blocks|key|179518|text|我使用的解决方案如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|179519|Pattern+numberPat+=+Pattern.compile("\\d%2B");
Matcher+matcher1+=+numberPat.matcher(line);

Pattern+stringPat+=+Pattern.compile("What+is+the+square+of",+Pattern.CASE_INSENSITIVE);
Matcher+matcher2+=+stringPat.matcher(line);

if+(matcher1.find()+&&+matcher2.find())
{
++++int+number+=+Integer.parseInt(matcher1.group());++++++++++++++++++++
++++pw.println(number+%2B+"+squared+=+"+%2B+(number+*+number));
}|code-block|syntax|javascript|179520|我确信这不是一个完美的解决方案，但它满足了我的需求。谢谢大家的帮助。:)|179521|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The solution I went with looks like this:

<pre><code>Pattern numberPat = Pattern.compile("\\d+");
Matcher matcher1 = numberPat.matcher(line);

Pattern stringPat = Pattern.compile("What is the square of", Pattern.CASE_INSENSITIVE);
Matcher matcher2 = stringPat.matcher(line);

if (matcher1.find() &amp;&amp; matcher2.find())
{
 int number = Integer.parseInt(matcher1.group()); 
 pw.println(number + " squared = " + (number * number));
}
</code></pre>

I'm sure it's not a perfect solution, but it suited my needs. Thank you all for the help. :)

blocks|key|177611|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|177612|str.matches(".*\\d.*");|code-block|syntax|javascript|177613|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|$]]

try this 

<pre><code>str.matches(".*\\d.*");
</code></pre>

blocks|key|176936|text|如果你想从输入字符串中提取第一个数字，你可以这样做-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|176937|public+static+String+extractNumber(final+String+str)+{++++++++++++++++
++++
++++if(str+==+null+%7C%7C+str.isEmpty())+return+"";
++++
++++StringBuilder+sb+=+new+StringBuilder();
++++boolean+found+=+false;
++++for(char+c+:+str.toCharArray()){
++++++++if(Character.isDigit(c)){
++++++++++++sb.append(c);
++++++++++++found+=+true;
++++++++}+else+if(found){
++++++++++++//+If+we+already+found+a+digit+before+and+this+char+is+not+a+digit,+stop+looping
++++++++++++break;++++++++++++++++
++++++++}
++++}
++++
++++return+sb.toString();
}|code-block|syntax|javascript|176938|示例：|176939|176940|对于输入"123abc"，上述方法将返回123。|blockquote|176941|176942|对于"abc1000def"，1000。|176943|对于"555abc45"，555。|176944|对于"abc"，将返回空字符串。|176945|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|X|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|Z|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|11|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|12|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|13|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|U|$]]

If you want to extract the first number out of the input string, you can do-
<pre><code>public static String extractNumber(final String str) { 
 
 if(str == null || str.isEmpty()) return &quot;&quot;;
 
 StringBuilder sb = new StringBuilder();
 boolean found = false;
 for(char c : str.toCharArray()){
 if(Character.isDigit(c)){
 sb.append(c);
 found = true;
 } else if(found){
 // If we already found a digit before and this char is not a digit, stop looping
 break; 
 }
 }
 
 return sb.toString();
}
</code></pre>
Examples:
<blockquote>
For input &quot;123abc&quot;, the method above will return 123.
For &quot;abc1000def&quot;, 1000.
For &quot;555abc45&quot;, 555.
For &quot;abc&quot;, will return an empty string.
</blockquote>

blocks|key|2785222|text|我认为它比regex更快。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2785223|public+final+boolean+containsDigit(String+s)+{
++++boolean+containsDigit+=+false;

++++if+(s+!=+null+&&+!s.isEmpty())+{
++++++++for+(char+c+:+s.toCharArray())+{
++++++++++++if+(containsDigit+=+Character.isDigit(c))+{
++++++++++++++++break;
++++++++++++}
++++++++}
++++}

++++return+containsDigit;
}|code-block|syntax|javascript|2785224|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|$]]

I think it is faster than regex .

<pre><code>public final boolean containsDigit(String s) {
 boolean containsDigit = false;

 if (s != null &amp;&amp; !s.isEmpty()) {
 for (char c : s.toCharArray()) {
 if (containsDigit = Character.isDigit(c)) {
 break;
 }
 }
 }

 return containsDigit;
}
</code></pre>

blocks|key|175655|text|s=s.replaceAll("[*a-zA-Z]",+"")取代了所有字母|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|175656|s=s.replaceAll("[*0-9]",+"")将替换所有数字|175657|如果你做以上两个替换，你将得到所有特殊字符的字符串|175658|如果您只想从String+s=s.replaceAll("[%5E0-9]",+"")中提取整数|175659|如果您只想从String+s=s.replaceAll("[%5Ea-zA-Z]",+"")中提取字母表|175660|快乐编码:)|175661|entityMap^0|0|V|0|0|S|0|0|6|Z|0|6|12|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|X|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|N|3|O|5|6|7|14|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|Q|$]]

