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

您可以使用拆分方法:

public class Demo {
    public static void main(String args[]) {
        String str = "004-034556";

        if ((str.contains("-"))) {
            String[] temp = str.split("-");
            for (String part:temp) {
                System.out.println(part);
            }
        }
        else {
            System.out.println(str + " does not contain \"-\".");
        }
    }
}

其他回答

String s="004-034556";
for(int i=0;i<s.length();i++)
{
    if(s.charAt(i)=='-')
    {
        System.out.println(s.substring(0,i));
        System.out.println(s.substring(i+1));
    }
}

正如大家所提到的,split()是在您的案例中可能使用的最佳选项。另一种方法可以是使用substring()。

String str="004-034556"
String[] sTemp=str.split("-");// '-' is a delimiter

string1=004 // sTemp[0];
string2=034556//sTemp[1];

这些要求为解释留下了空间。我建议写一个方法,

public final static String[] mySplit(final String s)

其封装了该功能。当然,您可以使用String.split(..),如实现的其他答案中所述。

您应该为输入字符串以及期望的结果和行为编写一些单元测试。

优秀的考生应包括:

 - "0022-3333"
 - "-"
 - "5555-"
 - "-333"
 - "3344-"
 - "--"
 - ""
 - "553535"
 - "333-333-33"
 - "222--222"
 - "222--"
 - "--4555"

通过定义相应的测试结果,您可以指定行为。

例如,如果“-333”应在[,333]中返回,或者如果它是一个错误。“333-333-33”是否可以在[333333-33]或[3333-333,33]中分开,或者这是一个错误?等等

String s = "TnGeneral|DOMESTIC";
String a[]=s.split("\\|");
System.out.println(a.toString());
System.out.println(a[0]);
System.out.println(a[1]);

输出:

TnGeneral
DOMESTIC

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>
 * &nbsp;&nbsp;<code>separate("12-34", '-') = "12", "34"</code><br>
 * &nbsp;&nbsp;<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;
}