次のコードでハッシュマップを反復した後に取得した値にアクセスできるかどうか疑問に思っています(マップに格納されていることはわかっていますが、ループの外で使用したい)。キーと対応する値は、ループ内で繰り返されます。同時ハッシュマップは、値を取得してループ外で使用するのに役立ちますか?
ありがとうございました。
public static void main(String[] args) {
Map<String, List<String>> maleMap = new LinkedHashMap<String, List<String>>();
Map<String, List<String>> femaleMap = new LinkedHashMap<String, List<String>>();
try {
Scanner scanner = new Scanner(new FileReader(.txt));
while (scanner.hasNextLine()) {
String nextLine = scanner.nextLine();
String[] column = nextLine.split(":");
if (column[0].equals("male") && (column.length == 4)) {
maleMap.put(column[1],
Arrays.asList(column[2], column[3]));
} else if (column[0].equals("female") && (column.length == 4)) {
femaleMap.put(column[1],
Arrays.asList(column[2], column[3]));
}
}
Set<Entry<String, List<String>>> entries = maleMap.entrySet();
Iterator<Entry<String, List<String>>> entryIter = entries
.iterator();
while (entryIter.hasNext()) {
Map.Entry entry = (Map.Entry) entryIter.next();
Object key = entry.getKey(); // Get the key from the entry.
List<String> value = (List<String>) entry.getValue();
Object value1 = " ";
Object value2 = " ";
int counter = 0;
for (Object listItem : (List) value) {
Writer writer = null;
Object Name = key;
Object Age = null;
Object ID = null;
if (counter == 0) {// first pass assign value to value1
value1 = listItem;
counter++;// increment for next pass
} else if (counter == 1) {// second pass assign value to
// value2
value2 = listItem;
counter++;// so we dont keep re-assigning listItem for
// further iterations
}
}
System.out.println(key + ":" + value1 + "," + value2);
scanner.close();
Writer writer = null;
Object Name = key;
Object Age = value1;
Object ID = value2;
try {
String filename = ".txt";
FileWriter fw = new FileWriter(filename, true);
fw.write("# Table" + Name + "\n" + "map:"+ Name + " a d2rq:ClassMap;" + "\n"
+ " dataStorage map:database;" + "\n"+ "Pattern " +"\""+ Name + "/@@"+ Age +
"." + ID + "@@\";" + "\n"+ " class :" + Name +";"+"\n"+ " ."+"\n");//
fw.close();
} catch (Exception e) {
e.printStackTrace();
}
}