<code>s=s.replaceAll("[*a-zA-Z]", "")</code> replaces all alphabets

<code>s=s.replaceAll("[*0-9]", "")</code> replaces all numerics

if you do above two replaces you will get all special charactered string

If you want to extract only integers from a <code>String s=s.replaceAll("[^0-9]", "")</code>

If you want to extract only Alphabets from a <code>String s=s.replaceAll("[^a-zA-Z]", "")</code>

Happy coding :)

blocks|key|2785408|text|我找不到一个正确的模式。请遵循下面的指南，获得一个小而甜蜜的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2785409|String+regex+=+"(.)*(\\d)(.)*";++++++
Pattern+pattern+=+Pattern.compile(regex);
String+msg+=+"What+is+the+square+of+10?";
boolean+containsNumber+=+pattern.matcher(msg).matches();|code-block|syntax|javascript|2785410|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|$]]

I could not find a single pattern correct.
Please follow below guide for a small and sweet solution.

<pre><code>String regex = "(.)*(\\d)(.)*"; 
Pattern pattern = Pattern.compile(regex);
String msg = "What is the square of 10?";
boolean containsNumber = pattern.matcher(msg).matches();
</code></pre>

blocks|key|2785310|text|对于“检查字符串是否包含Java中的数字”，下面的代码就足够了|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2785311|Pattern+p+=+Pattern.compile("([0-9])");
Matcher+m+=+p.matcher("Here+is+ur+string");

if(m.find()){
++++System.out.println("Hello+"%2Bm.find());
}|code-block|syntax|javascript|2785312|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|$]]

The code below is enough for "Check if a String contains numbers in Java"

<pre><code>Pattern p = Pattern.compile("([0-9])");
Matcher m = p.matcher("Here is ur string");

if(m.find()){
 System.out.println("Hello "+m.find());
}
</code></pre>

blocks|key|175600|text|Pattern+p+=+Pattern.compile("(([A-Z].*[0-9])");
Matcher+m+=+p.matcher("TEST+123");
boolean+b+=+m.find();
System.out.println(b);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|175601|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>Pattern p = Pattern.compile("(([A-Z].*[0-9])");
Matcher m = p.matcher("TEST 123");
boolean b = m.find();
System.out.println(b);
</code></pre>

blocks|key|176901|text|尝试以下模式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|176902|.matches("[a-zA-Z+]*\\d%2B.*")|code-block|syntax|javascript|176903|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|$]]

Try the following pattern:

<pre><code>.matches("[a-zA-Z ]*\\d+.*")
</code></pre>

blocks|key|2785415|text|下面的代码片段将告诉您字符串是否包含数字|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2785416|str.matches(".*\\d.*")
or
str.matches(.*[0-9].*)|code-block|syntax|javascript|2785417|例如的|2785418|String+str+=+"abhinav123";

str.matches(".*\\d.*")+or+str.matches(.*[0-9].*)++will+return+true+

str+=+"abhinav";

str.matches(".*\\d.*")+or+str.matches(.*[0-9].*)++will+return+false|2785419|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|$]]

Below code snippet will tell whether the String contains digit or not
<pre><code>str.matches(&quot;.*\\d.*&quot;)
or
str.matches(.*[0-9].*)
</code></pre>
For example
<pre><code>String str = &quot;abhinav123&quot;;

str.matches(&quot;.*\\d.*&quot;) or str.matches(.*[0-9].*) will return true 

str = &quot;abhinav&quot;;

str.matches(&quot;.*\\d.*&quot;) or str.matches(.*[0-9].*) will return false
</code></pre>

blocks|key|2785350|text|public+String+hasNums(String+str)+{
++++++++char[]+nums+=+{+'0',+'1',+'2',+'3',+'4',+'5',+'6',+'7',+'8',+'9'+};
++++++++char[]+toChar+=+new+char[str.length()];
++++++++for+(int+i+=+0;+i+<+str.length();+i%2B%2B)+{
++++++++++++toChar[i]+=+str.charAt(i);
++++++++++++for+(int+j+=+0;+j+<+nums.length;+j%2B%2B)+{
++++++++++++++++if+(toChar[i]+==+nums[j])+{+return+str;+}
++++++++++++}
++++++++}
++++++++return+"None";
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2785351|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public String hasNums(String str) {
 char[] nums = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' };
 char[] toChar = new char[str.length()];
 for (int i = 0; i &lt; str.length(); i++) {
 toChar[i] = str.charAt(i);
 for (int j = 0; j &lt; nums.length; j++) {
 if (toChar[i] == nums[j]) { return str; }
 }
 }
 return "None";
 }
</code></pre>

blocks|key|179582|text|当我在这里寻找一种在Kotlin语言中查找字符串中的数字的方法时，我将在这里将我的发现留给其他想要特定于Kotlin的解决方案的人。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|179583|查找字符串是否包含数字：|179584|val+hasDigits+=+sampleString.any+{+it.isDigit()+}|code-block|syntax|javascript|179585|查找字符串是否仅包含数字：|179586|val+hasOnlyDigits+=+sampleString.all+{+it.isDigit()+}|179587|从字符串中提取数字：|179588|val+onlyNumberString+=+sampleString.filter+{+it.isDigit()+}|179589|entityMap^0|A|6|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|13|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|U|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|V|$]]

