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请问字符串包含有关问题,等

热度:478   发布时间:2016-04-24 17:44:33.0
请教字符串包含问题,急等
我的网页上有一个输入框,如果会员输入的号码是CUS7100001~CUS7250000之间的任何一个,就返回true,否则返回false,请问用java类怎么写??谢谢!

------解决方案--------------------
public static boolean validate(String strInput){
return strInput.compareTo( "CUS7100001 ")> =0 && strInput.compareTo( "CUS7250000 ") <=0;
}
------解决方案--------------------
public boolean check(String s){
int pos = s.indexOf( "CUS ");
String numStr = s.substring(pos , s.length()-1);
int num = Integer.parseInt(numStr);
if(num > = 7100001 && num <= 7250000) return true;
else return false;
}
------解决方案--------------------
just use the String 's compareTo() method

compareTo
public int compareTo(String anotherString)
Compares two strings lexicographically. The comparison is based on the Unicode value of each character in the strings. The character sequence represented by this String object is compared lexicographically to the character sequence represented by the argument string. The result is a negative integer if this String object lexicographically precedes the argument string. The result is a positive integer if this String object lexicographically follows the argument string. The result is zero if the strings are equal; compareTo returns 0 exactly when the equals(Object) method would return true.
This is the definition of lexicographic ordering. If two strings are different, then either they have different characters at some index that is a valid index for both strings, or their lengths are different, or both. If they have different characters at one or more index positions, let k be the smallest such index; then the string whose character at position k has the smaller value, as determined by using the < operator, lexicographically precedes the other string. In this case, compareTo returns the difference of the two character values at position k in the two string -- that is, the value:

this.charAt(k)-anotherString.charAt(k)

If there is no index position at which they differ, then the shorter string lexicographically precedes the longer string. In this case, compareTo returns the difference of the lengths of the strings -- that is, the value:
this.length()-anotherString.length()


Specified by:
compareTo in interface Comparable <String>
Parameters:
anotherString - the String to be compared.
Returns:
the value 0 if the argument string is equal to this string; a value less than 0 if this string is lexicographically less than the string argument; and a value greater than 0 if this string is lexicographically greater than the string argument.


public class Test
{
public static void main(String args[]) {
Test test = new Test();
System.out.println(test.checkOutTheInput( "CUS7100001 "));
System.out.println(test.checkOutTheInput( "CUS7100000 "));
System.out.println(test.checkOutTheInput( "CUS7250000 "));
System.out.println(test.checkOutTheInput( "CUS7250001 "));

}

private boolean checkOutTheInput(String str) {
if (str.compareTo(new String( "CUS7100001 ")) > = 0 && str.compareTo(new String( "CUS7250000 ")) <= 0) {
return true;
}
return false;
}
}
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