Question

在我的应用中，我在标签和链接之间存在多对多关系，如下所示：

Tags <<-->> Links

我正在尝试返回与具有当前活动标记的链接相关的标记列表，但不包含在活动标记中。

我还想获得具有“其他”标签的链接数量的计数，这些标签需要受活动标签的限制。

使用下面的代码，我可以返回“其他”标签和链接数，但返回的计数是每个标签的所有链接。

我希望能够使用与我用于构建子查询的方法类似的方法来计算链接，但我正在努力使其工作。我已经尝试使用count NSExpression中生成的子查询，但是在评估子查询时会出现这种错误。

// Test array of tag names
self.activeTagArray = [@[@"tag1", @"tag2"] mutableCopy];

NSFetchRequest *fetchRequest = [NSFetchRequest fetchRequestWithEntityName:[Tag entityName]];

// We want to exclude the tags that are already active
NSPredicate *activeTagsPredicate = [NSPredicate predicateWithFormat:@"NOT ANY name IN %@", self.activeTagArray];

// Build subquery string to identify links that have all of the active tags in their tag set
NSString __block *subquery = @"SUBQUERY(links, $link, ";

[self.activeTagArray enumerateObjectsUsingBlock:^(id tagName, NSUInteger index, BOOL *stop) {
    if (index == self.activeTagArray.count - 1) {
        subquery = [subquery stringByAppendingString:[NSString stringWithFormat:@"SUBQUERY($link.tags, $tag, $tag.name = '%@') != NULL", tagName]];
    } else {
        subquery = [subquery stringByAppendingString:[NSString stringWithFormat:@"SUBQUERY($link.tags, $tag, $tag.name = '%@') != NULL AND ", tagName]];
    }
}];

subquery = [subquery stringByAppendingString:@") != NULL"];

NSLog(@"Subquery : %@", subquery);

NSPredicate *noTagsPredicate = [NSPredicate predicateWithFormat:subquery];


// Create a predicate array
NSArray *predicateArray = @[noTagsPredicate, activeTagsPredicate, userPredicate];
NSPredicate *compoundPredicate = [NSCompoundPredicate andPredicateWithSubpredicates:predicateArray];

fetchRequest.predicate = compoundPredicate;
fetchRequest.relationshipKeyPathsForPrefetching = @[@"links"];

// Set up the count expression
NSExpression *countExpression = [NSExpression expressionForFunction: @"count:" arguments:@[[NSExpression expressionForKeyPath: @"links.href"]]];


NSExpressionDescription *expressionDescription = [[NSExpressionDescription alloc] init];

expressionDescription.name = @"counter";
expressionDescription.expression = countExpression;
expressionDescription.expressionResultType = NSInteger32AttributeType;

fetchRequest.propertiesToFetch = @[@"name", expressionDescription];

// Sort by the tag name
NSSortDescriptor *sortDescriptor = [NSSortDescriptor sortDescriptorWithKey:@"name" ascending:YES];
fetchRequest.sortDescriptors = @[sortDescriptor];

fetchRequest.resultType = NSDictionaryResultType;

NSError *error = nil;

NSArray *resultsArray = [self.managedObjectContext executeFetchRequest:fetchRequest error:&error];
if (error) {
    NSLog(@"Error : %@", [error localizedDescription]);
}

NSMutableArray *allTags = [[NSMutableArray alloc] init];
for (NSDictionary *tagDict in resultsArray) {
    NSLog(@"Tag name : %@, Link Count : %@", tagDict[@"name"], tagDict[@"counter"]);
    [allTags addObject:tagDict[@"name"]];
}

[allTags addObjectsFromArray:self.activeTagArray];

非常感谢任何帮助！

Answer 1

如果我理解你的问题，下面的谓词会抓取你的“其他标签”，即与链接相关的所有标签，这些标签与所有给定的“活动标签”相关：

NSArray *activeTags = @[@"tag1", @"tag2"];

NSFetchRequest *request = [NSFetchRequest fetchRequestWithEntityName:@"Tag"];

NSPredicate *p1 = [NSPredicate predicateWithFormat:@"NOT name in %@", activeTags];
NSPredicate *p2 = [NSPredicate predicateWithFormat:@"SUBQUERY(links, $l, SUBQUERY($l.tags, $t, $t.name IN %@).@count = %d).@count > 0",
                   activeTags, [activeTags count]];
NSPredicate *predicate = [NSCompoundPredicate andPredicateWithSubpredicates:@[p1, p2]];
[request setPredicate:predicate];

现在是“技巧”：谓词p2的左侧是

SUBQUERY(links, $l, SUBQUERY($l.tags, $t, $t.name IN %@).@count = %d).@count

这正是应该包含在结果中的链接数，所以我们可以从该谓词创建表达式描述：

NSExpression *countExpression = [(NSComparisonPredicate *)p2 leftExpression];

NSExpressionDescription *expressionDescription = [[NSExpressionDescription alloc] init];
expressionDescription.name = @"counter";
expressionDescription.expression = countExpression;
expressionDescription.expressionResultType = NSInteger32AttributeType;

[request setResultType:NSDictionaryResultType];
[request setPropertiesToFetch:@[@"name", expressionDescription]];

生成的SQLite查询非常复杂。首先获取链接可能是明智的：

NSFetchRequest *linkRequest = [NSFetchRequest fetchRequestWithEntityName:@"Link"];
NSPredicate *linkPredicate = [NSPredicate predicateWithFormat:@"SUBQUERY(tags, $t, $t.name IN %@).@count = %d",
                              activeTags, [activeTags count]];
[linkRequest setPredicate:linkPredicate];
NSArray *activeLinks = [context executeFetchRequest:linkRequest error:&error];

并在单独的步骤中获取标记，这可以通过上面的代码完成，其中只有谓词p2被更简单的子查询替换

NSPredicate *p2 = [NSPredicate predicateWithFormat:@"SUBQUERY(links, $l, $l IN %@).@count > 0", activeLinks];

使用现有子查询过滤多对多计数表达式

1 个答案: