linux - 在调用 curl 的脚本中使用变量

Question

我试图在以下脚本中将 1.2.3.4 替换为变量 $wanip 的内容。

wanip="4.3.2.1"
echo $wanip
content=$(curl --insecure -H "X-DNSimple-Token: foo:bar" -H "Accept: application/json" -H "Content-Type: application/json" -X PUT -d "{\"record\": {\"name\": \"foo\",\"content\": \"1.2.3.4\"}}" https://acme.com/records/123)
echo $content

如果我从字面上用 $wanip* 替换 1.2.3.4，当我运行脚本时我会收到一条消息:“消息”:“解析 JSON 时遇到问题”。

Answer 1

尝试添加一个抽象层:

#!/bin/bash
wnip="4.3.2.1"
echo $wanip
command="curl --insecure -H 'X-DNSimple-Token: foo:bar' -H 'Accept: application/json' -H 'Content-Type: application/json' -X PUT -d '{\"record\": {\"name\": \"foo\",\"content\": \"${wnip}\"}}' https://acme.com/records/123"
echo $command
content=$($command)
echo $content