方法一:使用contains方法 Java 的String类提供了一个名为contains的方法,可以非常方便地判断一个字符串是否包含另一个字符串。 示例代码 publicclassSubstringCheck{publicstaticvoidmain(String[]args){StringmainString="Hello, World!";StringsubString="World";booleanisSubstring=mainString.contains(subString);System....
String -- check contains --> Substring section Found Substring -- display message --> "The string contains the substring." section Not Found Substring -- display message --> "The string does not contain the substring." 总结 在Java中,String类提供了contains()方法来判断一个字符串是否包含另一...
Write a Java program to perform a case-insensitive check to see if one string contains another. Write a Java program to find and print all starting indices of a substring within a string. Write a Java program to determine the longest common substring between two input strings. Java Code Edit...
Example 1: Java substring() With Only Start Index classMain{publicstaticvoidmain(String[] args){ String str1 ="program";// 1st character to the last character System.out.println(str1.substring(0));// program // 4th character to the last characterSystem.out.println(str1.substring(3));/...
Learn to check if a string contains the specified substring in Java using theString.contains()API. 1.String.contains()API TheString.contains(substring)searches a specified substring in the current string. It returnstrueif and only if thesubstringis found in the given string otherwise returnsfalse...
JDK 7 中的substring 上述问题在JDK 7中得到了解决。JDK 7中,substring方法会在堆中创建一个新的数组。 源码 //JDK 7 /** * Allocates a new {@codeString} that contains characters from a subarray * of the character array argument. The {@codeoffset} argument is the ...
String str = "test.txt.backup";int index = str.lastIndexOf("."); //结果为9 常用于处理文件后缀名或路径截取。contains方法 判断字符串是否包含指定子串，返回布尔值。底层调用indexOf实现，可读性更好。示例：String str = "Java is fun";boolean result = str.contains("fun"); // true 适用于简单...
public static void main(String[] args){ } //虽说是固定的,但其实args可以变,args是arguments(参数)的缩写,可以改为任意变量 //中括号的位置也可以变,可以放在参数的后面 //但是习惯上中括号放在前面,后面参数用args 输出语句: System.out.println("Hello,World!");输出之后换行 ...
Scanner userInteraction = new Scanner(System.in);String userInput = "";while (true) { System.out.println( "Welcome New Player! What will your name be?"); userInput = userInteraction.nextLine(); if (userInput.matches("[a-zA-Z]+")) { // check that the input only contains letters ...
PathMatcher pathMatcher=newAntPathMatcher();//这是我们的请求路径 需要被匹配(理解成匹配controller吧 就很容易理解了)String requestPath="/user/list.htm?username=aaa&departmentid=2&pageNumber=1&pageSize=20";//请求路径//路径匹配模版String patternPath="/user/list.htm**";assertTrue(pathMatcher.match...