blocks|key|2458258|text|使用此正则表达式进行搜索：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2458259|.(?=.{4})|code-block|syntax|javascript|2458260|RegEx演示|offset|length|2458261|不同的是，.将匹配任何字符，而不仅仅是一个数字，就像在正则表达式中一样。|style|CODE|2458262|Java代码：|BOLD|2458263|inputPhoneNum+=+inputPhoneNum.replaceAll(".(?=.{4})",+"*");|2458264|但是，如果您的意图是在最后4位之前屏蔽所有的数字，那么请使用：|2458265|.(?=(?:\D*\d){4})|2458266|或者在Java中：|2458267|inputPhoneNum+=+inputPhoneNum.replaceAll("\\d(?=(?:\\D*\\d){4})",+"*");|2458268|(?=(?:\\D*\\d){4})是一种正前瞻，它断言前面至少有4位数字，可以用0或更多的非数字分隔。|2458269|RegEx演示2|2458270|entityMap|0|LINK|mutability|MUTABLE|url|https://regex101.com/r/ldMIjK/1|1|https://regex101.com/r/ldMIjK/3^0|0|0|0|7|0|0|5|1|0|0|7|0|0|B|8|0|0|0|0|0|I|0|0|8|1|0^^$0|@$1|2|3|4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1H|8|@]|9|@$I|1I|J|1J|1|1K]]|A|$]]|$1|K|3|L|5|6|7|1L|8|@$I|1M|J|1N|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1O|8|@$I|1P|J|1Q|M|Q]]|9|@]|A|$]]|$1|R|3|S|5|D|7|1R|8|@]|9|@]|A|$E|F]]|$1|T|3|U|5|6|7|1S|8|@$I|1T|J|1U|M|Q]]|9|@]|A|$]]|$1|V|3|W|5|D|7|1V|8|@]|9|@]|A|$E|F]]|$1|X|3|Y|5|6|7|1W|8|@]|9|@]|A|$]]|$1|Z|3|10|5|D|7|1X|8|@]|9|@]|A|$E|F]]|$1|11|3|12|5|6|7|1Y|8|@$I|1Z|J|20|M|N]]|9|@]|A|$]]|$1|13|3|14|5|6|7|21|8|@]|9|@$I|22|J|23|1|24]]|A|$]]|$1|15|3|-4|5|6|7|25|8|@]|9|@]|A|$]]]|16|$17|$5|18|19|1A|A|$1B|1C]]|1D|$5|18|19|1A|A|$1B|1E]]]]

Use this regex for searching:

<pre><code>.(?=.{4})
</code></pre>

<a href="https://regex101.com/r/ldMIjK/1" rel="noreferrer">RegEx Demo</a>

Difference is that <code>.</code> will match any character not just a digit as in your regex.

Java code:

<pre><code>inputPhoneNum = inputPhoneNum.replaceAll(".(?=.{4})", "*");
</code></pre>

<hr>

However if your intent is to mask all digits before last 4 digits then use:

<pre><code>.(?=(?:\D*\d){4})
</code></pre>

Or in Java:

<pre><code>inputPhoneNum = inputPhoneNum.replaceAll("\\d(?=(?:\\D*\\d){4})", "*");
</code></pre>

<code>(?=(?:\\D*\\d){4})</code> is a positive lookahead that asserts presence of at least 4 digits ahead that may be separated by 0 or more non-digits.

<a href="https://regex101.com/r/ldMIjK/3" rel="noreferrer">RegEx Demo 2</a>

blocks|key|822559|text|我不擅长RegEx，但我认为您应该通过消除-事件来规范电话号码：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|822560|+++inputPhoneNum+=+inputPhoneNum.replace("-","").replaceAll("\\d(?=\\d{4})",+"*");|code-block|syntax|javascript|822561|entityMap^0|L|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

I'm not good in RegEx but I think you should normalize the phone numbers by getting rid of <code>-</code>occurences : 

<pre><code> inputPhoneNum = inputPhoneNum.replace("-","").replaceAll("\\d(?=\\d{4})", "*");
</code></pre>

blocks|key|822589|text|尝试使用2替换所有非数字或%2B为空，然后使用您的regex：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|822590|"%2B1-333-444-5678".replaceAll("[%5E\\d\\%2B]",+"").replaceAll("\\d(?=\\d{4})",+"*");|code-block|syntax|javascript|822591|输出|822592|%2B*******5678|822593|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Try to use two replace all non digit or + with empty then use your regex :

<pre><code>"+1-333-444-5678".replaceAll("[^\\d\\+]", "").replaceAll("\\d(?=\\d{4})", "*");
</code></pre>

Output

<pre><code>+*******5678
</code></pre>

blocks|key|822606|text|我觉得这应该管用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|822607|".*\\d(?=\\d{4})","*"|code-block|syntax|javascript|822608|您可以尝试使用本网站通过点击和试用创建。|offset|length|822609|entityMap|0|LINK|mutability|MUTABLE|url|https://regex101.com/^0|0|0|7|3|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

