java - 使用枚举来清晰地表示错误消息——这是好的做法吗？

Question

我想将我的错误消息和内容合并到一个文件中，并尽可能使我的代码更具可读性。

这是我的枚举文件中的示例:

public enum ZipErrorType {

// START: define exception messages (alphabetical order)
EMPTY_FILE_NAME_IN_LIST {
    public String toString() {
        return "One or more null/empty filename(s) found";
    }
},

FILE_DOESNT_EXIST {
    public String who(String sThisFile) {
        return "[" + sThisFile + "] does not exist";
    }
},

FILE_LIST_IS_NULL {
    public String toString() {
        return "File list is null/empty";
    }
},

FILENAME_NOT_ABSOLUTE {
    public String who(String sThisFile) {
        return "[" + sThisFile + "] is not absolute";
    }
},

MUST_BE_DIR {
    public String who(String sThisFile) {
        return "[" + sThisFile + "] must be a directory";
    }
},

MUST_BE_FILE {
    public String who(String sThisFile) {
        return "[" + sThisFile + "] must be a file";
    }
},

NULL_OR_EMPTY {
    public String who(String sThisFile) {
        return "[" + sThisFile + "] is null/empty";
    }
},

OUTPUT_FILE_ALREADY_EXISTS {
    public String who(String sThisFile) {
        return "[" + sThisFile + "] already exists";
    }
},

OUTPUT_FILENAME_EMPTY {
    public String toString() {
        return "Output filename is null/empty";
    }
},

OUTPUT_PATH_EMPTY {
    public String toString() {
        return "Output path is null/empty";
    }
},
// END: define exception messages

NONE {};

public String who(String sThisFile) { return ""; }
}

然后在我的程序中我有这样的代码:

private static ZipErrorType getFileErrorsIfAny(String sFilename, boolean shouldBeFile) {

    // check if given filename is absolute
    File file = new File(sFilename);
    if (!file.isAbsolute()) {
        return ZipErrorType.FILENAME_NOT_ABSOLUTE;
    }

    // check if file exists
    if (!file.exists()) {
        return ZipErrorType.FILE_DOESNT_EXIST;
    }

    // check if corresponding file is a file when it shouldn't be...
    if (file.isFile() && !shouldBeFile) {
        return ZipErrorType.MUST_BE_DIR;
    }
    // ...or a directory when it should be a file
    else if (file.isDirectory() && shouldBeFile) {
        return ZipErrorType.MUST_BE_FILE;
    }

    return ZipErrorType.NONE;
}

...以及我如何使用枚举的示例:

    // check input files
    for (String sFile : files) {
        if (sFile == null || sFile.trim().length() == 0) {
            throw new NullPointerException("One or more filename is null/empty");
        }

        errorIfAny = getFileErrorsIfAny(sFile.trim(), true); 
        if (!errorIfAny.equals(ZipErrorType.NONE)) {
            throw new ZipInputException(errorIfAny.who(sFile.trim()));
        }
    }

现在我知道仅凭这些代码片段很难判断，但从一般角度来看，这样可以吗？我正在做的事情不值得麻烦吗？有什么方法可以改进吗？

Answer 1

我建议使用简单的字符串模板而不是枚举来构建错误消息。像这样:

String EMPTY_FILE_NAME_IN_LIST_TEMPLATE = "One or more null/empty filename(s) found";
String FILE_DOESNT_EXIST_TEMPLATE = "[ %s ] does not exist";
String FILE_LIST_IS_NULL_TEMPLATE = "File list is null/empty";
String FILENAME_NOT_ABSOLUTE_TEMPLATE = "[ %s ] is not absolute";
String MUST_BE_DIR_TEMPLATE = "[ %s ] must be a directory";
String MUST_BE_FILE_TEMPLATE = "[ %s ] must be a file";
String NULL_OR_EMPTY_TEMPLATE = "[ %s ] is null/empty";
String OUTPUT_FILE_ALREADY_EXISTS_TEMPLATE = "[ %s ] already exists";
String OUTPUT_FILENAME_EMPTY_TEMPLATE = "Output filename is null/empty";
String OUTPUT_PATH_EMPTY_TEMPLATE = "Output path is null/empty";

然后，使用 String.format(template, sFilename) 构建实际消息。

您还可以考虑直接从 getFileErrorsIfAny() 方法中抛出异常:

File file = new File(sFilename);
if (!file.isAbsolute()) {
    throw new ZipInputException(String.format(FILENAME_NOT_ABSOLUTE_TEMPLATE, sFilename));
}

对我来说看起来更干净、更紧凑。