...

Answer 2

您可以使用此正则表达式：

String REGEX = "(^[a-zA-Z](?=.*\\d{3,})(?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%])(?=\\s+).{10,})";
String INPUT = "your password";
Pattern pattern = Pattern.compile(REGEX);
Matcher matcher = pattern.matcher(INPUT);
System.out.println("matches(): "+matcher.matches());


^[a-zA-Z]           # Start with a letter
(?=.*\\d{3,})       # at least three digits must occur
(?=.*[a-z])         # a lower case letter must occur at least once
(?=.*[A-Z])         # an upper case letter must occur at least once
(?=.*[@#$%])        # a special character must occur at least once
(?=\\s+)            # a space must occur at least once
.{10,}              # anything, at least ten places though
$                   # end-of-string

要使用此功能，请阅读您的密码（来自档案，摇摆元素......）并在validate(String pass)内调用validate方法，然后针对正则表达式进行检查。

Answer 3

switch (myapp_config.thisDevice == 'desktop') {
  case (primaryNavHeight > pageWrapperHeight):
    // do stuff
    $('#page-wrapper').css("min-height", primaryNavHeight + "px");
    break;
  case (primaryNavHeight < pageWrapperHeight):
    // do stuff
    $('#page-wrapper').css("min-height", primaryNavHeight + "px");
    break;
}

//在这里＆＃34;输入＆＃34;是一个扫描仪名称，您可以使用任何名称作为OBAMA。

Scanner input=new Scanner(System.in);
System.out.print("Enter Number");

//在这里你可以为＆＃34; x＆＃34;。

指定键盘值

int x=input.nextInt();

然后你可以打印出来。

用Java控制键盘输入

3 个答案: