我想弄清楚如何從我的字符串中刪除找到的匹配項(xiàng)。所以我的代碼示例目前看起來(lái)像這樣：public void checkText() { String helper = "985, 913, 123, SomeotherText, MoreText, MoreText"; Pattern pattern = Pattern.compile("\\b\\d{3}"); Matcher matcher = pattern.matcher(helper); while (matcher.find()) { String newtext = "Number: " + matcher.group() + "\n"+ newtext; helper.replaceAll(matcher.group(),""); } newtext = newtext + "________________\n"+ helper; editText.setText(newtext);}所以我的輸入字符串是：985, 913, 123, SomeotherText, MoreText, MoreText運(yùn)行代碼后我想看到的是：Number: 985Number: 913Number: 123________________________SomeotherText, MoreText, MoreText誰(shuí)能告訴我我當(dāng)前的代碼有什么問(wèn)題？
2 回答
飲歌長(zhǎng)嘯
TA貢獻(xiàn)1951條經(jīng)驗(yàn) 獲得超3個(gè)贊
您可以在代碼中更新一些內(nèi)容：
您應(yīng)該將更換的退貨設(shè)置為
helper如果您只替換為空字符串，則您的字符串將
, , ,在替換中以逗號(hào)和后續(xù)空格開(kāi)頭您可以初始化變量
String newtext = "";
查看Java 演示
您的代碼可能如下所示：
String helper = "985, 913, 123, SomeotherText, MoreText, MoreText";
Pattern pattern = Pattern.compile("\\b\\d{3}");
Matcher matcher = pattern.matcher(helper);
String newtext = "";
while (matcher.find()) {
newtext = "Number: " + matcher.group() + "\n"+ newtext;
helper = helper.replaceAll(matcher.group() + ", ","");
}
newtext = newtext + "________________\n"+ helper;
System.out.println(newtext);
結(jié)果：
Number: 123
Number: 913
Number: 985
________________
SomeotherText, MoreText, MoreText
慕的地10843
TA貢獻(xiàn)1785條經(jīng)驗(yàn) 獲得超8個(gè)贊
由于您已經(jīng)在使用該類，因此Matcher您還可以使用該方法Matcher.appendReplacement進(jìn)行替換：
public void checkText() {
String helper = "985, 913, 123, SomeotherText, MoreText, MoreText";
Pattern pattern = Pattern.compile("\\b\\d{3}, ");
Matcher matcher = pattern.matcher(helper);
StringBuffer sb = new StringBuffer();
while (matcher.find()) {
System.out.println("Number:"+matcher.group());
matcher.appendReplacement(sb, "");
}
matcher.appendTail(sb);
System.out.println(sb.toString());
}
添加回答
舉報(bào)
0/150
提交
取消