 Which algorithm for hashing name, firstName and birth-date of a person
By : user2954258
Date : March 29 2020, 07:55 AM
should help you out If you want to search for a person knowing only those credentials, you could store the SHA-1 in the database(or MD5 for speed, unless you have like a quadrillion people to sample). The hash will be worthless, as it stores no information about the person, but it can work for searching a database. You just want to make sure that the three pieces of information match, so it would be safe to just concatenate them: code :
user.hash = SHA1(user.firstName + user.DOB + user.lastName)
hash = SHA1(query.firstName + query.DOB + query.lastName)
for user in database:
if user.hash == hash:
return user
1
---------------------------------------------------
340,282,366,920,938,463,463,374,607,431,768,211,456
0.0000000000000000000000000000000000000293873 %
|
 Efficent way to sort 'Person.firstName' under first alphabetically letter
By : user7873777
Date : March 29 2020, 07:55 AM
hop of those help? I iterate through a long list of names, which I assign each to a 'Person' object. A Person object has a firstName and a lastName. I am looking for a efficent way to sort all the firstNames beginning with the letter A through Z. I have come up with this solutions but it is not efficent, or there is surely an easier way than creating '24' arrays for each letter of the alphabet. , You could do something like this: code :
NSArray *names = @[@"Bob", @"Book2", @"August", @"Zulu", @"Agnoz"];
NSMutableDictionary *buckets = [[NSMutableDictionary alloc] init];
for (NSString *name in names) {
NSString *index = [name substringToIndex:1];
if (!buckets[index]) {
buckets[index] = [[NSMutableArray alloc] init];
}
[buckets[index] addObject:name];
}
|
 I need to get the whole row for all duplicate entries efficiently
By : user3847257
Date : March 29 2020, 07:55 AM
like below fixes the issue internet! I'm pretty new to SQL and I need to get all the rows with duplicate information in certain fields and have them display right next to their other duplicates (group by duplicates). , I would do something like this: code :
with cte as (
SELECT *
, count(*) over (partition by B, C, D, E, F, G) as cnt
, dense_rank() over (order by B, C, D, E, F, G) as grp
FROM STI.[dbo].[tbl_Consignee]
)
select *
from cte
where cnt > 1
order by grp
|
 Java QueryDSL 3.6.0 Projection.bean(Person.class, person.firstName, person.count())
By : Yuki taka
Date : March 29 2020, 07:55 AM
it should still fix some issue I chained .as method on top of .count to groupByCount attribute as shown below. This way I was able to avoid writing n! constructors. code :
personRepo.getQueryDSL().createQuery(person).groupBy(person.isCitizen)
.list(Projections.bean(Alarm.class, person.isCitizen,
person.count().as(person.groupByCount));
|
 There is no argument given that corresponds to the required formal parameter 'firstName' of 'Person.Person(string, strin
By : user3535446
Date : March 29 2020, 07:55 AM
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» 
code :
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