Put variable in bash for loop - bash

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.