有没有什么好的方法将用户输入的一个字符串中以@开始的找出来,加上链接,但如果@以email的形式出现的就忽略,保留原来的形式。
第一个要求用正则表达式就可以实现但第二个好像有点困难啊....public static void main(String[] args) {
String reg="@.*?(?=[.,:;!?\\s#@。,:;!?])";
Pattern pattern=Pattern.compile(reg);
System.out.println(pattern.toString());
Matcher matcher=pattern.matcher("fsafaf加大@梵净山,金佛山@就佛网 杰弗里斯[email protected] fsla");
StringBuffer sb=new StringBuffer();
while(matcher.find()){
System.out.println(matcher.group());
matcher.appendReplacement(sb, "<a href='abc'>"+matcher.group()+"</a>");
}
matcher.appendTail(sb);
System.out.println(sb);
}
第一个要求用正则表达式就可以实现但第二个好像有点困难啊....public static void main(String[] args) {
String reg="@.*?(?=[.,:;!?\\s#@。,:;!?])";
Pattern pattern=Pattern.compile(reg);
System.out.println(pattern.toString());
Matcher matcher=pattern.matcher("fsafaf加大@梵净山,金佛山@就佛网 杰弗里斯[email protected] fsla");
StringBuffer sb=new StringBuffer();
while(matcher.find()){
System.out.println(matcher.group());
matcher.appendReplacement(sb, "<a href='abc'>"+matcher.group()+"</a>");
}
matcher.appendTail(sb);
System.out.println(sb);
}
String reg = "([a-zA-Z0-9-.]*)@(.*?)([,:;!?\\s#@。,:;!?]|$|\\.(?:com|org|net))";
Pattern pattern = Pattern.compile(reg);
System.out.println(pattern.toString());
Matcher matcher = pattern
.matcher("fsafaf加大@梵净山,金佛山@就佛网 杰弗里斯[email protected] fsla;@sina.com;ssd@sina");
StringBuffer sb = new StringBuffer();
while (matcher.find()) {
String begin = matcher.group(1);
String middle = matcher.group(2);
String end = matcher.group(3); // 判断email
if (!begin.equals("")) {
boolean isEmail = middle.length() != 0;
for (int i = 0; i < middle.length(); i++) {
if (!Character.isLetterOrDigit(middle.charAt(i))) {
isEmail = false;
break;
}
}
if (isEmail) {
if (end.length() > 1) {
//确认是email
continue;
}
}
} matcher.appendReplacement(sb, begin + "<a href='abc'>" + "@"
+ middle + "</a>" + end);
}
matcher.appendTail(sb);
System.out.println(sb); }
还是4楼的更有技术含量些。\(^o^)/