blocks|key|477900|text|你的问题还不清楚，但这可能对你的案子有用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|477901|String+str="NH3PO4";

++++int+lastChar=str.lastIndexOf("O");//replace+"O"+into+a+param+if+needed
++++String+toParse=str.substring(lastChar%2B1);
++++System.out.println("toParse="%2BtoParse);
++++try{
++++++++System.out.println("after+parse,+"+%2BInteger.parseInt(toParse));
++++}
++++catch+(NumberFormatException+ex){
++++++++System.out.println(toParse+%2B"+can+not+be+parsed+to+int");
++++}
}|code-block|syntax|javascript|477902|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|$]]

Your question is not clear, but this may work for your case. 

<pre><code>String str="NH3PO4";

 int lastChar=str.lastIndexOf("O");//replace "O" into a param if needed
 String toParse=str.substring(lastChar+1);
 System.out.println("toParse="+toParse);
 try{
 System.out.println("after parse, " +Integer.parseInt(toParse));
 }
 catch (NumberFormatException ex){
 System.out.println(toParse +" can not be parsed to int");
 }
}
</code></pre>

blocks|key|1279137|text|将字符串转换为char数组：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279138|String+molecule+=+"H2SO4+";
char[]+moleculeArray+=+str.toCharArray();
for(int+i+=+0;+i+<+moleculeArray.length;+i%2B%2B){
++++if(Character.isLetter(moleculeArray[i])){
++++++++//Huston+we+have+a+character!
++++++++if(i%2B1+<+moleculeArray.length+&&+Character.isDigit(moleculeArray[i%2B1])+{
++++++++++++int+digit+=+Character.getNumericValue(Character.isDigit(moleculeArray[i%2B1]);
++++++++++++//It+has+a+digit+do+something+extra!
++++++++}
++++}
}|code-block|syntax|javascript|1279139|然后遍历数组并使用Character.isDigit(c)和Character.isLetter(c)|1279140|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|$]]

Convert the String to a char array:

<pre><code>String molecule = "H2SO4 ";
char[] moleculeArray = str.toCharArray();
for(int i = 0; i &lt; moleculeArray.length; i++){
 if(Character.isLetter(moleculeArray[i])){
 //Huston we have a character!
 if(i+1 &lt; moleculeArray.length &amp;&amp; Character.isDigit(moleculeArray[i+1]) {
 int digit = Character.getNumericValue(Character.isDigit(moleculeArray[i+1]);
 //It has a digit do something extra!
 }
 }
}
</code></pre>

then iterate over the array and use Character.isDigit(c) and Character.isLetter(c)

blocks|key|477859|text|我认为您需要将字符串拆分为char数组，然后在该数组中搜索'o‘：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|477860|String+str+=+"H2SO4";+
char[]+charArray+=+str.toCharArray();|code-block|syntax|javascript|477861|然后得到:+H，2，S，O，4，你可以在这个数组中搜索"O“。|477862|希望能帮上忙！|477863|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I think you need to split your String to char array' and then search the 'o' in that array:

<pre><code>String str = "H2SO4"; 
char[] charArray = str.toCharArray();
</code></pre>

Then you got : [H, 2, S, O, 4], and you can search the "O" in this array.

Hope it helped!

blocks|key|477869|text|您可以使用正则表达式。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|477870|++++++.*O([0-9]%2B).*+|code-block|syntax|javascript|477871|并使用组提取数字处理字符O。|477872|请在这里了解更多：|477873|http://docs.oracle.com/javase/tutorial/essential/regex/groups.html|offset|length|477874|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|0|0|1U|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]]|$1|G|3|H|5|6|7|X|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Y|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|Z|8|@]|9|@$M|10|N|11|1|12]]|A|$]]|$1|O|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|L]]]]

you can use regular expression. 

<pre><code> .*O([0-9]+).* 
</code></pre>

And use group to extract the number proceeding character O. 

