Skip to main content

GO-API PROJECT STRUCTURE

GOAPI Tutorial

In this post, We will create repository structure for our GOLANG APIs project.
We have directory structure like below for this project. We will follow this structure for this project.
  
 
├├── goapi
│   ├── api-frontend
│   ├── db
│   │   ├── db.sql
│   │   └── model.sql
│   ├── docs
│   ├── README.md
│   └── server
│       ├── bin
│       │   └── main
│       ├── Makefile
│       └── src
│           └── test.com
│               └── goapi
│                   └── app
│                       ├── apis
│                       │   └── student.go
│                       ├── cmd
│                       │   └── main
│                       │       └── main.go
│                       ├── db
│                       │   ├── client_db.go
│                       │   └── student_db.go
│                       ├── model
│                       │   └── student.go
│                       ├── router.go
│                       └── utils
│                           ├── constant.go
│                           └── rest_util.go

goapi: This is the root directory for this project.
api-front-end: This directory contains front-end code and related utilities
db: This is containing sql script and sample data of data base
docs: This contains documentation of this project
server: This directory contains two directories bin and src
bin: bin directory contain executable binary of this project
src: This directory contains source code of this project. src directory has apis, cmd, db, model and utils directories.
apis: This directories contain apis definitions of this project
cmd: contains main method to start server
db: This contains db connection and query
model: This contains request and response structure
utils: This contains helper functions
router.go: contains apis routs and path

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Labels:

Comments

Post a Comment

Popular posts from this blog

SSC Practice Problem Solution

SSC Practice Problem Solution In this post, we will discuss solution of  SSC Practice Problem . Question 1  The efficiency of (A + B) = 100/24 = (25/6)% The efficiency of (A + B + C) = 100/8 = (25/2)% The efficiency of C = 25/2 – 25/6 = (50/6)% Hence, C can alone finish this job in = 100/(50/6) = 12 days Answer: 12 days option (1) Question 2  Original price = Rs. 6000; Price after discount = 6000 – 1200 = Rs. 4800; Price after raising service contract = 4800 + 480 = Rs. 5280 Answer: Rs. 5280 option (2) Question 3  Suppose the received money by A, B, and C is respectively 5x, 6x, and 9x. 5x = 450; => x = 90; Hence, the total money = 20x = 20*90 = Rs. 1800; Answer: Rs. 1800 option (2) Question 4 Suppose the Cost price of the bag= Rs. x; Hence, x + 0.15x = 230; => x = Rs. 200; Selling price after selling on 20% = 200 + 20% of 200 = Rs. 240 Answer: Rs. 240 option (3) Question 5  The required percentage ...
Post a Comment
Read more

Average

Average In this post, We will learn about average.  Average of any collection of data is the sum of data divide by number of data. Average = (Sum of observations /  Number of observations) Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.  problems The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is (CAT 2018) As specified in the question mean of x and y and z is 80 x+y+z=3*80 mean of x, y, z,u,v is 75 x+y+z+u+v=75*5 2u=x+y 2v=y+z x+y+z+((x+y+y+z)/2)=75*5 y+3(x+y+z)=75*10 y=750-720 y=30 since x+y+z=240 y=30 so x+z=210 since x>=z so minimum value of x=210/2 = 105 Problem2: In first 5 inings sehwag has scored at average of 40. In sixth ining He had scored 52 what is His average after 6th ining. one simple approach is getting total score divided by no of inin...
Post a Comment
Read more

Complex Number

Complex Number While solving the equation of the form  a x 2 + b x + c = 0 , the roots of the equations can take three forms which are as follows: Two Distinct Real Roots Similar Root No Real roots (Complex Roots) The introduction of complex numbers in the 16th century made it possible to solve the equation  x 2 + 1 = 0 . The roots of the equation are of the form  x = ± − 1 ‾ ‾ ‾ √  and no real roots exist. Thus, with the introduction of complex numbers, we have Imaginary roots. We denote  − 1 ‾ ‾ ‾ √  with the symbol i, where i denotes Iota (Imaginary number). An equation of the form z= a+ib, where a and b are real numbers, is defined to be a complex number. The real part is denoted by Re z = a and the imaginary part is denoted by Im z = b. Algebraic Operation on Complex numbers: Addition of two complex numbers Subtraction of two complex number Multiplication of two complex number Division of two complex number Power of Iota (i) ...
Post a Comment
Read more

Logarithm

Logarithm When a x   = N , then we say that x = logarithm of N to the base a and write it as x = log a N . In simple words, it represents the power to which a number must be raised. Let me expand on that by giving a simpler example. If we are asked, what would be the result if ‘a’ is multiplied with itself ‘b’ times; then your answer would be x = a*a*a*…. b (times). This can also be written as a^b. This is also known as ‘a raised to the power of b’ If we are asked, which number multiplied with itself ‘b’ times, will result in a; then you are asked for the value of x such that x*x*x*… b(times) = a => x^b = a => x = a^(1/b) This is also known as ‘bth root of a’ If swe are asked, how many times should you multiply ‘a’ with itself to get ‘b’, that is where the concept of logarithm comes into the picture. You are asked for the value of ‘x’ such that a*a*a…. x (times) = b => x = Log  a  b This is also known as ‘ Log b to the base a’ ...
Post a Comment
Read more