如何使用bash在CURL请求中使用变量? [英] How to use a variable in a CURL request with bash?

问题描述

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:

I'm using a bash CURL script to connect to the Cloudflare APIv4. The goal is to update an A-record. My script:

   # 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:

当我在脚本中调用current_ip变量时，不会将其打印出来.输出将是"content" : ""而不是"content" : "1.2.3.4".

The current_ip variable is not printed when I call it in my script. The output will be "content" : "" and not "content" : "1.2.3.4".

我使用了其他 stackoverflow帖子我正在尝试遵循他们的示例，但我认为我仍然在做错事，只是无法弄清楚是什么. :(

I used other stackoverflow posts and I'm trying to follow their examples but I think I'm still doing something wrong, just can't figure out what. :(

推荐答案

为此使用 jq ，正如查尔斯·达菲(Charles Duffy)的答案所暗示的，这是一个非常好的主意.但是，如果您不能或不想安装jq，则可以使用普通的POSIX shell进行操作.

Using jq for this, as Charles Duffy's answer suggests, is a very good idea. However, if you can't or do not want to install jq here is what you can do with plain 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

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