Find out more here:

<a href="http://docs.oracle.com/javase/tutorial/essential/regex/groups.html" rel="nofollow">http://docs.oracle.com/javase/tutorial/essential/regex/groups.html</a>

blocks|key|1279222|text|有多种方法，它们都非常简单，而且都取自关于字符串的正式文档。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1279223|//Find+character+index+of+O,+and+parse+next+character+as+int:
int+index+=+str.indexOf("O");
Integer.parseInt(str.subString(index,+index%2B1));

//loop+trough+char+array+and+check+each+char
char[]+arr+=+str.toCharArray();
for(char+ch+:+arr){
++++if(isNumber(ch){..}
}
boolean+isNumber(char+ch){
++++//read+below
}|code-block|syntax|javascript|1279224|请参阅ascii表和这里|1279225|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/lang/String.html|1|https://stackoverflow.com/questions/4968323/java-parse-int-value-from-a-char^0|J|A|0|0|0|A|2|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

There are multiple ways, all of them very simple and taken from <a href="http://docs.oracle.com/javase/7/docs/api/java/lang/String.html" rel="nofollow noreferrer">Official documentation on String</a>.

<pre><code>//Find character index of O, and parse next character as int:
int index = str.indexOf("O");
Integer.parseInt(str.subString(index, index+1));

//loop trough char array and check each char
char[] arr = str.toCharArray();
for(char ch : arr){
 if(isNumber(ch){..}
}
boolean isNumber(char ch){
 //read below
}
</code></pre>

refer to ascii table and <a href="https://stackoverflow.com/questions/4968323/java-parse-int-value-from-a-char">here</a>

blocks|key|1279247|text|为了解析O之后的每个字符，可以使用以下代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279248|char+[]+lettersArray+=+source.toCharArray();
for(int+i+=0+;i<lettersArray.length;i%2B%2B){
++if(Character.isLetter(lettersArray[i])){
++++++if(lettersArray[i]=='O'){
++++++++++try{
+++++++++++++int+a+=+Interger.parseInteger(lettersArray[i].toString());
++++++++++}catch(Exception+e){
++++++++++++++e.printStackTrace();
++++++++++}
++++++}
+++}
}|code-block|syntax|javascript|1279249|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|$]]

In order to parse every character after exactly O you can use the follow code:

<pre><code>char [] lettersArray = source.toCharArray();
for(int i =0 ;i&lt;lettersArray.length;i++){
 if(Character.isLetter(lettersArray[i])){
 if(lettersArray[i]=='O'){
 try{
 int a = Interger.parseInteger(lettersArray[i].toString());
 }catch(Exception e){
 e.printStackTrace();
 }
 }
 }
}
</code></pre>

blocks|key|40512|text|final+int+numAfterO;
final+String+strNum;
final+int+oIndex+=+chemicalName.lastIndexOf("O");
if+(oIndex+>=+0+&&+chemicalName.length()+>=+oIndex+%2B+2)+{
++++strNum+=+chemicalName.subString(oIndex,+oIndex+%2B+1);
}+else+{
++++strNum+=+null;
}
if+(strNum+!=+null)+{
++++try+{
++++++++numAfterO+=+Integer.parseInt(strNum);
++++}+catch+(NumberFormatException+e)+{
++++++++numAfter0+=+-1;
++++}
}
android.util.Log.d("MYAPP",+"The+number+after+the+last+O+is+"+%2B+numberAfterO);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|40513|我想这就是你想要的。|unstyled|40514|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>final int numAfterO;
final String strNum;
final int oIndex = chemicalName.lastIndexOf("O");
if (oIndex &gt;= 0 &amp;&amp; chemicalName.length() &gt;= oIndex + 2) {
 strNum = chemicalName.subString(oIndex, oIndex + 1);
} else {
 strNum = null;
}
if (strNum != null) {
 try {
 numAfterO = Integer.parseInt(strNum);
 } catch (NumberFormatException e) {
 numAfter0 = -1;
 }
}
android.util.Log.d("MYAPP", "The number after the last O is " + numberAfterO);
</code></pre>

I assume this is what you want.

blocks|key|477915|text|这是一个解析整个分子结构的东西。这一个解析整数>9。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|477916|public+static+class+Molecule+{
++++private+List<MoleculePart>+parts+=+new+ArrayList<MoleculePart>();

