我需要将字符串数组转换为字符串的Java代码。


当前回答

试试数组。toString重载方法。

否则,试试下面的通用实现:

public static void main(String... args) throws Exception {

    String[] array = {"ABC", "XYZ", "PQR"};

    System.out.println(new Test().join(array, ", "));
}

public <T> String join(T[] array, String cement) {
    StringBuilder builder = new StringBuilder();

    if(array == null || array.length == 0) {
        return null;
    }

    for (T t : array) {
        builder.append(t).append(cement);
    }

    builder.delete(builder.length() - cement.length(), builder.length());

    return builder.toString();
}

其他回答

你需要从数组列表中生成字符串的代码,

Iterate through all elements in list and add it to your String result

你可以用两种方式:使用String作为结果或使用StringBuffer/StringBuilder。

例子:

String result = "";
for (String s : list) {
    result += s;
}

...但由于性能原因,这不是一个好的实践。更好的方法是使用StringBuffer(线程安全)或StringBuilder,它们更适合添加字符串

你可以这样做,给定一个基本类型的数组a:

StringBuffer result = new StringBuffer();
for (int i = 0; i < a.length; i++) {
   result.append( a[i] );
   //result.append( optional separator );
}
String mynewstring = result.toString();
String[] strings = new String[25000];
for (int i = 0; i < 25000; i++) strings[i] = '1234567';

String result;
result = "";
for (String s : strings) result += s;
//linear +: 5s

result = "";
for (String s : strings) result = result.concat(s);
//linear .concat: 2.5s

result = String.join("", strings);
//Java 8 .join: 3ms

Public String join(String delimiter, String[] s)
{
    int ls = s.length;
    switch (ls)
    {
        case 0: return "";
        case 1: return s[0];
        case 2: return s[0].concat(delimiter).concat(s[1]);
        default:
            int l1 = ls / 2;
            String[] s1 = Arrays.copyOfRange(s, 0, l1); 
            String[] s2 = Arrays.copyOfRange(s, l1, ls); 
            return join(delimiter, s1).concat(delimiter).concat(join(delimiter, s2));
    }
}
result = join("", strings);
// Divide&Conquer join: 7ms

如果你没有选择,只能使用Java 6或7,那么你应该使用分治连接。

public class ArrayToString
{   
    public static void main(String[] args)
    {
        String[] strArray = new String[]{"Java", "PHP", ".NET", "PERL", "C", "COBOL"};
        
        String newString = Arrays.toString(strArray);
        
        newString = newString.substring(1, newString.length()-1);

        System.out.println("New New String: " + newString);
    }
}

我喜欢使用谷歌的番石榴Joiner,例如:

Joiner.on(", ").skipNulls().join("Harry", null, "Ron", "Hermione");

将产生与以下相同的String:

new String("Harry, Ron, Hermione");

ETA: Java 8现在也有类似的支持:

String.join(", ", "Harry", "Ron", "Hermione");

不能看到跳过空值的支持,但这很容易解决。