blocks|key|759916|text|Java+8+lambda表达式。既快速又简单。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|759917|boolean+allLetters+=+someString.chars().allMatch(Character::isLetter);|code-block|syntax|javascript|759918|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|$]]

Java 8 lambda expressions. Both fast and simple.

<pre><code>boolean allLetters = someString.chars().allMatch(Character::isLetter);
</code></pre>

blocks|key|975413|text|或者，如果您正在使用Apache+Commons，则可以使用[StringUtils.isAlpha()]。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|975414|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html^0|U|N|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]]]]

Or if you are using Apache Commons, <a href="http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html" rel="noreferrer">[StringUtils.isAlpha()]</a>.

blocks|key|975578|text|我使用了这个正则表达式(".*[a-zA-Z]%2B.*")。使用if+not语句，它将避免所有在任何类型的其他字符之前、结尾或之间有字母的表达式。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|975579|String+strWithLetters+=+"123AZ456";
if(!+Pattern.matches(".*[a-zA-Z]%2B.*",+str1))
+return+true;
else+return+false|code-block|syntax|javascript|975580|entityMap^0|B|H|V|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

I used this regex expression <code>(".*[a-zA-Z]+.*")</code>. With <code>if not</code> statement it will avoid all expressions that have a letter before, at the end or between any type of other character.

<pre><code>String strWithLetters = "123AZ456";
if(! Pattern.matches(".*[a-zA-Z]+.*", str1))
 return true;
else return false
</code></pre>

blocks|key|759881|text|private+boolean+isOnlyLetters(String+s){
++++char+c='+';
++++boolean+isGood=false,+safe=isGood;
++++int+failCount=0;
++++for(int+i=0;i<s.length();i%2B%2B){
++++++++c+=+s.charAt(i);
++++++++if(Character.isLetter(c))
++++++++++++isGood=true;
++++++++else{
++++++++++++isGood=false;
++++++++++++failCount%2B=1;
++++++++}
++++}
++++if(failCount==0+&&+s.length()>0)
++++++++safe=true;
++++else
++++++++safe=false;
++++return+safe;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|759882|我知道这里有点挤。我在我的程序中使用了它，并感觉到了与人们分享它的愿望。它可以判断字符串中是否有任何字符不是字母。如果你想要一些容易澄清和回顾的东西，可以使用它。|unstyled|759883|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>private boolean isOnlyLetters(String s){
 char c=' ';
 boolean isGood=false, safe=isGood;
 int failCount=0;
 for(int i=0;i&lt;s.length();i++){
 c = s.charAt(i);
 if(Character.isLetter(c))
 isGood=true;
 else{
 isGood=false;
 failCount+=1;
 }
 }
 if(failCount==0 &amp;&amp; s.length()&gt;0)
 safe=true;
 else
 safe=false;
 return safe;
}
</code></pre>

I know it's a bit crowded. I was using it with my program and felt the desire to share it with people. It can tell if any character in a string is not a letter or not. Use it if you want something easy to clarify and look back on.

blocks|key|759981|text|下面是更快的方法。考虑到字母只有a-z，A-Z。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|759982|public+static+void+main(+String[]+args+){+
++++++++System.out.println(bestWay("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
++++++++System.out.println(isAlpha("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));

++++++++System.out.println(bestWay("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
++++++++System.out.println(isAlpha("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
++++}

++++public+static+boolean+bettertWay(String+name)+{
++++++++char[]+chars+=+name.toCharArray();
++++++++long+startTimeOne+=+System.nanoTime();
++++++++for(char+c+:+chars){
++++++++++++if(!(c>=65+&&+c<=90)&&!(c>=97+&&+c<=122)+){
++++++++++++++++System.out.println(System.nanoTime()+-+startTimeOne);
++++++++++++++++++++return+false;
++++++++++++}
++++++++}
++++++++System.out.println(System.nanoTime()+-+startTimeOne);
++++++++return+true;
++++}


++++public+static+boolean+isAlpha(String+name)+{
++++++++char[]+chars+=+name.toCharArray();
++++++++long+startTimeOne+=+System.nanoTime();
++++++++for+(char+c+:+chars)+{
++++++++++++if(!Character.isLetter(c))+{
++++++++++++++++System.out.println(System.nanoTime()+-+startTimeOne);
++++++++++++++++return+false;
++++++++++++}
++++++++}
++++++++System.out.println(System.nanoTime()+-+startTimeOne);
++++++++return+true;
++++}|code-block|syntax|javascript|759983|运行时间以纳秒为单位计算。它可能会因系统而异。|759984|5748//bettertWay+without+numbers
true
89493+//isAlpha+without++numbers
true
3284+//bettertWay+with+numbers
false
22989+//isAlpha+with+numbers
false|759985|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|$]]

Faster way is below. Considering letters are only a-z,A-Z.

<pre><code>public static void main( String[] args ){ 
 System.out.println(bestWay("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
 System.out.println(isAlpha("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));

 System.out.println(bestWay("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
 System.out.println(isAlpha("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
 }

 public static boolean bettertWay(String name) {
 char[] chars = name.toCharArray();
 long startTimeOne = System.nanoTime();
 for(char c : chars){
 if(!(c&gt;=65 &amp;&amp; c&lt;=90)&amp;&amp;!(c&gt;=97 &amp;&amp; c&lt;=122) ){
 System.out.println(System.nanoTime() - startTimeOne);
 return false;
 }
 }
 System.out.println(System.nanoTime() - startTimeOne);
 return true;
 }


 public static boolean isAlpha(String name) {
 char[] chars = name.toCharArray();
 long startTimeOne = System.nanoTime();
 for (char c : chars) {
 if(!Character.isLetter(c)) {
 System.out.println(System.nanoTime() - startTimeOne);
 return false;
 }
 }
 System.out.println(System.nanoTime() - startTimeOne);
 return true;
 }
</code></pre>

Runtime is calculated in nano seconds. It may vary system to system.

<pre><code>5748//bettertWay without numbers
true
89493 //isAlpha without numbers
true
3284 //bettertWay with numbers
false
22989 //isAlpha with numbers
false
</code></pre>

blocks|key|760003|text|检查这个，我猜这对你有帮助，因为它在我的项目中工作，所以一旦你检查了这段代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|760004|if(!+Pattern.matches(".*[a-zA-Z]%2B.*[a-zA-Z]",+str1))
+{
+++String+not+contain+only+character;
+}
else+
{
++String+contain+only+character;
}|code-block|syntax|javascript|760005|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|$]]

Check this,i guess this is help you because it's work in my project so once you check this code

<pre><code>if(! Pattern.matches(".*[a-zA-Z]+.*[a-zA-Z]", str1))
 {
 String not contain only character;
 }
else 
{
 String contain only character;
}
</code></pre>

blocks|key|975374|text|尝试使用正则表达式：String.matches|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|975375|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/lang/String.html#matches(java.lang.String)^0|A|E|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]]]]

Try using regular expressions: <a href="https://docs.oracle.com/javase/8/docs/api/java/lang/String.html#matches(java.lang.String)" rel="nofollow noreferrer">String.matches</a>

blocks|key|975503|text|++++++++String+expression+=+"%5E[a-zA-Z]*$";
++++++++CharSequence+inputStr+=+str;
++++++++Pattern+pattern+=+Pattern.compile(expression);
++++++++Matcher+matcher+=+pattern.matcher(inputStr);
++++++++if(matcher.matches())
++++++++{
++++++++++++++//if+pattern+matches+
++++++++}
++++++++else
++++++++{
+++++++++++++//if+pattern+does+not+matches
++++++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|975504|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> String expression = "^[a-zA-Z]*$";
 CharSequence inputStr = str;
 Pattern pattern = Pattern.compile(expression);
 Matcher matcher = pattern.matcher(inputStr);
 if(matcher.matches())
 {
 //if pattern matches 
 }
 else
 {
 //if pattern does not matches
 }
</code></pre>

blocks|key|760019|text|public+boolean+isAlpha(String+name)
{
++++String+s=name.toLowerCase();
++++for(int+i=0;+i<s.length();i%2B%2B)
++++{
++++++++if((s.charAt(i)>='a'+&&+s.charAt(i)<='z'))
++++++++{
++++++++++++continue;
++++++++}
++++++++else
++++++++{
+++++++++++return+false;
++++++++}
++++}
++++return+true;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|760020|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 boolean isAlpha(String name)
{
 String s=name.toLowerCase();
 for(int i=0; i&lt;s.length();i++)
 {
 if((s.charAt(i)&gt;='a' &amp;&amp; s.charAt(i)&lt;='z'))
 {
 continue;
 }
 else
 {
 return false;
 }
 }
 return true;
}
</code></pre>

blocks|key|975692|text|感觉好像我们需要的是找出字符是否只是字母。这是你如何解决它的方法-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|975693|Character.isAlphabetic(c)+|code-block|syntax|javascript|975694|帮助检查字符串的字符是否为字母。其中c是|975695|char+c+=+s.charAt(elementIndex);|975696|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|$]]

