我想通过分隔符“-”将字符串“004-034556”拆分为两个字符串:
part1 = "004";
part2 = "034556";
这意味着第一个字符串将包含“-”之前的字符,第二个字符串将包括“-”之后的字符。
我还想检查字符串中是否有“-”。
我想通过分隔符“-”将字符串“004-034556”拆分为两个字符串:
part1 = "004";
part2 = "034556";
这意味着第一个字符串将包含“-”之前的字符,第二个字符串将包括“-”之后的字符。
我还想检查字符串中是否有“-”。
当前回答
我查看了所有答案,发现所有答案都是第三方许可或基于正则表达式的。
下面是我使用的一个很好的哑实现:
/**
* Separates a string into pieces using
* case-sensitive-non-regex-char-separators.
* <p>
* <code>separate("12-34", '-') = "12", "34"</code><br>
* <code>separate("a-b-", '-') = "a", "b", ""</code>
* <p>
* When the separator is the first character in the string, the first result is
* an empty string. When the separator is the last character in the string the
* last element will be an empty string. One separator after another in the
* string will create an empty.
* <p>
* If no separators are set the source is returned.
* <p>
* This method is very fast, but it does not focus on memory-efficiency. The memory
* consumption is approximately double the size of the string. This method is
* thread-safe but not synchronized.
*
* @param source The string to split, never <code>null</code>.
* @param separator The character to use as splitting.
* @return The mutable array of pieces.
* @throws NullPointerException When the source or separators are <code>null</code>.
*/
public final static String[] separate(String source, char... separator) throws NullPointerException {
String[] resultArray = {};
boolean multiSeparators = separator.length > 1;
if (!multiSeparators) {
if (separator.length == 0) {
return new String[] { source };
}
}
int charIndex = source.length();
int lastSeparator = source.length();
while (charIndex-- > -1) {
if (charIndex < 0 || (multiSeparators ? Arrays.binarySearch(separator, source.charAt(charIndex)) >= 0 : source.charAt(charIndex) == separator[0])) {
String piece = source.substring(charIndex + 1, lastSeparator);
lastSeparator = charIndex;
String[] tmp = new String[resultArray.length + 1];
System.arraycopy(resultArray, 0, tmp, 1, resultArray.length);
tmp[0] = piece;
resultArray = tmp;
}
}
return resultArray;
}
其他回答
// This leaves the regexes issue out of question
// But we must remember that each character in the Delimiter String is treated
// like a single delimiter
public static String[] SplitUsingTokenizer(String subject, String delimiters) {
StringTokenizer strTkn = new StringTokenizer(subject, delimiters);
ArrayList<String> arrLis = new ArrayList<String>(subject.length());
while(strTkn.hasMoreTokens())
arrLis.add(strTkn.nextToken());
return arrLis.toArray(new String[0]);
}
可以使用Split():
import java.io.*;
public class Splitting
{
public static void main(String args[])
{
String Str = new String("004-034556");
String[] SplittoArray = Str.split("-");
String string1 = SplittoArray[0];
String string2 = SplittoArray[1];
}
}
否则,您可以使用StringTokenizer:
import java.util.*;
public class Splitting
{
public static void main(String[] args)
{
StringTokenizer Str = new StringTokenizer("004-034556");
String string1 = Str.nextToken("-");
String string2 = Str.nextToken("-");
}
}
这:
String[] out = string.split("-");
应该做你想做的事。string类有许多方法可用于字符串操作。
查看javadoc上String类中的split()方法。
https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#split(java.lang.String)
String data = "004-034556-1212-232-232";
int cnt = 1;
for (String item : data.split("-")) {
System.out.println("string "+cnt+" = "+item);
cnt++;
}
这里有许多拆分字符串的示例,但我很少优化代码。
import java.io.*;
public class BreakString {
public static void main(String args[]) {
String string = "004-034556-1234-2341";
String[] parts = string.split("-");
for(int i=0;i<parts.length;i++) {
System.out.println(parts[i]);
}
}
}