Update all.sql

Author: erjan

sql-practicecom/medium/all.sql

@@ -18,31 +18,20 @@ Show patient_id and first_name from patients where their first_name start and en
 at least 6 characters long.
 
 select patient_id, first_name from patients where first_name like 's%s' and len(first_name)>=6
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 Show patient_id, first_name, last_name from patients whos diagnosis is 'Dementia'.
 
 Primary diagnosis is stored in the admissions table.
 
 select p.patient_id, first_name, last_name from patients p join admissions a on p.patient_id = a.patient_id
 where diagnosis = 'Dementia'
 
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 Display every patient's first_name.
 Order the list by the length of each name and then by alphbetically
 
 select first_name from patients p order by len(first_name) , first_name
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 Show the total amount of male patients and the total amount of female patients in the patients table.
 Display the two results in the same row.
 
@@ -51,12 +40,6 @@ select
   sum(case when gender = 'F' then 1 end) as female_count 
 from patients;
 
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 Show first and last name, allergies from patients which have allergies to either 'Penicillin' or 'Morphine'. Show results 
 ordered ascending by allergies then by first_name then by last_name.
@@ -66,182 +49,133 @@ select first_name,last_name,allergies
  from patients where allergies in ('Penicillin' , 'Morphine')
 
  order by allergies, first_name,last_name
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 Show patient_id, diagnosis from admissions. Find patients admitted multiple times for the same diagnosis.
 
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 select 
 patient_id, diagnosis from admissions
 group by diagnosis,patient_id
 having count(*)>1
 
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 Show the city and the total number of patients in the city.
 Order from most to least patients and then by city name ascending.
 
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 select 
 
 city, count(patient_id) as x from patients
 group by city
 order by x desc, city
 
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 Show first name, last name and role of every person that is either patient or doctor.
 The roles are either "Patient" or "Doctor"
 
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 select 
 
  first_name,last_name,'Patient' from patients 
  union all
 
  select 
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  first_name,last_name,'Doctor' from doctors
 
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 Show all allergies ordered by popularity. Remove NULL values from query.
 
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 select 
 allergies, count(*) as total_diagnosis
  from patients  where allergies is not null  group by allergies order by total_diagnosis desc
 
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+Show all patient's first_name, last_name, and birth_date who were born in the 1970s decade. Sort the 
+list starting from the earliest birth_date.
 
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+select first_name,last_name, birth_date from patients 
+where year(birth_date) between 1970 and 1979
+order by birth_date 
 
 
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+We want to display each patient's full name in a single column. Their last_name in all upper 
+letters must appear first, then first_name in all lower case letters. Separate the last_name 
+and first_name with a comma. Order the list by the first_name in decending order
+EX: SMITH,jane
 
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+select 
+concat(upper(last_name), ',', lower(first_name)) as new_name_format from patients
+order by first_name desc
 
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+Show the province_id(s), sum of height; where the total sum of its patient's height is greater than or equal to 7,000.
 
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+select 
+province_id, sum(height) 
+from patients group by province_id
+having sum(height) >=7000
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+Show the difference between the largest weight and smallest weight for patients with the last name 'Maroni'
 
 
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+select 
+max(weight) - min(weight)
+from patients where last_name = 'Maroni'
 
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+Show all of the days of the month (1-31) and how many
+admission_dates occurred on that day. Sort by the day with
+most admissions to least admissions.
 
+select 
 
+day(admission_date) as  day_number,
+count(admission_date) as number_of_admissions
 
+from admissions
 
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+group by day(admission_date)
+order by number_of_admissions desc
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+Show all columns for patient_id 542's most recent admission_date.
 
+select * from  admissions a where a.patient_id = 542 order by admission_date desc limit 1
 
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+SELECT *
+FROM admissions
+GROUP BY patient_id
+HAVING
+  patient_id = 542
+  AND max(admission_date)
 
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+Show patient_id, attending_doctor_id, and diagnosis for admissions that match one of the two criteria:
+1. patient_id is an odd number and attending_doctor_id is either 1, 5, or 19.
+2. attending_doctor_id contains a 2 and the length of patient_id is 3 characters.
 
 
+select 
 
+patient_id, attending_doctor_id, diagnosis
 
+from  admissions 
 
+where mod(patient_id,2) = 1 and attending_doctor_id in(1,5,19) 
 
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+or attending_doctor_id like '%2%' and len(patient_id) = 3
 
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+Show first_name, last_name, and the total number of admissions attended for each doctor.
 
+Every admission has been attended by a doctor.
 
+select 
 
+first_name, last_name, count(*)
+from doctors d join  admissions a on d.doctor_id = a.attending_doctor_id
 
+group by first_name,last_name
 
 
 
@@ -250,44 +184,51 @@ allergies, count(*) as total_diagnosis
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+For each doctor, display their id, full name, and the first and last admission date they attended.
 
+select 
 
+d.doctor_id, concat(first_name, ' ',last_name) as full_name, min(admission_date) first_admission_date, max(admission_date) last_admission_date
+from doctors d join admissions a on d.doctor_id = a.attending_doctor_id
 
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+group by d.doctor_id, first_name,last_name
 
 
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+Display the total amount of patients for each province. Order by descending.
 
+select 
 
+p2.province_name, count(p.patient_id) as x
+from patients p join province_names p2 on p.province_id = p2.province_id
+group by p2.province_name
 
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+order by x desc
 
 
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+For every admission, display the patient's full name, their admission 
+diagnosis, and their doctor's full name who diagnosed their problem.
 
+select 
 
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+concat(p.first_name, ' ',p.last_name) as patient_name, diagnosis, 
+concat(d.first_name, ' ', d.last_name) as doctor_name
+from admissions a join patients p on p.patient_id = a.patient_id join doctors d 
+on d.doctor_id = a.attending_doctor_id
 
 
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+display the number of duplicate patients based on their first_name and last_name.
+select 
+first_name, last_name, count(*) as num_of_duplicates
+from patients 
 
+group by first_name,last_name
+having count(*) >1
 
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+order by num_of_duplicates desc