Tuple Relational Calculus:
This is a ‘non-procedural query language’, whereas ‘Relational Algebra’ is a ‘procedural query language’.
The TRC describes the desired information without giving a specific procedure for obtaining the result.
The query expression in this language is :
{ t / P(t) }
Consider the following relation:
R1 : FACULTY-1
| FNo | FName | DNo | Qual | Salary |
| 22 | Riya | 21 | Ph.D | 55000 |
| 24 | Priya | 22 | M. Tech | 50000 |
| 25 | Zayn | 22 | M. Tech | 42000 |
| 27 | Harry | 23 | M. Tech | 28000 |
| 30 | Sia | 23 | M. Tech | 32000 |
| 33 | Max | 24 | Ph.D | 53000 |
| 35 | Fred | 24 | Ph.D | 52000 |
| 37 | Diva | 25 | M. Tech | 26000 |
| 39 | Ben | 25 | M. Tech | 24000 |
| 40 | Trent | 25 | M. Tech | 34000 |
Example-1 : Find the Faculty No, Faculty name, DNo, Qual and salaries of all the faculty, who are drawing more than 30000.
{ t / t ϵ Faculty-1 ˄ t[Salary] > 30000 }
Output :
| FNo | FName | DNo | Qual | Salary |
| 22 | Riya | 21 | Ph.D | 55000 |
| 24 | Priya | 22 | M. Tech | 50000 |
| 25 | Zayn | 22 | M. Tech | 42000 |
| 30 | Sia | 23 | M. Tech | 32000 |
| 33 | Max | 24 | Ph.D | 53000 |
| 35 | Fred | 24 | Ph.D | 52000 |
| 40 | Trent | 25 | M. Tech | 34000 |
Example-2 : Find the Faculty No and Faculty name of all the faculty, who are drawing more than 40000.
{ t / Ǝ s ϵ Faculty-1 ( t[FNo] = s[FNo] ˄
t[ FName] = s{FName] ˄ s[Salary] > 40000 }
Output :
| FNo | FName |
| 22 | Riya |
| 24 | Priya |
| 25 | Zayn |
| 33 | Max |
| 35 | Fred |
Example-3 : Find the names of students who are studying in departments.
{ t / Ǝ s ϵ STUD ( t[SName] = s[SName] ˄
Ǝ u ϵ DEPT ( s[DNo] = u[DNo] ˄ u[DName] = ‘IT’ ) ) }
Consider the following relations :
| R1 : STUD | R2 : DEPT | ||||
| SNo | SName | DNo | DNo | DName | |
| 21 | Riya | 21 | 21 | CSE | |
| 23 | Priya | 21 | 22 | IT | |
| 26 | Zayn | 22 | 23 | ECE | |
| 28 | Harry | 22 | 24 | ME | |
| 30 | Sia | 23 | 25 | EEE | |
Output : SName
Zayn
Harry
Example-4 : Find the name of the department where ‘pqr’ is studying.
{ t / Ǝ s ϵ DEPT ( t[DName] = s[DName] ˄
Ǝ u ϵ STUD ( s[DNo] = u[DNo] ˄ u[SName] = ‘pqr’ ) ) }
Output : DName
CSE
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