Question

例如：输入String是

Input =  "Hello there, How are you? Fine!! @xyz"

，输出为

Output = "Hello there myWord How are you myWord Fine myWord myWord  myWord xyz"

我尝试使用Pattern类和Matcher类，但它只替换了一种类型的模式和str.replace(".","myWord");

Answer 1

您可以使用\\s*[^\\w\\s]\\s*

\\s*：[^\\w\\s]表示一个或多个空格

^：\\w无法捕获\\s和\\w

a-zA-Z0-9_表示\\s

String s="Hello there, How are you? Fine!! @xyz"; System.out.println(s.replaceAll("\\s*[^\\w\\s]\\s*", " myWord "));平均空间

Hello there myWord How are you myWord Fine myWord  myWord  myWord xyz

输出：

[]

为了避免任何其他特殊字符，不应该替换它，只需将它们添加到\\s*[^\\w\\s:;\\[\\]]\\s*例如const regex = /\s*[^\w\s\]\[;]\s*/g; const str = `Hello there, How are you? Fine!! ; @xyz []`; const subst = ` myWord `; const result = str.replace(regex, subst); console.log(result);中，如@ brso05所示

演示

<script src="https://cdnjs.cloudflare.com/ajax/libs/jspdf/1.3.2/jspdf.min.js"></script>
    <script src="https://cdnjs.cloudflare.com/ajax/libs/html2canvas/0.4.1/html2canvas.min.js"></script>

Answer 2

您可以这样做：

String str = "Hello there, How are you? Fine!! @xyz";
    str = str.replaceAll("[\\{\\.\\?!\"><',/\\\\\\|@#$%^&\\*~`\\-_=+\\}]", "myWord");
    System.out.println(str);

输出：

你好，你好吗？你好吗？我是谁？我是谁？我是谁？我爱你好！

用“myWord”

2 个答案: