如何在 bash 的 CURL 请求中使用变量？

Question

目标：

我正在使用 bash CURL 脚本连接到 Cloudflare APIv4。目标是更新 A 记录。我的脚本：

   # Get current public IP
      current_ip=curl --silent ipecho.net/plain; echo

   # Update A record
      curl -X PUT "https://api.cloudflare.com/client/v4/zones/ZONEIDHERE/dns_records/DNSRECORDHERE" \
        -H "X-Auth-Email: EMAILHERE" \
        -H "X-Auth-Key: AUTHKEYHERE" \
        -H "Content-Type: application/json" \
        --data '{"id":"ZONEIDHERE","type":"A","name":"example.com","content":"'"${current_ip}"'","zone_name":"example.com"}'

问题：

当我在我的脚本中调用它时，current_ip 变量没有被打印出来。输出将是 "content" : "" 而不是 "content" : "1.2.3.4".

我使用了 other Whosebug 帖子，我正在尝试遵循他们的示例，但我认为我仍然做错了什么，只是无法弄清楚是什么。 :(

Answer 1

从 shell 脚本编辑 JSON 的可靠方法是使用 jq:

# set shell variables with your contents
email="yourEmail"
authKey="yourAuthKey"
zoneid="yourZoneId"
dnsrecord="yourDnsRecord"

# make sure we show errors; --silent without --show-error can mask problems.
current_ip=$(curl --fail -sS ipecho.net/plain) || exit

# optional: template w/ JSON content that won't change
json_template='{"type": "A", "name": "example.com"}'

# build JSON with content that *can* change with jq
json_data=$(jq --arg zoneid "$zoneid" \
               --arg current_ip "$current_ip" \
               '.id=$zoneid | .content=$current_ip' \
               <<<"$json_template")

# ...and submit
curl -X PUT "https://api.cloudflare.com/client/v4/zones/$zoneid/dns_records/$dnsrecord" \
  -H "X-Auth-Email: $email" \
  -H "X-Auth-Key: $authKey" \
  -H "Content-Type: application/json" \
  --data "$json_data"

Answer 2

正如 Charles Duffy 的回答所暗示的那样，为此使用 jq 是一个非常好的主意。但是，如果您不能或不想安装 jq，您可以使用普通 POSIX shell.

#!/bin/sh
set -e

current_ip="$(curl --silent --show-error --fail ipecho.net/plain)"
echo "IP: $current_ip"

# Update A record
curl -X PUT "https://api.cloudflare.com/client/v4/zones/ZONEIDHERE/dns_records/DNSRECORDHERE" \
    -H "X-Auth-Email: EMAILHERE" \
    -H "X-Auth-Key: AUTHKEYHERE" \
    -H "Content-Type: application/json" \
    --data @- <<END;
{
    "id": "ZONEIDHERE",
    "type": "A",
    "name": "example.com",
    "content": "$current_ip",
    "zone_name": "example.com"
}
END

如何在 bash 的 CURL 请求中使用变量？

How to use a variable in a CURL request with bash?

api

bash

shell

curl

cloudflare