java - 在 Java 中将子字符串转换为 Int

Question

List<Book> myLibrary = new ArrayList <Book>(312); 
public BookLibrary (Scanner inFile){
    while(inFile.hasNextLine()){
     String aBook = inFile.nextLine();
     String title = aBook.substring(0, 32).trim();
     String language = aMovie.substring(33, 40).trim();
     String rating = aMovie.substring(41, 50).trim();
     int hardCover = Integer.parseInt(aBook.substring(51, 55).trim()); 
     int paperBack = Integer.parseInt(aBook.substring(56));
     myLibrary.add(new  Book (title, language, rating,
     hardCover, paperBack));
     count++;
    }

这里只是一些示例代码。我试图找出为什么当我尝试运行我的程序时收到此运行时错误。我的程序中没有编译器错误。

Exception in thread "main" java.lang.NumberFormatException: For input string: ""
    at java.lang.NumberFormatException.forInputString(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at MovieCollection.<init>(BookLibrary.java:21)

Answer 1

文件格式有错误(缺少某些值)或索引有错误(子字符串方法的参数)，请将此方法调用纠正为适当的:

substring(51, 55)
substring(56)

现在其中一个返回空字符串。

如果hardCower或papperBack可以为空，则使用try/catch构造:

int hardCover = 0;
int paperBack = 0;
try{
  hardCover = Integer.parseInt(aBook.substring(51, 55).trim());
}catch(NumberFormatException nfe){}

try{
  paperBack = Integer.parseInt(aBook.substring(56));
}catch(NumberFormatException nfe){}

或进行检查

String toInt1 = aBook.substring(51, 55).trim();
if(toInt1.length > 0){
  hardCover = Integer.parseInt(toInt1);
}