如何从两个不同的数组中找到共同的对象并打印另一个对象？
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【中文标题】如何从两个不同的数组中找到共同的对象并打印另一个对象？【英文标题】：How to find the common object from two different arrays and print another object? 【发布时间】：2018-02-14 12:34:54 【问题描述】：我有 2 个数组从 2 个不同的 API 接收
预订数组
addrLine1 = "Al Thanyah Fifth, Dubai, United Arab Emirates";
addrLine2 = " ";
apntDate = "14 Feb 2018";
apntDt = "2018-02-14 15:31:30";
apntTime = "03:31 pm";
apptLat = "25.071102142334";
apptLong = "55.142993927002";
"appt_duration" = "";
bid = 555;
bookType = 1;
cancelAmt = 30;
"cat_id" = 591da979227951a10f004ad7;
"cat_name" = "Car Wash";
cid = 21;
"coupon_discount" = 0;
"customer_notes" = "";
"distance_met" = 120;
email = "fawasfais@gmail.com";
expireTime = "";
fname = Fawas;
"job_imgs" = 0;
"job_start_time" = "1970-01-01 00:00:00";
"job_timer" = "";
pPic = "https://s3.amazonaws.com/iserve/ProfileImages/20170720110011AM.png";
"payment_type" = 1;
phone = 568573570;
"price_per_min" = 10;
"pro_notes" = "";
services = "5923f21c2279518a1661cb09,5923f22d2279513f1861cb09";
status = 2;
statusMsg = "Provider Accepted.";
timer = "";
"timer_status" = "";
"visit_amount" = 60;
选定的服务数组
"price_per_unit" = 30;
"sub_cat_id" = 5923f21c2279518a1661cb09;
"sub_cat_name" = SUV;
unit = 1;
,
"price_per_unit" = 45;
"sub_cat_id" = 5923f22d2279513f1861cb09;
"sub_cat_name" = Sedan;
unit = 1;
"price_per_unit" = 65;
"sub_cat_id" = 5923f23e2279518a1661cb09;
"sub_cat_name" = Bus;
unit = 1;
,
"price_per_unit" = 75;
"sub_cat_id" = 5923f24f2279513f1861cb09;
"sub_cat_name" = Lorry;
unit = 1;
我想将上述两个数组与第一个数组中的“services”键和第二个数组中的“sub_cat_id”进行比较，最后打印包含车辆的公共服务 id 的 price_per_unit。
示例 - 5923f21c2279518a1661cb09,5923f22d2279513f1861cb09 id 在第一个数组中找到，因此打印第二个数组中两个 id 的 price_per_units - 30、45
有什么建议吗？ （如果任何代码 sn-ps 需要回答 cmets 中提到的这个问题，我将使用所需的代码编辑和更新问题。
This is how I get the second array from the server
-(void)getServiceTypes
NSDictionary *queryParams;
queryParams = @
@"ent_sess_token":flStrForStr([[NSUserDefaults standardUserDefaults] objectForKey:KDAcheckUserSessionToken]),
@"ent_dev_id":flStrForStr([[NSUserDefaults standardUserDefaults] objectForKey:kPMDDeviceIdKey]),
@"ent_catid":_dictAppointmentDetails[@"cat_id"]
;
NetworkHandler *handler = [NetworkHandler sharedInstance];
[handler composeRequestWithMethod:@"getSubCategory"
paramas:queryParams
onComplition:^(BOOL succeeded, NSDictionary *response)
if (succeeded)
[self getServiceTypesResponse:response];
];
NSLog(@"param%@",queryParams);
-(void)getServiceTypesResponse:(NSDictionary *)response
NSLog(@"GETSERVICETYPES%@",response);
serviceTypeDict = response[@"data"];
NSLog(@"SERVICETYPEDIC%@",serviceTypeDict);
serviceTypeArray = response[@"data"];
NSLog(@"SERVICETYPEARRAY%@",serviceTypeArray);
serviceTypeIDArray = [[serviceTypeArray valueForKey:@"sub_cat_id"]copy];
NSLog(@"SERVICETYPEIDARRAY%@",serviceTypeIDArray);
谢谢
【问题讨论】：
您是否为上述 JSON 数据创建了模型类？ 第一个是一个对象，它有一个名为“服务”的数组，对吧？ @ibnetariq 第一个数组的对象是“services”，它有两个值等于第二个数组的对象键“sub_cat_id” 你需要检查第一个数组的每个对象和第二个数组的每个对象吗？ @ibnetariq 我想将上述两个数组与第一个数组中的“services”键和第二个数组中的“sub_cat_id”进行比较，最后在单元格标签中打印包含车辆的公共服务 ID 的 price_per_unit表格视图 【参考方案1】：让它在操场上工作：
let booking: [String: Any] = [
"services": "5923f21c2279518a1661cb09,5923f22d2279513f1861cb09",
"status":2
]
let selectedServices: [[String: Any]] = [
[
"price_per_unit": 30,
"sub_cat_id": "5923f21c2279518a1661cb09",
"sub_cat_name": "SUV",
"unit": 1
],
[
"price_per_unit": 45,
"sub_cat_id": "5923f22d2279513f1861cb09",
"sub_cat_name": "Sedan",
"unit": 1
],
[
"price_per_unit": 65,
"sub_cat_id": "5923f23e2279518a1661cb09",
"sub_cat_name": "Bus",
"unit": 1
],
[
"price_per_unit": 75,
"sub_cat_id": "5923f24f2279513f1861cb09",
"sub_cat_name": "Lorry",
"unit": 1
]
]
if let bookingServices = booking["services"] as? String
let services: [String] = bookingServices.components(separatedBy: ",")
let commonServices = selectedServices.filter( services.contains( $0["sub_cat_id"] as? String ?? "" ) )
print(commonServices)
Objective-C 等效项：
NSDictionary *booking = @
@"services": @"5923f21c2279518a1661cb09,5923f22d2279513f1861cb09",
@"status": @2
;
NSArray *selectedServices = @[
@
@"price_per_unit": @30,
@"sub_cat_id": @"5923f21c2279518a1661cb09",
@"sub_cat_name": @"SUV",
@"unit": @1
,
@
@"price_per_unit": @45,
@"sub_cat_id": @"5923f22d2279513f1861cb09",
@"sub_cat_name": @"Sedan",
@"unit": @1
,
@
@"price_per_unit": @65,
@"sub_cat_id": @"5923f23e2279518a1661cb09",
@"sub_cat_name": @"Bus",
@"unit": @1
,
@
@"price_per_unit": @75,
@"sub_cat_id": @"5923f24f2279513f1861cb09",
@"sub_cat_name": @"Lorry",
@"unit": @1
];
NSString *servicesString = [booking valueForKey: @"services"];
NSArray *bookingServices = [servicesString componentsSeparatedByString:@","];
NSMutableArray *commonServices = [[NSMutableArray alloc] init];
for (NSDictionary *selectedService in selectedServices)
NSString *sub_cat_id = selectedService[@"sub_cat_id"];
if([bookingServices containsObject: sub_cat_id])
[commonServices addObject: selectedService];
NSLog(@"%@", commonServices);
【讨论】：
抱歉没看到。 检查更新的答案。 这行得通！谢谢 :) 使用谓词方法过滤速度很快，如另一个答案中所述。也尝试使用谓词。【参考方案2】：我认为你应该使用 NSPredicate，因为这是快速枚举。仅两行代码
我在这里使用你的演示内容。
NSDictionary *maindic = @@"price_per_min" :@"10",
@"services":@"5923f21c2279518a1661cb09,5923f22d2279513f1861cb09",
;
NSDictionary *dic1 =@@"price_per_unit" :@"30",
@"sub_cat_id":@"5923f21c2279518a1661cb09",
;
NSDictionary *dic2 =@@"price_per_unit" :@"45",
@"sub_cat_id":@"5923f22d2279513f1861cb09",
;
NSDictionary *dic3 =@@"price_per_unit" :@"65",
@"sub_cat_id":@"5923f23e2279518a1661cb09",
;
NSArray *mainArr= @[dic1,dic2,dic3];
NSPredicate *predicate1 = [NSPredicate predicateWithFormat:@"sub_cat_id == '5923f21c2279518a1661cb09'"];
NSPredicate *predicate2 = [NSPredicate predicateWithFormat:@"sub_cat_id == '5923f22d2279513f1861cb09'"];
NSPredicate *predicate = [NSCompoundPredicate orPredicateWithSubpredicates:@[predicate1, predicate2]];
NSArray *resultArray = [mainArr filteredArrayUsingPredicate:predicate];
它会在resultArray中返回
"price_per_unit" = 30;
"sub_cat_id" = 5923f21c2279518a1661cb09;
,
"price_per_unit" = 45;
"sub_cat_id" = 5923f22d2279513f1861cb09;
【讨论】：
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