我想将String转换为Character类的对象数组，但我无法执行转换。我知道我可以使用toCharArray()方法将String转换为原始数据类型“char”的数组，但它无法将String转换为Character类型的对象数组。 / p>
我将如何这样做？
答案 0 :(得分：166)
使用此：
String str = "testString";
char[] charArray = str.toCharArray();
Character[] charObjectArray = ArrayUtils.toObject(charArray);
答案 1 :(得分：54)
一张java-8的衬垫：
String str = "testString";
//[t, e, s, t, S, t, r, i, n, g]
Character[] charObjectArray =
str.chars().mapToObj(c -> (char)c).toArray(Character[]::new);
它的作用是：
IntStream个字符（您可能还想查看codePoints()）Character（您需要强制转换为实际上它是char，然后Java会将其自动封装到Character）toArray() 答案 2 :(得分：32)
为什么不自己写一点方法
public Character[] toCharacterArray( String s ) {
if ( s == null ) {
return null;
}
int len = s.length();
Character[] array = new Character[len];
for (int i = 0; i < len ; i++) {
array[i] = new Character(s.charAt(i));
}
return array;
}
答案 3 :(得分：4)
我希望下面的代码能为您提供帮助。
String s="Welcome to Java Programming";
char arr[]=s.toCharArray();
for(int i=0;i<arr.length;i++){
System.out.println("Data at ["+i+"]="+arr[i]);
}
它正常工作，输出是：
Data at [0]=W
Data at [1]=e
Data at [2]=l
Data at [3]=c
Data at [4]=o
Data at [5]=m
Data at [6]=e
Data at [7]=
Data at [8]=t
Data at [9]=o
Data at [10]=
Data at [11]=J
Data at [12]=a
Data at [13]=v
Data at [14]=a
Data at [15]=
Data at [16]=P
Data at [17]=r
Data at [18]=o
Data at [19]=g
Data at [20]=r
Data at [21]=a
Data at [22]=m
Data at [23]=m
Data at [24]=i
Data at [25]=n
Data at [26]=g
答案 4 :(得分：2)
在这种情况下，您必须编写自己的方法。使用循环并使用charAt(i)获取每个字符，并使用Character[]将其设置为arrayname[i] = string.charAt[i]数组。
答案 5 :(得分：2)
String#toCharArray会返回char数组，您拥有的数组为Character。在大多数情况下，使用char或Character与autoboxing一样无关紧要。你的情况下的问题是数组没有自动装箱，我建议你使用char（char[]）数组。
答案 6 :(得分：2)
此方法将String作为参数并返回Character Array
/**
* @param sourceString
* :String as argument
* @return CharcterArray
*/
public static Character[] toCharacterArray(String sourceString) {
char[] charArrays = new char[sourceString.length()];
charArrays = sourceString.toCharArray();
Character[] characterArray = new Character[charArrays.length];
for (int i = 0; i < charArrays.length; i++) {
characterArray[i] = charArrays[i];
}
return characterArray;
}
答案 7 :(得分：1)
另一种方法。
String str="I am a good boy";
char[] chars=str.toCharArray();
Character[] characters=new Character[chars.length];
for (int i = 0; i < chars.length; i++) {
characters[i]=chars[i];
System.out.println(chars[i]);
}
答案 8 :(得分：0)
如果您正在使用JTextField，那么它可能会有所帮助..