Feels as if our need is to find whether the character are only alphabets.
Here's how you can solve it-
<pre><code>Character.isAlphabetic(c) 
</code></pre>
helps to check if the characters of the string are alphabets or not.
where c is
<pre><code>char c = s.charAt(elementIndex);
</code></pre>

blocks|key|760154|text|如果只允许使用ASCII字母，则可以使用字符类\p{Alpha}。(这相当于[\p{Lower}\p{Upper}]或[a-zA-Z]。)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|760155|boolean+allLettersASCII+=+str.matches("\\p{Alpha}*");|code-block|syntax|javascript|760156|要允许所有Unicode字母，请使用字符类\p{L}+(或等效的\p{IsL})。|760157|boolean+allLettersUnicode+=+str.matches("\\p{L}*");|760158|请参阅Pattern+documentation。|760159|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/9/docs/api/java/util/regex/Pattern.html^0|N|9|12|K|1N|8|0|0|L|5|W|7|0|0|3|7|3|L|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|Q|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

To allow only ASCII letters, the character class <code>\p{Alpha}</code> can be used. (This is equivalent to <code>[\p{Lower}\p{Upper}]</code> or <code>[a-zA-Z]</code>.)
<pre><code>boolean allLettersASCII = str.matches(&quot;\\p{Alpha}*&quot;);
</code></pre>
For allowing all Unicode letters, use the character class <code>\p{L}</code> (or equivalently, <code>\p{IsL}</code>).
<pre><code>boolean allLettersUnicode = str.matches(&quot;\\p{L}*&quot;);
</code></pre>
See the <a href="https://docs.oracle.com/javase/9/docs/api/java/util/regex/Pattern.html" rel="nofollow noreferrer"><code>Pattern</code> documentation</a>.

blocks|key|760232|text|我发现了一种简单的方法来检查一个字符串是否所有的数字都是字母。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|760233|public+static+boolean+isStringLetter(String+input)+{

++++boolean+b+=+false;
++++for+(int+id+=+0;+id+<+input.length();+id%2B%2B)+{
++++++++if+('a'+<=+input.charAt(id)+&&+input.charAt(id)+<=+'z')+{
++++++++++++b+=+true;
++++++++}+else+if+('A'+<=+input.charAt(id)+&&+input.charAt(id)+<=+'Z')+{
++++++++++++b+=+true;
++++++++}+else+{
++++++++++++b+=+false;
++++++++}
++++}
++++return+b;
}|code-block|syntax|javascript|760234|我希望它能帮助任何正在寻找这种方法的人。|760235|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|$]]

I found an easy of way of checking a string whether all its digit is letter or not.
<pre><code>public static boolean isStringLetter(String input) {

 boolean b = false;
 for (int id = 0; id &lt; input.length(); id++) {
 if ('a' &lt;= input.charAt(id) &amp;&amp; input.charAt(id) &lt;= 'z') {
 b = true;
 } else if ('A' &lt;= input.charAt(id) &amp;&amp; input.charAt(id) &lt;= 'Z') {
 b = true;
 } else {
 b = false;
 }
 }
 return b;
}
</code></pre>
I hope it could help anyone who is looking for such method.

blocks|key|760041|text|使用StringUtils.isAlpha()方法，它将使您的工作变得简单。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|760042|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Use StringUtils.isAlpha() method and it will make your life simple.

The idea is to have a String read and to verify that it does not contain any numeric characters. So something like "smith23" would not be acceptable.

Check if String contains only letters

Java

这个想法是读取一个字符串，并验证它不包含任何数字字符。所以像"smith23“这样的东西是不能接受的。

问检查字符串是否只包含字母
EN

回答 13

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问检查字符串是否只包含字母EN

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问检查字符串是否只包含字母
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