I think this should work

<pre><code>".*\\d(?=\\d{4})","*"
</code></pre>

You can try creating by hit and trial using <a href="https://regex101.com/" rel="nofollow noreferrer">this website</a>.

blocks|key|822623|text|如果您不想使用regex，另一种解决方案是从端到开始使用StringBuilder循环通过String，然后追加前4位数字，然后再追加*+(只需按正常方式追加任何非数字字符)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|822624|public+static+String+lastFour(String+s)+{
++++++++StringBuilder+lastFour+=+new+StringBuilder();
++++++++int+check+=+0;
++++++++for+(int+i+=+s.length()+-+1;+i+>=+0;+i--)+{
++++++++++++if+(Character.isDigit(s.charAt(i)))+{
++++++++++++++++check%2B%2B;
++++++++++++}
++++++++++++if+(check+<=+4)+{
++++++++++++++++lastFour.append(s.charAt(i));
++++++++++++}+else+{
++++++++++++++++lastFour.append(Character.isDigit(s.charAt(i))+?+"*"+:+s.charAt(i));
++++++++++++}
++++++++}
++++++++return+lastFour.reverse().toString();
++++}|code-block|syntax|javascript|822625|在网上试试！|822626|entityMap|0|LINK|mutability|MUTABLE|url|https://ideone.com/jfV5Ac^0|7|5|S|D|19|6|1V|1|0|0|0|6|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|D|@$9|15|A|16|1|17]]|E|$]]|$1|M|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

If you don't want to use <code>regex</code>, an alternate solution would be to loop through the <code>String</code> with a <code>StringBuilder</code> from end to start, and append the first 4 digits and then <code>*</code> after that (and just append any non-digit characters as normal)

<pre><code>public static String lastFour(String s) {
 StringBuilder lastFour = new StringBuilder();
 int check = 0;
 for (int i = s.length() - 1; i &gt;= 0; i--) {
 if (Character.isDigit(s.charAt(i))) {
 check++;
 }
 if (check &lt;= 4) {
 lastFour.append(s.charAt(i));
 } else {
 lastFour.append(Character.isDigit(s.charAt(i)) ? "*" : s.charAt(i));
 }
 }
 return lastFour.reverse().toString();
 }
</code></pre>

<a href="https://ideone.com/jfV5Ac" rel="nofollow noreferrer">Try it online!</a>

blocks|key|822632|text|这就是我所用的，它可能是有用的，只是掩盖了一些数字在提供的数字。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|822633|+/*
+*+mask+mobile+number+.
+*/
public++String+maskMobileNumber(String+mobile)+{
++++final+String+mask+=+"*******";
++++mobile+=+mobile+==+null+?+mask+:+mobile;
++++final+int+lengthOfMobileNumber+=+mobile.length();
++++if+(lengthOfMobileNumber+>+2)+{
++++++++final+int+maskLen+=+Math.min(Math.max(lengthOfMobileNumber+/+2,+2),+6);
++++++++final+int+start+=+(lengthOfMobileNumber+-+maskLen)+/+2;
++++++++return+mobile.substring(0,+start)+%2B+mask.substring(0,+maskLen)+%2B+mobile.substring(start+%2B+maskLen);
++++}
++++return+mobile;
}|code-block|syntax|javascript|822634|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is what I used, it may be useful, just masks some digits in the provided number
<pre><code> /*
 * mask mobile number .
 */
public String maskMobileNumber(String mobile) {
 final String mask = &quot;*******&quot;;
 mobile = mobile == null ? mask : mobile;
 final int lengthOfMobileNumber = mobile.length();
 if (lengthOfMobileNumber &gt; 2) {
 final int maskLen = Math.min(Math.max(lengthOfMobileNumber / 2, 2), 6);
 final int start = (lengthOfMobileNumber - maskLen) / 2;
 return mobile.substring(0, start) + mask.substring(0, maskLen) + mobile.substring(start + maskLen);
 }
 return mobile;
}
</code></pre>

I need to mask the phone number. it may consist of the digits, + (for country code) and dashes. The country code may consist of 1 or more digits. I have created such kind of regular expression to mask all the digits except the last 4:

<pre><code>inputPhoneNum.replaceAll("\\d(?=\\d{4})", "*");
</code></pre>

For such input: +13334445678

I get this result: +*******5678

However, it doesn't work for such input: +1-333-444-5678
In particular, it returns just the same number without any change. While the desired output is masking all the digits except for the last 4, plus sign and dashes.
That is why I was wondering how I can change my regular expression to include dashes? I would be grateful for any help!

Masking phone number Java

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我要掩盖电话号码。它可以由数字、+(代表国家代码)和破折号组成。国家代码可以由1位或更多位数组成。我创建了这样的正则表达式来掩盖所有的数字，但最后的4位除外：inputPhoneNum.replaceAll("\\d(?=\\d{4})", "*");用于这样的输入：+13334445678我得到的结果是：+*5678...

问屏蔽电话号码Java
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问屏蔽电话号码JavaEN