将String数组转换为int值并将其存储在int数组中

时间：2015-04-27 13:17:13

标签： java arrays string file int

我有这部分代码：

int[] myIntArray = {0x2e80,0x008c,0x0993,0x09c5,0x058b,0x4c9c,0x0390,0x1e96,0x0989,0x0ac4,0x4cad,0x0d93,0x09c5,0x0a84,0x0591,0x04c5,0x058b,0x4c9c,0x0390,0x1ec5,0x0d87,0x0589,0x0591,0x0580,0x1fc4,0x4cb2,0x0591,0x048a,0x1991,0x4c84,0x4c8d,0x1988,0x0e89,0x09c5,0x0e90,0x18c5,0x1e80,0x0d96,0x038b,0x0d87,0x0080,0x4c86,0x038b,0x0a8c,0x0880,0x0286,0x09c5,0x058b,0x4c9c,0x0390,0x1ec5,0x0392,0x02c5,0x1c8a,0x1b80,0x1e96,0x4c9c,0x0390,0x4c86,0x0d8b,0x028a,0x18c5,0x0e80,0x4c96,0x1986,0x0f80,0x1f96,0x0a90,0x00c5,0x0397,0x4c8d,0x0d95,0x1c9c,0x42e5};
for (int i=0; i<=73; i++){
  String s1=Decrypt(k,myIntArray[i]);
  String s2= s1.substring(2,6);
  String s=convertHexToString(s2);
  System.out.print(s);
}

从数组中获取十六进制值并对其执行一些操作。它的工作正常，因为我想要。

我想做同样的事情，但我想从文件中读取值并对其执行相同的操作，我试过这个：

String token1 = "";
Scanner inFile1 = new Scanner(new    File("chipertext.txt")).useDelimiter(",\\s*");
List<String> temps = new ArrayList<String>();
while (inFile1.hasNext()) {
  token1 = inFile1.next();
  temps.add(token1);
}
inFile1.close();
String[] tempsArray = temps.toArray(new String[74]);
int[] myIntArray = new int[tempsArray.length];

for (int i = 0; i < tempsArray.length; i++) {
  myIntArray[i] = Integer.parseInt(tempsArray[i]);
}
for (int i=0; i<=73; i++){
  String s1=Decrypt(k,myIntArray[i]);
  String s2= s1.substring(2,6);
  String s=convertHexToString(s2);
  System.out.print(s);
}

但是我收到了这个错误：

Exception in thread "main" java.lang.NumberFormatException: For input string: "0x2e80  0x2e80
0x008c
0x0993
0x09c5
0x058b
0x4c9c
0x0390
0x1e96
0x0989
0x0ac4
0x4cad
0x0d93
0x09c5
0x0a84
0x0591
0x04c5
0x058b
0x4c9c
0x0390
0x1ec5
0x0d87
0x0589
0x0591
0x0580"

存储在文件中的值如下：

0x2e80
0x008c
0x0993
0x09c5
0x058b
0x4c9c
0x0390
0x1e96
0x0989
0x0ac4
0x4cad
0x0d93
0x09c5
0x0a84
0x0591
0x04c5
0x058b
0x4c9c
0x0390
0x1ec5
0x0d87
0x0589
0x0591
0x0580

我认为这意味着该字符串不能存储为整数吗？那怎么办呢？以及它之前如何存储在整数数组中？！我不知道有人可以帮帮我吗？

工作代码 

String token1 = "";
Scanner inFile1 = new Scanner(new File("chipertext.txt"));
List<String> temps = new ArrayList<String>();
while (inFile1.hasNext()) {
token1 = inFile1.next();
temps.add(token1);
}
inFile1.close();
String[] tempsArray = temps.toArray(new String[73]);
int[] myIntArray = new int[tempsArray.length];
for (int i = 0; i < tempsArray.length; i++) {
myIntArray[i] = Integer.parseInt(tempsArray[i].substring(2), 16);
}
for (int i=0; i<=73; i++){
   String s1=Decrypt(k,myIntArray[i]);
   String s2= s1.substring(2,6);
   String s=convertHexToString(s2);
   System.out.print(s);
}

谢谢大家的帮助!!!

5 个答案:

答案 0 :(得分：1)

这样可以使您的所有数字格式相同

String[] tempsArray = temps.toArray(new String[74]);
int[] myIntArray = new int[tempsArray.length];
for (int i = 0; i < tempsArray.length; i++) {
  myIntArray[i] = Integer.parseInt(tempsArray[i].substring(2), 16);
  System.out.println(myIntArray[i]);
}

如果它不起作用，则打印出进入循环的内容，看看你是否正确解析文件。从您解析的文件中获取的字符串数组中打印一个项目，然后将其输入到该文件中。

    String hex = "0x4c9c";
    int value = Integer.parseInt(hex.substring(2), 16);
    System.out.println(value);

打印出文件解析的输出，很可能它与您期望的不匹配。 

List<String> temps = new ArrayList<String>();
while (inFile1.hasNext()) {
  token1 = inFile1.next();
  temps.add(token1);

  System.out.println(token1); //Check this output. Is it a hex string?
}
inFile1.close();