As I was redirected here searching for a method to find digits in string in <code>Kotlin</code> language, I'll leave my findings here for other folks wanting a solution specific to Kotlin.
Finding out if a string contains digit:
<pre><code>val hasDigits = sampleString.any { it.isDigit() }
</code></pre>
Finding out if a string contains only digits:
<pre><code>val hasOnlyDigits = sampleString.all { it.isDigit() }
</code></pre>
Extract digits from string:
<pre><code>val onlyNumberString = sampleString.filter { it.isDigit() }
</code></pre>

blocks|key|179483|text|你可以试试这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|179484|String+text+=+"ddd123.0114cc";
++++String+numOnly+=+text.replaceAll("\\p{Alpha}","");
++++try+{
++++++++double+numVal+=+Double.valueOf(numOnly);
++++++++System.out.println(text+%2B"+contains+numbers");
++++}+catch+(NumberFormatException+e){
++++++++System.out.println(text%2B"+not+contains+numbers");
++++}+++++|code-block|syntax|javascript|179485|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|$]]

You can try this

<pre><code>String text = "ddd123.0114cc";
 String numOnly = text.replaceAll("\\p{Alpha}","");
 try {
 double numVal = Double.valueOf(numOnly);
 System.out.println(text +" contains numbers");
 } catch (NumberFormatException e){
 System.out.println(text+" not contains numbers");
 } 
</code></pre>

blocks|key|177620|text|由于您不仅想要查找一个数字，而且还想要提取它，您应该编写一个小函数来完成这项工作。一个字母接一个字母地走，直到你找到一个数字。啊，我刚刚在stackoverflow上找到了必要的代码：find+integer+in+string。看一下被接受的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|177621|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/9742680/given-a-string-find-the-first-embedded-occurrence-of-an-integer^0|2K|M|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

As you don't only want to look for a number but also extract it, you should write a small function doing that for you. Go letter by letter till you spot a digit. Ah, just found the necessary code for you on stackoverflow: <a href="https://stackoverflow.com/questions/9742680/given-a-string-find-the-first-embedded-occurrence-of-an-integer">find integer in string</a>. Look at the accepted answer.

blocks|key|175673|text|.matches(".*\\d%2B.*")只适用于数字，而不适用于其他符号，如//或*等。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|175674|entityMap^0|0|K|12|2|15|1|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

<code>.matches(".*\\d+.*")</code> only works for numbers but not other symbols like <code>//</code> or <code>*</code> etc.

blocks|key|177005|text|ASCII位于UNICODE的开头，因此您可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|177006|(x+>=+97+&&+x+<=+122)+%7C%7C+(x+>=+65+&&+x+<=+90)+//+97+==+'a'+and+65+=+'A'|code-block|syntax|javascript|177007|我相信你可以计算出其他的值。|177008|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

ASCII is at the start of UNICODE, so you can do something like this:

<pre><code>(x &gt;= 97 &amp;&amp; x &lt;= 122) || (x &gt;= 65 &amp;&amp; x &lt;= 90) // 97 == 'a' and 65 = 'A'
</code></pre>

I'm sure you can figure out the other values...

I'm writing a program where the user enters a String in the following format: 

<pre><code>"What is the square of 10?"
</code></pre>

<ol>
<li>I need to check that there is a number in the String </li>
<li>and then extract just the number. </li>
<li>If i use <code>.contains("\\d+")</code> or <code>.contains("[0-9]+")</code>, the program can't find a number in the String, no matter what the input is, but <code>.matches("\\d+")</code>will only work when there is only numbers.</li>
</ol>

What can I use as a solution for finding and extracting?

Check if a String contains numbers Java

Kotlin

Java

我正在编写一个程序，其中用户输入以下格式的字符串："What is the square of 10?"我需要检查字符串中是否有一个数字，然后只提取这个数字。如果我使用.contains("\\d+")或.contains("[0-9]+")，则无论输入什么，程序都无法在字符串中找到数字，但.matches("\\d+")仅在只有数字时才起作用。我可以使用什么作为查找和提取的解决方案？

问检查字符串是否包含数字Java
EN

回答 16

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问检查字符串是否包含数字JavaEN

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问检查字符串是否包含数字Java
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