ios - NSPredicate 将 NIL 字符串作为任何字符处理

Question

我有一个 NSPredicate 看起来像这样:

NSPredicate *predicateSearch;

    if ([category_array count]>0 && [location_array count]>0) {
        predicateSearch =[NSPredicate predicateWithFormat:@"category_id IN %@ AND location_id IN %@ AND company_title CONTAINS %@",category_array,location_array,searchStr];
    }
    else if ([category_array count]>0){

         predicateSearch =[NSPredicate predicateWithFormat:@"category_id IN %@ AND company_title CONTAINS %@",category_array,searchStr];
    }
    else if([location_array count]>0){

        predicateSearch = [NSPredicate predicateWithFormat:@"location_id IN %@ AND company_title CONTAINS %@",location_array,searchStr];
    }

    [fetchRequest setPredicate:predicateSearch];

由于 NSPredicate 无法处理 Nil 值，我使用此 If 语句来格式化谓词并将 Nil 值处理为'任何'。这里的问题是，如果 searchStr 为 Nil，我还希望将其处理为“ANY”。在 Nil 的情况下，我试图将其设置为空字符串，但它不起作用。有什么方法可以在不对每种情况使用任何 If 语句的情况下实现这一点(9 个 if 语句太多了)？

Answer 1

使用复合谓词。例如，根据您发布的内容:

NSPredicate *categoryPredicate = (category_array.count > 0)
    ? [NSPredicate predicateWithFormat:@"category_id IN %@", category_array]
    : [NSPredicate predicateWithValue:YES];

NSPredicate *locationPredicate = (category_array.count > 0)
    ? [NSPredicate predicateWithFormat:@"location_id IN %@", location_array]
    : [NSPredicate predicateWithValue:YES];

NSPredicate *searchPredicate = (searchStr.length > 0)
    ? [NSPredicate predicateWithFormat:@"company_title CONTAINS %@", searchStr]
    : [NSPredicate predicateWithValue:YES];

NSCompoundPredicate *predicate =
   [NSCompoundPredicate
    andPredicateWithSubpredicates:@[                                            
                                    categoryPredicate,
                                    locationPredicate,
                                    searchPredicate,
                                   ]
   ];