【发布时间】:2019-12-15 21:50:10
【问题描述】:
假设我有一个变量 -
[
{
"outer_key_1" = [
{
"ip_cidr" = "172.16.6.0/24"
"range_name" = "range1"
},
{
"ip_cidr" = "172.16.7.0/24"
"range_name" = "range2"
},
{
"ip_cidr" = "172.17.6.0/24"
"range_name" = "range3"
},
{
"ip_cidr" = "172.17.7.0/24"
"range_name" = "range4"
},
]
},
{
"outer_key_2" = [
{
"ip_cidr" = "172.16.5.0/24"
"range_name" = "range5"
},
{
"ip_cidr" = "172.17.5.0/24"
"range_name" = "range6"
},
]
},
]
我可以合并列表中的地图来获得这个输出 -
{
"outer_key_1" = [
{
"ip_cidr" = "172.16.6.0/24"
"range_name" = "range1"
},
{
"ip_cidr" = "172.16.7.0/24"
"range_name" = "range2"
},
{
"ip_cidr" = "172.17.6.0/24"
"range_name" = "range3"
},
{
"ip_cidr" = "172.17.7.0/24"
"range_name" = "range4"
},
]
"outer_key_2" = [
{
"ip_cidr" = "172.16.5.0/24"
"range_name" = "range5"
},
{
"ip_cidr" = "172.17.5.0/24"
"range_name" = "range6"
},
]
}
我已经使用
result = merge(variable[0], variable[1])
但是当我尝试这个时
result = merge(variable[*])
我收到一个错误提示
调用函数“合并”失败:参数必须是映射或对象,得到 “元组”。
为什么我使用 splat 运算符时合并失败? 有没有更好的方法按照上面的要求合并列表中的地图?
【问题讨论】:
标签: for-loop maps terraform hcl