blocks|key|1704359|text|Long.parseLong(data)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1704360|和catch+exception，它处理减号。|unstyled|1704361|虽然数字的数量是有限的，但这实际上创建了一个可以使用的数据变量，我想这是最常见的用例。|1704362|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>Long.parseLong(data)
</code></pre>

and catch exception, it handles minus sign. 

Although the number of digits is limited this actually creates a variable of the data which can be used, which is, I would imagine, the most common use-case.

blocks|key|1917912|text|您必须允许超过一个数字(+%2B符号)，如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1917913|String+regex+=+"[0-9]%2B";+
String+data+=+"23343453";+
System.out.println(data.matches(regex));|code-block|syntax|javascript|1917914|entityMap^0|D|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|$]]

You must allow for more than a digit (the <code>+</code> sign) as in:

<pre><code>String regex = "[0-9]+"; 
String data = "23343453"; 
System.out.println(data.matches(regex));
</code></pre>

blocks|key|1917981|text|您还可以使用Apache+Commons中的NumberUtil.isNumber(String+str)|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1917982|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/math/NumberUtils.html#isDigits(java.lang.String)^0|M|V|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]]]]

You can also use <a href="https://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/math/NumberUtils.html#isDigits(java.lang.String)" rel="noreferrer">NumberUtil.isNumber(String str)</a> from Apache Commons

blocks|key|1704494|text|还有一个尚未发布的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1704495|String+regex+=+"\\p{Digit}%2B";+//+uses+POSIX+character+class|code-block|syntax|javascript|1704496|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|$]]

One more solution, that hasn't been posted, yet:

<pre><code>String regex = "\\p{Digit}+"; // uses POSIX character class
</code></pre>

blocks|key|1918228|text|就性能而言，使用正则表达式的代价很高。尝试将字符串解析为长值是低效和不可靠的，而且可能不是您所需要的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1918229|我的建议是简单地检查每个字符是否是一个数字，使用Java+8+lambda表达式可以有效地完成以下操作：|1918230|boolean+isNumeric+=+someString.chars().allMatch(x+->+Character.isDigit(x));|code-block|syntax|javascript|1918231|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Using regular expressions is costly in terms of performance. Trying to parse string as a long value is inefficient and unreliable, and may be not what you need.

What I suggest is to simply check if each character is a digit, what can be efficiently done using Java 8 lambda expressions:

<pre><code>boolean isNumeric = someString.chars().allMatch(x -&gt; Character.isDigit(x));
</code></pre>

blocks|key|1704831|text|我们可以使用Pattern.compile("[0-9]%2B.[0-9]%2B")或Pattern.compile("\\d%2B.\\d%2B")。它们具有相同的含义。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1704832|模式0-9表示数字。与'\d‘相同。'%2B‘表示它出现的次数更多。“”整型或浮点型。|1704833|尝试以下代码：|1704834|import+java.util.regex.Pattern;

++++public+class+PatternSample+{

++++++++public+boolean+containNumbersOnly(String+source){
++++++++++++boolean+result+=+false;
++++++++++++Pattern+pattern+=+Pattern.compile("[0-9]%2B.[0-9]%2B");+//correct+pattern+for+both+float+and+integer.
++++++++++++pattern+=+Pattern.compile("\\d%2B.\\d%2B");+//correct+pattern+for+both+float+and+integer.

++++++++++++result+=+pattern.matcher(source).matches();
++++++++++++if(result){
++++++++++++++++System.out.println("\""+%2B+source+%2B+"\""++%2B+"+is+a+number");
++++++++++++}else
++++++++++++++++System.out.println("\""+%2B+source+%2B+"\""++%2B+"+is+a+String");
++++++++++++return+result;
++++++++}

++++++++public+static+void+main(String[]+args){
++++++++++++PatternSample+obj+=+new+PatternSample();
++++++++++++obj.containNumbersOnly("123456.a");
++++++++++++obj.containNumbersOnly("123456+");
++++++++++++obj.containNumbersOnly("123456");
++++++++++++obj.containNumbersOnly("0123456.0");
++++++++++++obj.containNumbersOnly("0123456a.0");
++++++++}

++++}|code-block|syntax|javascript|1704835|输出：|1704836|"123456.a"+is+a+String
"123456+"+is+a+String
"123456"+is+a+number
"0123456.0"+is+a+number
"0123456a.0"+is+a+String|1704837|entityMap^0|6|W|13|S|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|10|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|11|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|R|5|L|7|13|8|@]|D|@]|E|$M|N]]|$1|S|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|T|$]]

We can use either <code>Pattern.compile("[0-9]+.[0-9]+")</code> or <code>Pattern.compile("\\d+.\\d+")</code>. They have the same meaning.

the pattern [0-9] means digit. The same as '\d'.
'+' means it appears more times.
'.' for integer or float.

