以预定义的格式转换为NSString并转换

Question

我需要以预定义的格式转换为字符串格式，然后转换回原始字典。字符串必须是预定格式的。例如：

我所做的非常乏味：

将字典转换为字符串：

+ (NSString *)cipherTextWithMessage:(Message *)message {
    NSString *entity = @"Message";
    NSString *entityID = message.entityID;
    NSString *roomID = message.room.entityID;
    NSString *userID = message.sender.userID;
    NSString *datetime = [[self private_dateFormatter] stringFromDate:message.datetime];
    NSNumber *selfDestructive = message.selfDestructive;
    NSString *status = message.status;
    NSString *type = message.type;
    NSString *originalBlobID = message.originalBlobID;
    NSString *thumbnailBlobID = message.thumbnailBlobID;
    NSString *text = message.text;
    NSString *posx = [Format posStringWithNumber:message.posx];
    NSString *posy = [Format posStringWithNumber:message.posy];
    NSString *link = message.link;
    NSString *plainText = [NSString stringWithFormat:@"entity=%@ entityID=%@ roomID=%@ userID=%@ datetime=%@ selfDestructive=%@ status=%@ type=%@ originalBlobID=%@ thumbnailBlobID=%@ posx=%@ posy=%@ link=%@ text=%@", entity, entityID, roomID, userID, datetime, selfDestructive, status, type, originalBlobID, thumbnailBlobID, posx, posy, link, text];
    return [self private_encrypt:plainText];
}

将字符串转换为字典

+ (NSMutableDictionary *)infoWithCipherText:(NSString *)cipherText {
    NSString *plainText = [self private_decrypt:cipherText];
    if ([plainText hasPrefix:[NSString stringWithFormat:@"entity=%@", @"Message"]]) {
        NSArray *array = [plainText componentsSeparatedByString:@" "];
        NSString *entity = [((NSString *)array[0]) substringFromIndex:@"entity=".length];
        NSString *entityID = [((NSString *)array[1]) substringFromIndex:@"entityID=".length];
        NSString *roomID = [((NSString *)array[2]) substringFromIndex:@"roomID=".length];
        NSString *userID = [((NSString *)array[3]) substringFromIndex:@"userID=".length];
        NSString *datetime = [((NSString *)array[4]) substringFromIndex:@"datetime=".length];
        NSString *selfDestructive = [((NSString *)array[5]) substringFromIndex:@"selfDestructive=".length];
        NSString *status = [((NSString *)array[6]) substringFromIndex:@"status=".length];
        NSString *type = [((NSString *)array[7]) substringFromIndex:@"type=".length];
        NSString *originalBlobID = [((NSString *)array[8]) substringFromIndex:@"originalBlobID=".length];
        NSString *thumbnailBlobID = [((NSString *)array[9]) substringFromIndex:@"thumbnailBlobID=".length];

        NSString *posx = [((NSString *)array[10]) substringFromIndex:@"posx=".length];
        NSString *posy = [((NSString *)array[11]) substringFromIndex:@"posy=".length];
        NSString *link = [((NSString *)array[12]) substringFromIndex:@"link=".length];

        NSString *pref = [NSString stringWithFormat:@"entity=%@ entityID=%@ roomID=%@ userID=%@ datetime=%@ selfDestructive=%@ status=%@ type=%@ originalBlobID=%@ thumbnailBlobID=%@ posx=%@ posy=%@ link=%@ text=", entity, entityID, roomID, userID, datetime, selfDestructive, status, type, originalBlobID, thumbnailBlobID, posx, posy, link];
        NSString *text = [plainText substringFromIndex:pref.length];
        NSMutableDictionary *info = [NSMutableDictionary dictionary];
        info[@"entity"] = entity;
        info[@"entityID"] = entityID;
        info[@"roomID"] = roomID;
        info[@"userID"] = userID;
        info[@"datetime"] = [[self private_dateFormatter] dateFromString:datetime];
        info[@"selfDestructive"] = [NSNumber numberWithBool:selfDestructive.boolValue];
        info[@"status"] = status;
        info[@"type"] = type;
        info[@"originalBlobID"] = originalBlobID;
        info[@"thumbnailBlobID"] = thumbnailBlobID;
        info[@"posx"] = [Format posWithPosString:posx];
        info[@"posy"] = [Format posWithPosString:posy];
        info[@"link"] = link;
        info[@"text"] = text;
        return info;
    }
}

1. 对于第一个功能，我可能会写入dictionary[@"entityID"] = message.entityID。因此，这仍在将字典转换为字符串问题。

2. 我不想为我所有的实体写这个，有什么简单的方法吗？

3. 最后一个属性是"文本"，它可以包含任何字符，包括'='和空白。这就是为什么我将其放在最后并构造'前缀字符串。但这很乏味。

4. 订单无关紧要。但是，如果按字母顺序排列，那将是一件好事。

Answer 1

类似的东西：

字典的字典：

NSMutableString *result = [NSMutableString string];
NSDictionary *dict = ... // your dictionary
for (NSString *key in [dict allKeys]) {
    id value = dict[key];
    if (result.length) {
        [result appendString:@" "];
    }
    [result appendFormat:@"%@=%@", key, [value description]];
}
// result has the desired string

字典字符串：

NSMutableDictionary dict = [NSMutableDictionary dictionary];
NSString *text = ... // the string created from the above
NSArray *array = [plainText componentsSeparatedByString:@" "];
for (NSString *pair in array) {
    NSArray *parts = [pair componentsSeparatedByString:@"="];
    dict[parts[0]] = parts[1];
}
// dict is now populated with the keys and values

Answer 2

尚不清楚您最终要做什么，但在我看来，您可能会过度思考。我在这里要做的是将字典直接转换为plist-因为此功能是在构建的（例如，这就是writeToFile:所做的）。您可以写入文件并读取文件为NSSTRING，PRESTO，您的字典有一个XML版本。

另外，如果您致电NSKeyedArchiver archivedDataWithRootObject:，则可以单个步骤将字典转换为NSDATA，适合加密并发送到另一台计算机

无论哪种方式，反向过程都同样微不足道。

Answer 3

您的" 将字典转换为字符串" block没有字典，所以我不知道如何帮助您，因为您的消息对象可以是每个事物。。

您可以使用" 字符串到字典"转换：

+ (NSMutableDictionary *)infoWithCipherText:(NSString *)cipherText {
    NSString *plainText = [self private_decrypt:cipherText];
    if ([plainText hasPrefix:[NSString stringWithFormat:@"entity=%@", @"Message"]]) {
        NSArray *array = [plainText componentsSeparatedByString:@" "];
        NSMutableDictionary *info = [NSMutableDictionary dictionary];
        for (NSString *keyvalString in array) {
            NSArray *keyvalArray = [keyvalString componentsSeparatedByString:@"="];
            if ([keyvalArray count] == 2) {
                info[[keyvalArray firstObject]] = [keyvalArray lastObject];
            }
        }
       return info;
    } else {
       return nil;
    }
}