将变量放入 bash for 循环

Question

I am making a post request in bash multiple times with a for loop like:

     for n in {1..5}; do curl -X POST   'http://localhost:8080/hello'   -H 'accept: */*'   -H 'Content-Type: application/json'   -d '{
    "a": 1,
    "b": 2
      }';done

Now the values a and b are the same so I wanted to generate random values for the variables.So I created two random values:

a= awk -v min=10 -v max=20 -v num=1 'BEGIN{srand(); for (i=1;i<=num;i++) printf ("%.5f\n",min+rand()*(max-min+1))}'
b= awk -v min=50 -v max=60 -v num=1 'BEGIN{srand(); for (i=1;i<=num;i++) printf ("%.5f\n",min+rand()*(max-min+1))}'

and tried to insert them into my post request like this:

     for n in {1..5}; do curl -X POST   'http://localhost:8080/hello'   -H 'accept: */*'   -H 'Content-Type: application/json'   -d '{
  "a": a,
  "b": b
}';done

but it didnt work I also tried:

 for (n in {1..5};a=    awk -v min=10 -v max=20 -v num=1 'BEGIN{srand(); for (i=1;i<=num;i++) printf ("%.5f\n",min+rand()*(max-min+1))}'; b=    awk -v min=50 -v max=60 -v num=1 'BEGIN{srand(); for (i=1;i<=num;i++) printf ("%.5f\n",min+rand()*(max-min+1))}'); do curl -X POST   'http://localhost:8080/hello'   -H 'accept: */*'   -H 'Content-Type: application/json'   -d '{
  "a": 1,
  "b": 2
}';done

but that also didnt work, what am I doing wrong?

Thanks in regard

Answer 1

You must perform some small corrections so you can use all of this in a script.

Next you can see the values being replaced (a and b). Adapt for what you need

a=`awk -v min=10 -v max=20 -v num=1 'BEGIN{srand(); for (i=1;i<=num;i++) printf ("%.5f\n",min+rand()*(max-min+1))}'`
b=`awk -v min=50 -v max=60 -v num=1 'BEGIN{srand(); for (i=1;i<=num;i++) printf ("%.5f\n",min+rand()*(max-min+1))}'`


for n in {1..5}; do 
  echo curl -X POST   'http://localhost:8080/hello'   -H 'accept: */*'   -H 'Content-Type: application/json'   -d "{

  \"a\": $a,

  \"b\": $b

}";done

The first problem is that you want a and b to have the result of awk command. For that you need to (for example) enclose the command between ``

To have the variables being substituted you cannot have them between '...' (single quotes). Otherwise they are not substituted. So you must use double quotes "..." to have $a and $b being substituted by their variable values.