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。