++++public+Molecule(String+s)+{
++++++++String+name+=+"";
++++++++String+amount+=+"";
++++++++for+(char+c+:+s.toCharArray())+{
++++++++++++if+(inBetween(c,+'A',+'Z'))+{
++++++++++++++++if+(!name.isEmpty())+//+first+iteration,+the+name+is+still+empty.+gotta+omit.
++++++++++++++++++++save(name,+amount);
++++++++++++++++name+=+"";+
++++++++++++++++amount+=+"";
++++++++++++++++name+%2B=+c;+//add+it+to+the+temporary+name
++++++++++++}
++++++++++++else+if+(inBetween(c,+'a',+'z'))
++++++++++++++++name+%2B=+c;+//if+it's+a+lowercase+letter,+add+it+to+the+temporary+name
++++++++++++else+if+(inBetween(c,+'0',+'9'))
++++++++++++++++amount+%2B=+c;
++++++++}
++++++++save(name,+amount);
++++}

++++public+String+toString()+{
++++++++String+s+=+"";
++++++++for+(MoleculePart+part+:+parts)
++++++++++++s+%2B=+part.toString();
++++++++return+s;
++++}

++++//add+part+to+molecule+structure+after+some+parsing
++++private+void+save(String+tmpMoleculename,+String+amount)+{
++++++++MoleculePart+part+=+new+MoleculePart();
++++++++part.amount+=+amount.isEmpty()+?+1+:+Integer.parseInt(amount);
++++++++part.element+=+Element.valueOf(tmpMoleculename);
++++++++parts.add(part);
++++}

++++private+static+boolean+inBetween(char+c,+char+start,+char+end)+{
++++++++return+(c+>=+start+&&+c+<=+end);
++++}
}

public+static+class+MoleculePart+{
++++public+Element+element;
++++public+int+amount;

++++public+String+toString()+{
++++++++return+element.name()+%2B+(amount+>+1+?+amount+:+"");
++++}
}

public+static+enum+Element+{
++++O,+S,+H,+C,+Se,+Ni+//add+as+many+as+you+like
}

public+static+void+main(String[]+args)+{
++++System.out.println(new+Molecule("Ni84OH43Se"));
}|code-block|syntax|javascript|477917|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|$]]

Here's something that parses an entire molecule structure. This one parses integers > 9 as well.

<pre><code>public static class Molecule {
 private List&lt;MoleculePart&gt; parts = new ArrayList&lt;MoleculePart&gt;();

 public Molecule(String s) {
 String name = "";
 String amount = "";
 for (char c : s.toCharArray()) {
 if (inBetween(c, 'A', 'Z')) {
 if (!name.isEmpty()) // first iteration, the name is still empty. gotta omit.
 save(name, amount);
 name = ""; 
 amount = "";
 name += c; //add it to the temporary name
 }
 else if (inBetween(c, 'a', 'z'))
 name += c; //if it's a lowercase letter, add it to the temporary name
 else if (inBetween(c, '0', '9'))
 amount += c;
 }
 save(name, amount);
 }

 public String toString() {
 String s = "";
 for (MoleculePart part : parts)
 s += part.toString();
 return s;
 }

 //add part to molecule structure after some parsing
 private void save(String tmpMoleculename, String amount) {
 MoleculePart part = new MoleculePart();
 part.amount = amount.isEmpty() ? 1 : Integer.parseInt(amount);
 part.element = Element.valueOf(tmpMoleculename);
 parts.add(part);
 }

 private static boolean inBetween(char c, char start, char end) {
 return (c &gt;= start &amp;&amp; c &lt;= end);
 }
}

public static class MoleculePart {
 public Element element;
 public int amount;

 public String toString() {
 return element.name() + (amount &gt; 1 ? amount : "");
 }
}

public static enum Element {
 O, S, H, C, Se, Ni //add as many as you like
}

public static void main(String[] args) {
 System.out.println(new Molecule("Ni84OH43Se"));
}
</code></pre>

I have string H2SO4 for example, how can i parse to int 4 after O? I can't use substring(4), because user can type for example NH3PO4 and there 4 is substring(5), so how can I parse any character that is exactly after O?

Thanks for help.

How to parse character after some character

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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例如，我有字符串H2SO4，如何在O之后解析到int 4？我不能使用子字符串(4)，因为用户可以输入例如NH3PO4，并且有4的子字符串(5)，那么我如何解析任何与O完全相同的字符呢？谢谢你帮忙。

问如何在某个字符之后解析字符
EN

回答 8

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问如何在某个字符之后解析字符EN

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问如何在某个字符之后解析字符
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