答案 1 :(得分：0)

您发布的代码有两种可能性

  1. 您将不同的十六进制代码存储在文件

    中

    //而不是
    0x2e80
    //作为
    2e80

    //并修改代码以指定基数
    myIntArray [i] = Integer.parseInt（tempsArray [i]，16）;

  2. 或将它们存储为当前

    //并修改代码以删除前缀并指定基数
    myIntArray [i] = Integer.parseInt（tempsArray [i] .substring（2），16）;

    3. 查看两种解决方案的代码段

    案例1的

    。）

    List<String> temps = new ArrayList<>();
temps.add("2e80");
String[] tempsArray = temps.toArray(new String[74]);
int parseInt = Integer.parseInt(tempsArray[0], 16);
System.out.println("parseInt = " + parseInt);
    案例2的

    。）

    List<String> temps = new ArrayList<>();
temps.add("0x2e80");
String[] tempsArray = temps.toArray(new String[74]);
int parseInt = Integer.parseInt(tempsArray[0].substring(2), 16);
System.out.println("parseInt = " + parseInt);

    两个片段的输出

    parseInt = 11904

    修改

    OPs代码的固定版本

    // import java.nio.charset.Charset;
// import java.nio.file.Files;
// import java.nio.file.Paths;
// import java.util.List;

List<String> lines = Files.readAllLines(Paths.get("chipertext.txt"), 
        Charset.defaultCharset());
int[] myIntArray = new int[lines.size()];
for (int i = 0; i < myIntArray.length; i++) {
    myIntArray[i] = Integer.parseInt(lines.get(i).substring(2), 16);
}

答案 2 :(得分：0)

请看Tomasz Nurkiewicz的答案：Parsing a Hexadecimal String to an Integer throws a NumberFormatException?

要恢复，您需要使用

Integer.parseInt(tempsArray[i], 16)

radix参数设置为16，告诉parseInt解析String的hexa值。在此之前，您需要删除每个＆＃34; 0x＆＃34;

答案 3 :(得分：0)

假设您正在存储 Protected Sub txtIdType_TextChanged(sender As Object, e As EventArgs) Dim txtb As TextBox = CType(FormViewPerson.FindControl("txtIdType"), TextBox) Dim txtb1 As TextBox = CType(FormViewPerson.FindControl("TextBoxIDCode"), TextBox) If (txtb.Text = "Leternjoftimi" OrElse txtb.Text = "KosovoIDCard" OrElse txtb.Text = "Licna karta") Then txtb1.MaxLength = 10 End If End Sub 以将其表示为十六进制数，您可以使用接受0x参数的Integer.parseInt方法。基数10表示小数，基数16表示十六进制。并且您可以使用子字符串String of String来删除radix噪音。

所以解决方案如下所示

0x

当我运行它时，我打印输出如下

 System.out.println(Integer.parseInt("-FF", 16) ); // prints -255 

        String [] hexaDecimalNumbers = {"0x2e80","0x008c","0x0993","0x09c5","0x058b","0x4c9c","0x0390","0x1e96","0x0989","0x0ac4","0x4cad","0x0d93","0x09c5","0x0a84","0x0591","0x04c5","0x058b","0x4c9c","0x0390","0x1ec5","0x0d87","0x0589","0x0591","0x0580","0x1fc4","0x4cb2","0x0591","0x048a","0x1991","0x4c84","0x4c8d","0x1988","0x0e89","0x09c5","0x0e90","0x18c5","0x1e80","0x0d96","0x038b","0x0d87","0x0080","0x4c86","0x038b","0x0a8c","0x0880","0x0286","0x09c5","0x058b","0x4c9c","0x0390","0x1ec5","0x0392","0x02c5","0x1c8a","0x1b80","0x1e96","0x4c9c","0x0390","0x4c86","0x0d8b","0x028a","0x18c5","0x0e80","0x4c96","0x1986","0x0f80","0x1f96","0x0a90","0x00c5","0x0397","0x4c8d","0x0d95","0x1c9c","0x42e5"};
        for(String hexaDecimal: hexaDecimalNumbers) {
            System.out.print(Integer.parseInt(hexaDecimal.substring(2), 16) + " , ");
        }

答案 4 :(得分：0)

而不是11904 , 140 , 2451 , 2501 , 1419 , 19612 , 912 , 7830 , 2441 , 2756 , 19629 , 3475 , 2501 , 2692 , 1425 , 1221 , 1419 , 19612 , 912 , 7877 , 3463 , 1417 , 1425 , 1408 , 8132 , 19634 , 1425 , 1162 , 6545 , 19588 , 19597 , 6536 , 3721 , 2501 , 3728 , 6341 , 7808 , 3478 , 907 , 3463 , 128 , 19590 , 907 , 2700 , 2176 , 646 , 2501 , 1419 , 19612 , 912 , 7877 , 914 , 709 , 7306 , 7040 , 7830 , 19612 , 912 , 19590 , 3467 , 650 , 6341 , 3712 , 19606 , 6534 , 3968 , 8086 , 2704 , 197 , 919 , 19597 , 3477 , 7324 , 17125 , 使用Integer.parseInt(tempsArray[i]);

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