下面给出的代码需要两个不同的输入，但是我只想传递单个输入，即文件夹“ test”的路径以及其余功能。还有要生成的final.tbl应该在相同的输入文件夹路径中生成：
public class Migrator {
private static final String KEY1 = "post_tran_id";
private static final String KEY2 = "post_tran_cust_id";
void migrate(String post_tran, String post_tran_cust) throws IOException {
Map<String, Map<String, String>> h1 = loadFile(post_tran, KEY1);
Map<String, Map<String, String>> h2 = loadFile(post_tran_cust, KEY2);
PrintStream out = new PrintStream("final.tbl");
for (Map.Entry<String, Map<String, String>> entry : h1.entrySet()) {
Map<String, String> data = entry.getValue();
String k = data.get(KEY2);
if (k != null && h2.containsKey(k)) {
print(out, KEY1, data.get(KEY1));
print(out, KEY2, data.get(KEY2));
// Print remaining rows in any order
for (String key : data.keySet()) {
if ( ! key.equals(KEY1) && ! key.equals(KEY2) ) {
print(out, key, data.get(key));
}
}
data = h2.get(k);
for (String key : data.keySet()) {
if ( ! key.equals(KEY2) ) {
print(out, key, data.get(key));
}
}
out.println(); // Record separator
}
}
}
private void print(PrintStream out, String key, String data) {
out.print("[name]");
out.print(key);
out.print("[/name]");
out.print("=");
out.print("[data]");
out.print(data);
out.print("[/data]");
out.println();
}
private Map<String, Map<String, String>> loadFile(String fileName, String key) throws
IOException {
Map<String, Map<String, String>> result = new HashMap<String, Map<String, String>>
();
BufferedReader br = new BufferedReader(new FileReader(fileName));
String line;
do {
Map<String, String> data = new HashMap<String, String>();
while ((line = br.readLine()) != null && !line.isEmpty()) {
data.put(getKey(line), getData(line));
}
result.put(data.get(key), data);
} while (line != null);
br.close();
return result;
}
private String getKey(String line) {
String[] tokens = line.split("=");
int length = tokens[0].length();
return tokens[0].substring(6, length - 7);
}
private String getData(String line) {
String[] tokens = line.split("=");
int length = tokens[1].length();
return tokens[1].substring(6, length - 7);
}
public static void main(String[] args) throws IOException { Migrator mg =
new Migrator();
mg.migrate("D:\\test\\post_tran.tbl",
"D:\\test\\post_tran_cust.tbl"); }
}
为了使您的migrate方法采用1个参数但能够使用许多路径，您始终可以将所有路径附加到一个字符串中，并在migrate方法中解析它们。
例子：
String appendedArgument = "D:\\test\\post_tran.tbl;D:\\test\\post_tran_cust.tbl";
请注意，用分号分隔两条路径。
然后，您可以调用方法：
mg.migrate(appendedArgument);
并在另一侧进行解析：
void migrate(String argument) throws IOException
{
String[] splitArgument.split(";");
String post_tran = splitArgument[0];
String post_tran_cust = splitArgument[1];
Map<String, Map<String, String>> h1 = loadFile(post_tran, KEY1);
Map<String, Map<String, String>> h2 = loadFile(post_tran_cust, KEY2);
}
使用这种方法，您可以根据需要向迁移方法中发送尽可能多的路径，这使您（在这种情况下）也可以将路径存储在您要存储final.tbl文件的位置。
这将使appendedArgument字符串看起来像：
String appendedArgument = "D:\\test\\;D:\\test\\post_tran.tbl;D:\\test\\post_tran_cust.tbl";
然后，您需要在migration方法中相应地对其进行解析。
本文收集自互联网，转载请注明来源。
如有侵权，请联系[email protected] 删除。
我来说两句