如何在bash方法中将参数传递给curl命令

我有一个执行curl命令的方法，它返回结果。我想将参数传递给此方法，但它不起作用

commande_mine() {
    local MY_OUTPUT=$(curl -X POST \
  http://localhost:5000/myapp \
  -H 'cache-control: no-cache' \
  -H 'content-type: application/json' \
  -d '{
        "filepath": "$1"
    }')
    echo $MY_OUTPUT
}

for f in "/Users/anthony/my files"/*
do 
    commande_mine $f >> test.txt
    break # break first time until everything works as expected
done

如何使用传入curl命令内函数的$f参数？

您可以使用：

commande_mine() {
  local MY_OUTPUT=$(curl -X POST \
  http://localhost:5000/myapp \
  -H 'cache-control: no-cache' \
  -H 'content-type: application/json' \
  -d '{
        "filepath": "'"$1"'"
    }')
    echo "$MY_OUTPUT"
}

并称之为：

for f in "/Users/anthony/my files"/*
do 
    commande_mine "$f"
    break
done > test.txt