Try following code:

<pre><code>import java.util.regex.Pattern;

 public class PatternSample {

 public boolean containNumbersOnly(String source){
 boolean result = false;
 Pattern pattern = Pattern.compile("[0-9]+.[0-9]+"); //correct pattern for both float and integer.
 pattern = Pattern.compile("\\d+.\\d+"); //correct pattern for both float and integer.

 result = pattern.matcher(source).matches();
 if(result){
 System.out.println("\"" + source + "\"" + " is a number");
 }else
 System.out.println("\"" + source + "\"" + " is a String");
 return result;
 }

 public static void main(String[] args){
 PatternSample obj = new PatternSample();
 obj.containNumbersOnly("123456.a");
 obj.containNumbersOnly("123456 ");
 obj.containNumbersOnly("123456");
 obj.containNumbersOnly("0123456.0");
 obj.containNumbersOnly("0123456a.0");
 }

 }
</code></pre>

Output:

<pre><code>"123456.a" is a String
"123456 " is a String
"123456" is a number
"0123456.0" is a number
"0123456a.0" is a String
</code></pre>

blocks|key|1918348|text|根据Oracle的Java文档：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1918349|private+static+final+Pattern+NUMBER_PATTERN+=+Pattern.compile(
++++++++"[\\x00-\\x20]*[%2B-]?(NaN%7CInfinity%7C((((\\p{Digit}%2B)(\\.)?((\\p{Digit}%2B)?)"+%2B
++++++++"([eE][%2B-]?(\\p{Digit}%2B))?)%7C(\\.((\\p{Digit}%2B))([eE][%2B-]?(\\p{Digit}%2B))?)%7C"+%2B
++++++++"(((0[xX](\\p{XDigit}%2B)(\\.)?)%7C(0[xX](\\p{XDigit}%2B)?(\\.)(\\p{XDigit}%2B)))"+%2B
++++++++"[pP][%2B-]?(\\p{Digit}%2B)))[fFdD]?))[\\x00-\\x20]*");
boolean+isNumber(String+s){
return+NUMBER_PATTERN.matcher(s).matches()
}|code-block|syntax|javascript|1918350|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|$]]

According to Oracle's Java Documentation:

<pre><code>private static final Pattern NUMBER_PATTERN = Pattern.compile(
 "[\\x00-\\x20]*[+-]?(NaN|Infinity|((((\\p{Digit}+)(\\.)?((\\p{Digit}+)?)" +
 "([eE][+-]?(\\p{Digit}+))?)|(\\.((\\p{Digit}+))([eE][+-]?(\\p{Digit}+))?)|" +
 "(((0[xX](\\p{XDigit}+)(\\.)?)|(0[xX](\\p{XDigit}+)?(\\.)(\\p{XDigit}+)))" +
 "[pP][+-]?(\\p{Digit}+)))[fFdD]?))[\\x00-\\x20]*");
boolean isNumber(String s){
return NUMBER_PATTERN.matcher(s).matches()
}
</code></pre>

blocks|key|1704909|text|尝试这部分代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1704910|void+containsOnlyNumbers(String+str)
{
++++try+{
++++++++Integer+num+=+Integer.valueOf(str);
++++++++System.out.println("is+a+number");
++++}+catch+(NumberFormatException+e)+{
++++++++//+TODO:+handle+exception
++++++++System.out.println("is+not+a+number");
++++}

}|code-block|syntax|javascript|1704911|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 part of code:

<pre><code>void containsOnlyNumbers(String str)
{
 try {
 Integer num = Integer.valueOf(str);
 System.out.println("is a number");
 } catch (NumberFormatException e) {
 // TODO: handle exception
 System.out.println("is not a number");
 }

}
</code></pre>

In Java for String class there is a method called matches, how to use this method to check if my string is having only digits using regular expression. I tried with below examples, but both of them returned me false as result.

<pre><code>String regex = "[0-9]";
String data = "23343453";
System.out.println(data.matches(regex));
</code></pre>

<hr>

<pre><code>String regex = "^[0-9]";
String data = "23343453";
System.out.println(data.matches(regex));
</code></pre>

How to check if a string contains only digits in Java

Java

在Java for String类中有一个名为matches的方法，如何使用该方法通过正则表达式检查我的字符串是否只有数字。我尝试了下面的例子，但它们都返回了false作为结果。String regex = "[0-9]";String data = "23343453";System.out.println(data...

问如何在Java中检查字符串是否只包含数字
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