public JTextField display;
String number=e.getActionCommand();
display.setText(display.getText()+number);
ch=number.toCharArray();
for( int i=0; i<ch.length; i++)
System.out.println("in array a1= "+ch[i]);
答案 9 :(得分：0)
String str = "somethingPutHere";
Character[] c = ArrayUtils.toObject(str.toCharArray());
答案 10 :(得分：0)
如果您不想依赖第三方API，则以下是JDK7或更低版本的工作代码。我没有像上面的其他解决方案那样实例化临时字符对象。 foreach循环更具可读性，请看自己：）
public static Character[] convertStringToCharacterArray(String str) {
if (str == null || str.isEmpty()) {
return null;
}
char[] c = str.toCharArray();
final int len = c.length;
int counter = 0;
final Character[] result = new Character[len];
while (len > counter) {
for (char ch : c) {
result[counter++] = ch;
}
}
return result;
}
答案 11 :(得分：0)
我在 java.io 中使用了 StringReader 类。其中一个函数UPDATE SAFETY_ADMIN.SAFETY_USERS tbl1
SET (SUPERVISOR_ID,DEPT_ID,USER_ID,EMPLOYEE_TYPE,EMPLOYEE_NAME,EMAIL,MODIFIED_BY,MODIFIED_ON) =
(
SELECT SAFETY_ADMIN.FN_GETSUPERVISORID(tbl2.SUPID),CAST(tbl2.DEPTID AS NUMBER(19)),UPPER(SUBSTR(tbl2.EMAIL,1,INSTR(tbl2.EMAIL,'@')-1)),tbl2.EMPTYPE,tbl2.EMPNAME,tbl2.EMAIL,SYS_CONTEXT('USERENV', 'OS_USER'),CURRENT_TIMESTAMP
FROM PS_LOAD.EMPLOYEEDATA tbl2
WHERE tbl1.EMPLOYEE_ID = CAST(tbl2.EMPID AS NUMBER(19)) AND
(
tbl1.SUPERVISOR_ID <> SAFETY_ADMIN.FN_GETSUPERVISORID(tbl2.SUPID) OR
tbl1.DEPT_ID <> CAST(tbl2.DEPTID AS NUMBER(19)) OR
tbl1.USER_ID <> UPPER(SUBSTR(tbl2.EMAIL,1,INSTR(tbl2.EMAIL,'@')-1)) OR
tbl1.EMPLOYEE_TYPE <> tbl2.EMPTYPE OR
tbl1.EMPLOYEE_NAME <> tbl2.EMPNAME OR
tbl1.EMAIL <> tbl2.EMAIL
)
) where exists
(
SELECT 1
FROM PS_LOAD.EMPLOYEEDATA tbl3
WHERE tbl1.EMPLOYEE_ID = CAST(tbl3.EMPID AS NUMBER(19)) AND
(
tbl1.SUPERVISOR_ID <> SAFETY_ADMIN.FN_GETSUPERVISORID(tbl3.SUPID) OR
tbl1.DEPT_ID <> CAST(tbl3.DEPTID AS NUMBER(19)) OR
tbl1.USER_ID <> UPPER(SUBSTR(tbl3.EMAIL,1,INSTR(tbl3.EMAIL,'@')-1)) OR
tbl1.EMPLOYEE_TYPE <> tbl3.EMPTYPE OR
tbl1.EMPLOYEE_NAME <> tbl3.EMPNAME OR
tbl1.EMAIL <> tbl3.EMAIL
)
将字符串的内容读入数组。
read(char[] cbuf)运行它会为您提供输出：
String str = "hello";
char[] array = new char[str.length()];
StringReader read = new StringReader(str);
try {
read.read(array); //Reads string into the array. Throws IOException
} catch (IOException e) {
e.printStackTrace();
}
for (int i = 0; i < str.length(); i++) {
System.out.println("array["+i+"] = "+array[i]);
}
答案 12 :(得分：0)
将字符串转换为字符数组，然后将字符数组转换回字符串
//Givent String
String given = "asdcbsdcagfsdbgdfanfghbsfdab";
//Converting String to Character Array(It's an inbuild method of a String)
char[] characterArray = given.toCharArray();
//returns = [a, s, d, c, b, s, d, c, a, g, f, s, d, b, g, d, f, a, n, f, g, h, b, s, f, d, a, b]
//Converting back Character array to String
int length = Arrays.toString(characterArray).replaceAll("[, ]","").length();
//First Way to get the string back
Arrays.toString(characterArray).replaceAll("[, ]","").substring(1,length-1)
//returns asdcbsdcagfsdbgdfanfghbsfdab
or
// Second way to get the string back
Arrays.toString(characterArray).replaceAll("[, ]","").replace("[","").replace("]",""))
//returns asdcbsdcagfsdbgdfanfghbsfdab
答案 13 :(得分：0)
String[] arr = { "abc", "cba", "dac", "cda" };
Map<Character, Integer> map = new HashMap<>();
String string = new String();
for (String a : arr) {
string = string.concat(a);
}
System.out.println(string);
for (int i = 0; i < string.length(); i++) {
if (map.containsKey(string.charAt(i))) {
map.put(string.charAt(i), map.get(string.charAt(i)) + 1);
} else {
map.put(string.charAt(i), 1);
}
}
System.out.println(map);
//输出{a=4, b=2, c=4, d=2}