无与伦比的闭合支架/支架

Question

我正试图在Bash中执行一个脚本，但是把这个错误扔了，

curl: (3) URL using bad/illegal format or missing URL
curl: (3) URL using bad/illegal format or missing URL
curl: (3) URL using bad/illegal format or missing URL
curl: (3) unmatched close brace/bracket in URL position 24:
UTC","user_id":"01234"}}

我尝试删除牙套，但不起作用，这是行，

response=$(curl  -X POST -H "Authorization: Bearer ${bearer_token}" -H "Content-Type: application/json" -d '{"cursus_user":{"begin_at":"'${start}'","cursus_id":"'${cursus_id}'","end_at":"'${end}'","user_id":"'${user}'"}}' "https://xxxxxx/xxxxxx.com")

任何人都知道问题在哪里？

[更新]

我真的看不到错误:(

'
{
    "cursus_user":
    {
        "begin_at":     "'${start}'",
        "cursus_id":    "'${cursus_id}'",
        "end_at":       "'${end}'",
        "user_id":      "'${user}'"
    }
}
'

Answer 1

这实际上是格式的评论。

两个提示：

• 使用 jq 生成json-将使您的所有报价正确使用，
• 请使用阵列可读性（我不喜欢水平滚动）

data=$(
    jq  --null-input \
        --compact-output \
        --arg begin_at  "$start" \
        --arg cursus_id "$cursus_id" \
        --arg end_at    "$end" \
        --arg user_id   "$user" \
        '{cursus_user: $ARGS.named}'
)

curl_opts=(
    -X POST
    -H "Authorization: Bearer ${bearer_token}"
    -H "Content-Type: application/json" 
    -d "$data"
)

response=$(curl "${curl_opts[@]}" "https://xxxxxx/xxxxxx.com")

Answer 2

shellCheck 标识curl命令中的几个未引用的变量。它甚至提供了校正的代码：

response=$(curl -X POST -H "Authorization: Bearer ${bearer_token}" -H "Content-Type: application/json" -d '{"cursus_user":{"begin_at":"'"${start}"'","cursus_id":"'"${cursus_id}"'","end_at":"'"${end}"'","user_id":"'"${user}"'"}}' "https://xxxxxx/xxxxxx.com")

使用 shellCheck 在使用Shell Code时，通常会节省很多时间。

Answer 3

添加其他双引号；说。，“'” $ {start}“'”而不是“'$ {start}'”

{
    "cursus_user":
    {
        "begin_at":     "'"${start}"'",
        "cursus_id":    "'"${cursus_id}"'",
        "end_at":       "'"${end}"'",
        "user_id":      "'"${user}"'"
    }
}