Question

我有两个数组，一个名为'arrayGO'（存储来自数据库的用户列表），另一个名为'idtGO'（存储一个用户ID列表）。 arrayGO 符合以下结构：

@ “安德烈-A”，@ “A线-A”，@ “布鲁诺-B”，@ “博拉-B”，@ “卡欧-C”

我有一个命令可以过滤这个数组，分隔破折号之后的字母，这样做组织数组：

NSMutableDictionary *dictionary = [NSMutableDictionary dictionary];
    NSMutableDictionary *dictionaryIDT = [NSMutableDictionary dictionary];

    for (NSString *string in arrayGO) {

        NSArray *components = [string componentsSeparatedByString:@"-"];
        NSString *key = components[1];
        NSMutableArray *tempArray = dictionary[key];

        if (!tempArray){
            tempArray = [NSMutableArray array];
        }

        [tempArray addObject:components[0]];
         dictionary[key] = tempArray;
    }

这个命令以我想要的方式返回数组，在这种情况下是：

A =（艾琳，安德烈）; B =（博拉，布鲁诺）; C =（卡欧）;

但是我需要对存储用户'ids'的变量做同样的事情，在这个cade中，数组'idtGO'符合以下结构：

@ “4”，@ “1”，@ “12”，@ “22”，@ “33”

我想要做的就是我对名称列表所做的一样，所以在这种情况下和id，我试图得到以下结果：

0 =（4,1）; 1 =（12,22）; 2 =（33）;

我试试这个：

NSMutableDictionary *dictionary = [NSMutableDictionary dictionary];
NSMutableDictionary *dictionaryIDT = [NSMutableDictionary dictionary];
int idtIndex = 0;
int idtKeys = 0;

for (NSString *string in arrayGO) {

    NSArray *components = [string componentsSeparatedByString:@"-"];

    NSString *key = components[1];
    NSString *keyIdt = [NSString stringWithFormat:@"%d", idtIndex];
    NSMutableArray *tempArray = dictionary[key];
    NSMutableArray *tempArrayidt = dictionaryIDT[keyIdt];

    if (!tempArray){

        tempArray = [NSMutableArray array];
        tempArrayidt = [NSMutableArray array];

        [tempArrayidt addObject:keyIdt];
        dictionaryIDT[keyIdt] = tempArrayidt;
        idtKeys++;
        idtIndex++;

    }else{

        idtIndex++;
    }

    [tempArray addObject:components[0]];
    dictionary[key] = tempArray;


}

如果有人帮我解决这个难题，我将不胜感激。

Answer 1

好伙计们我努力尝试并最终得到这个谜题的答案，下面的代码说明了如何：

NSMutableDictionary  *dictionary = [NSMutableDictionary dictionary];
NSMutableDictionary  *dictionaryIDT = [NSMutableDictionary dictionary];
     NSString *keyTemp = @"nulo";

    for (NSString *string in arrayGO) {

        NSArray *components = [string componentsSeparatedByString:@"-"];

        NSString *key = components[1];

        NSString *keyIdt = [NSString stringWithFormat:@"%@", idt[idtKeys]];
        NSString *keyIdtIndex = [NSString stringWithFormat:@"%d",idtIndex];
        NSMutableArray *tempArray = dictionary[key];
        NSMutableArray *tempArrayidt = dictionaryIDT[keyIdtIndex];

        if (!tempArray){

            tempArray = [NSMutableArray array];
            tempArrayidt = [NSMutableArray array];

        }

        [tempArray addObject:components[0]];
        dictionary[key] = tempArray;


            if([keyTemp isEqual: @"nulo"]){
                //Veririca se a variavel keyTemp é igual a zero, se ela for isso quer dizer que é a primeira vez que estamos aqui, e se for a variavel keyTemp Sera igual a key

                keyTemp = key;

                [tempArrayidt addObject:keyIdt];
                dictionaryIDT[keyIdtIndex] = tempArrayidt;
                idtKeys++;

            }else{
                //Se caimos aqui entao quer dizer que jaé a segunda vez que nos estamos caindo aqui, neste caso devemos verificar se a keyTemp ainda é a mesma


                if([key isEqual: keyTemp]){
                    //Se for igual é porque nada mudou, ainda estamos tratando aquela chave, entao...

                    [tempArrayidt addObject:keyIdt];
                    dictionaryIDT[keyIdtIndex] = tempArrayidt;
                    idtKeys++;

                }else{

                    //Se nao é porque a key mudou, entao agora nos devemos aumentar o idtIndex, redefinir o idtKeys e falar que a key é igual a keyTemp
                    idtIndex++;

                    keyIdtIndex = [NSString stringWithFormat:@"%d",[keyIdtIndex intValue] + 1];

                    idtKeys++;
                    keyTemp = key;

                    //Agora faremos o mesmo esquema de sempre...
                    [tempArrayidt addObject:keyIdt];
                    dictionaryIDT[keyIdtIndex] = tempArrayidt;

                }

            }


    }


NSLog(@"%@",dictionaryIDT);

过滤NSDictionary中的键

1 个答案: