Question

编辑：大量编辑，抱歉导致混淆！

我正在开发一个应用程序，其中我使用两种不同的方法从JSON WebService获得2个数组。

在第一个数组中，我获得了本地股票。在第二个数组中，我得到供应商库存。

虽然我认为图像可以更好地解释：

My awesome application diagram

因此，webservices获得了许多项目。数组1包含来自本地库存的一些项目，而数组2包含来自供应商的项目，以及添加的信息，如描述等。我想合并两个数组，将本地stock-array1中的现有项目更新为供应商库存数组2中项目的描述。

这是我想要实现的一个例子。所有产品都由唯一的标识符属性标识（在示例中未提及），并且我重写了isEqual和hash来补偿它。如果产品具有simliar ID，则认为它是相同的。

**Product**

@property id *internalID;
@property id *externalID ;
@property id *localStock;
@property id *supplierStock;
@property id *image;
@property id *productInfo;

现在Webservice 1会返回三种产品：

产品1

*internalID = ABCDEF
*externalID = (null)
*localStock = 15
*supplierStock = (null)
*image = (null)
*productInfo = (null);

产品2

*internalID = GHIJK
*externalID = (null)
*localStock = 13
*supplierStock = (null)
*image = niceImage.png
*productInfo = @"This Product is Awesome!";

产品3

*internalID = LMNOP
*externalID = (null)
*localStock = 7
*supplierStock = (null)
*image = (null)
*productInfo = (null);

Webservice 2返回四个产品，其中两个产品也在数组1中：

产品1

    *internalID = (null)
    *externalID = 123456
    *localStock = (null)
    *supplierStock = 12
    *image = external_product1image.jpg
    *productInfo = @"This product is also Awesome and in both local stock and supplier stock!";

产品4

    *internalID = (null)
    *externalID = 23456
    *localStock = (null)
    *supplierStock = 11
    *image = niceImage.png
    *productInfo = @"This Product is Awesome and only available from our supplier!";

产品3

    *internalID = (null)
    *externalID = 78901
    *localStock = (null)
    *supplierStock = 7
    *image = external)supplierimage.jpg
    *productInfo = @"This product is also Awesome and in both local stock and supplier stock!";

然后mergedArray应如下所示：

产品1

*internalID = ABCDEF
*externalID = 123456
*localStock = 15
*supplierStock = 12
*image = external_product1image.jpg
*productInfo = @"This product is also Awesome and in both local stock and supplier stock!";

//因此，阵列1中的产品1将其属性与阵列2中的产品1合并

产品2

*internalID = GHIJK
*externalID = (null)
*localStock = 13
*supplierStock = (null)
*image = niceImage.png
*productInfo = @"This Product is Awesome!";

产品3

*internalID = LMNOP
*externalID = 78901
*localStock = 7
*supplierStock = 7
*image = external)supplierimage.jpg
*productInfo = @"This product is also Awesome and in both local stock and supplier stock!";

产品4

*internalID = (null)
*externalID = 23456
*localStock = 13
*supplierStock = 11
*image = niceImage.png
*productInfo = @"This Product is Awesome and only available from our supplier!";

这是我使用的代码，但似乎有时会出现故障：

- (void) compareArrays:(id)sender metBreedte:(NSString *)breedte metHoogte:(NSString *)hoogte metDiameter:(NSString *)diameter
{
    NSMutableSet *getBandenSet = [NSMutableSet setWithArray:getBandenArray];      NSUInteger i = 0;

    while (i < [getBandenInfo1Array count]) {
        id getBandenInfo1Object = [getBandenInfo1Array objectAtIndex:i];
        if ([getBandenSet containsObject:getBandenInfo1Object])

        {

            BWBand *getBandenBand = [getBandenSet member:getBandenInfo1Object];

            BWBand *getBandenInfo1Band = getBandenInfo1Object;

            // Do stuff to the Banden (sync)
            getBandenBand.alternatievePrijs = getBandenInfo1Band.alternatievePrijs;
            getBandenBand.itemName = getBandenInfo1Band.itemName;
            getBandenBand.supplierStock = getBandenInfo1Band.supplierStock;
            getBandenBand.grossPrice = getBandenInfo1Band.grossPrice;
            getBandenBand.eancode = getBandenInfo1Band.eancode;
            getBandenBand.EMarked = getBandenInfo1Band.EMarked;
            getBandenBand.garagePrijs = getBandenInfo1Band.garagePrijs;
            getBandenBand.loadIndex = getBandenInfo1Band.loadIndex;
            getBandenBand.brand = getBandenInfo1Band.brand;
            getBandenBand.custPrice = getBandenInfo1Band.custPrice;
            getBandenBand.netPrice = getBandenInfo1Band.netPrice;
            getBandenBand.tyreLabel = getBandenInfo1Band.tyreLabel;
            getBandenBand.TyreLabelFuel = getBandenInfo1Band.TyreLabelFuel;
            getBandenBand.TyreLabelNoise = getBandenInfo1Band.TyreLabelNoise;
            getBandenBand.TyreLabelNoiseLevel = getBandenInfo1Band.TyreLabelNoiseLevel;
            getBandenBand.TyreLabelWet = getBandenInfo1Band.TyreLabelWet;
            getBandenBand.foto = getBandenInfo1Band.foto;
            NSMutableString *string1 = [NSMutableString stringWithString: getBandenBand.colorStock];
            NSString *newString = [string1 substringToIndex:[string1 length]-2];
            NSString *colorStock = [NSString stringWithFormat:@"%@-%@",newString,getBandenInfo1Band.supplierStock];
            getBandenBand.colorStock = colorStock;
           [getBandenSet removeObject:getBandenInfo1Object];
            [getBandenInfo1Array removeObjectIdenticalTo:getBandenInfo1Object];
        } else
        {i++;}

        mergedBandenArray = [NSMutableArray arrayWithArray:[getBandenArray arrayByAddingObjectsFromArray:getBandenInfo1Array]];
       [mergedBandenArray sortUsingDescriptors:
         [NSArray arrayWithObjects:
          [NSSortDescriptor sortDescriptorWithKey:@"colorStock" ascending:NO], [NSSortDescriptor sortDescriptorWithKey:@"brand" ascending:YES], nil]];}

可以使用NSPredicates完成所有这些吗？如果是这样，怎么样？

提前致谢！

Answer 1

以下算法不要求您为isEqual类覆盖hash和Product：

首先按照uniqueItemID键按升序对两个数组进行排序。
然后并行遍历两个阵列，并将两个阵列中的产品添加或合并到目标阵列。

如果两个数组都已排序，则可以轻松找到匹配和不匹配的产品，只需一个循环并行遍历两个数组。

以下代码有希望证明这一想法：

NSSortDescriptor *sortByID = [NSSortDescriptor sortDescriptorWithKey:@"uniqueItemID" ascending:YES];
NSArray *sorted1 = [array1 sortedArrayUsingDescriptors:@[sortByID]];
NSArray *sorted2 = [array2 sortedArrayUsingDescriptors:@[sortByID]];

NSUInteger i1 = 0, i2 = 0;
NSMutableArray *combined = [NSMutableArray array];
while (i1 < [sorted1 count] && i2 < [sorted2 count]) {
    Product *p1 = sorted1[i1];
    Product *p2 = sorted2[i2];
    switch ([p1.uniqueItemID compare:p2.uniqueItemID]) {
        case NSOrderedAscending:
            // p1 is in first array, but not in second:
            [combined addObject:p1];
            i1++;
            break;
        case NSOrderedDescending:
            // p2 is in second array, but not in first:
            [combined addObject:p2];
            i2++;
            break;
        case NSOrderedSame:
            // p1, p2 have same ID:
            ... update p1 with values from p2 ...
            [combined addObject:p1];
            i1++, i2++;
            break;
    }
}
// Add remaining products from first array:
while (i1 < [sorted1 count]) {
    [combined addObject:sorted1[i1]];
    i1++;
}
// Add remaining products from second array:
while (i2 < [sorted2 count]) {
    [combined addObject:sorted2[i2]];
    i2++;
}

Answer 2

1）。合并两个数组：

NSMutableSet *set = [NSMutableSet setWithArray:array1];
[set addObjectsFromArray:array2];

NSArray *array = [set allObjects];

2）。要在NSPredicates的帮助下删除重复项：

1. Filtering Fun with Predicates

2. Hello, NSPredicate

3）。合并＆amp;删除重复项：

NSArray *array1, *array2;
MSMutableArray *result = [array1 mutableCopy];
for (id object in array2)
{
  [result removeObject:object];  // remove duplicates if it's already there.
  [result addObject:object];
}

IOS：在更新“重复”属性的同时合并两个NSArrays

2 个答案: