Question

我有一个数据库，用户可以在其中提交有关他们采取了多少步骤的信息，我试图将其拉出来制作排行榜，但由于它是单独提交的，因此他们会单独提取。

SELECT step_count.steps, step_count.email, logins.FirstName, logins.LastName
  FROM step_count, logins
  WHERE step_count.email=logins.email

这是查询，这就是提取信息的方式

steps | email                          | FirstName | LastName

5000  | one@test.com                   | One       | onetest
300   | two@test.com                   | Two       | twotest
200   | one@test.com                   | One       | onetest

有没有办法将两者合并在一起以获得总步数？

Answer 1

您可以将SUM聚合函数与GROUP BY语句一起使用：

SELECT SUM(step_count.steps), 
       step_count.email, 
       logins.FirstName, 
       logins.LastName
FROM step_count, logins
WHERE step_count.email=logins.email
GROUP BY step_count.email, logins.FirstName, logins.LastName

通过这种方式，您可以按要聚合的字段进行分组，并为每个组求和一个数字字段（步骤）。

看看：mysql aggregate functions

您还应该查看INNER JOIN语句来连接表而不是使用WHERE子句。

Answer 2

您可以尝试此查询：

SELECT
    sum(step_count.steps) as sum_steps,
    step_count.email,
    logins.FirstName,
    logins.LastName
FROM
   step_count
   inner join logins on
       step_count.email=logins.email
group by
    step_count.email,
    logins.FirstName,
    logins.LastName

Answer 3

您可以在mysql中使用SUM（）函数。 - https://dev.mysql.com/doc/refman/5.7/en/group-by-functions.html#function_sum - https://www.tutorialspoint.com/mysql/mysql-sum-function.htm 上面提供了一些参考资料以获取更多信息，但您的查询看起来像

SELECT SUM(step_count.steps), step_count.email, logins.FirstName, logins.LastName
    FROM step_count, logins
    WHERE step_count.email=logins.email
    GROUP BY logins.email;

Answer 4

您可以通过指定要重新组合的密钥来使用+(void)callPostWithRawData:(NSDictionary *)dict withURL:(NSString *)strUrl withToken:(NSString *)strToken withBlock:(dictionary)block { NSError *error; NSData *jsonData = [NSJSONSerialization dataWithJSONObject:dict options:0 error:&error]; NSString *jsonString = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding]; AFURLSessionManager *manager = [[AFURLSessionManager alloc] initWithSessionConfiguration:[NSURLSessionConfiguration defaultSessionConfiguration]]; Please use below method. manager.responseSerializer = [AFHTTPResponseSerializer serializer]; NSMutableURLRequest *req = [[AFJSONRequestSerializer serializer] requestWithMethod:@"POST" URLString:[NSString stringWithFormat:@"%@/%@",WebserviceUrl,strUrl] parameters:nil error:nil]; req.timeoutInterval= [[[NSUserDefaults standardUserDefaults] valueForKey:@"timeoutInterval"] longValue]; [req setValue:strToken forHTTPHeaderField:@"Authorization"]; [req setValue:@"application/json" forHTTPHeaderField:@"Content-Type"]; [req setValue:@"application/json" forHTTPHeaderField:@"Accept"]; [req setHTTPBody:[jsonString dataUsingEncoding:NSUTF8StringEncoding]]; [[manager dataTaskWithRequest:req completionHandler:^(NSURLResponse * _Nonnull response, id _Nullable responseObject, NSError * _Nullable error) { if (!error) { if ([responseObject isKindOfClass:[NSData class]]) { NSHTTPURLResponse *httpResponse = (NSHTTPURLResponse *) response; if ((long)[httpResponse statusCode]==201) { NSMutableDictionary *dict = [[NSMutableDictionary alloc] init]; [dict setObject:@"201" forKey:@"Code"]; if ([httpResponse respondsToSelector:@selector(allHeaderFields)]) { NSDictionary *dictionary = [httpResponse allHeaderFields]; NSLog(@"%@",[dictionary objectForKey:@"Location"]); [dict setObject:[NSString stringWithFormat:@"%@",[dictionary objectForKey:@"Location"]] forKey:@"Id"]; block(dict); } } else if ((long)[httpResponse statusCode]==200) { //Leave Hours Calculate NSDictionary *serializedData = [NSJSONSerialization JSONObjectWithData:responseObject options:kNilOptions error:nil]; block(serializedData); } else{ } } else if ([responseObject isKindOfClass:[NSDictionary class]]) { block(responseObject); } } else { NSMutableDictionary *dict = [[NSMutableDictionary alloc] init]; [dict setObject:ServerResponceError forKey:@"error"]; block(dict); } }] resume]; }，并使用GROUP BY上的SUM()来获取groupBy中指定的密钥的总步数。

更多信息：https://dev.mysql.com/doc/refman/5.7/en/group-by-handling.html

将两个mysql行与一个重复的id组合在一起

4 个答案: