用jq连接json数组对象

Question

我在json文件下面:

{
"EventId": "60a0490c",
"Resources": [
    {
        "ResourceType": "AWS::STS::AssumedRole", 
        "ResourceName": "AutoScaling"
    }, 
    {
        "ResourceType": "AWS::IAM::Role", 
        "ResourceName": "arn:aws:iam:autoscaling.amazonaws.com/AWSServiceRoleForAutoScaling"
    }
    ]
}

我想连接Resources元素中的键值对,并在单行输出:

60a0490c,AutoScaling=AWS::STS::AssumedRole#AWS::IAM::Role=arn:aws:iam:autoscaling.amazonaws.com/AWSServiceRoleForAutoScaling

我试过它:

cat file.json | jq '.EventId + "," + (.Resources[] | join("="))' -r

它给我的输出为:

60a0490c,AutoScaling=AWS::STS::AssumedRole
60a0490c,AutoScaling=AWS::IAM::Role=arn:aws:iam:autoscaling.amazonaws.com/AWSServiceRoleForAutoScaling

它为每个键值对创建新条目,我希望它与不同的分隔符(#)在单行上

提前致谢.

Answer 1

你可以试试这个:

jq ' .EventId + "," + ([(.Resources[] | join("="))] | join("#"))' -r file
60a0490c,AWS::STS::AssumedRole=AutoScaling#AWS::IAM::Role=arn:aws:iam:autoscaling.amazonaws.com/AWSServiceRoleForAutoScaling

要使用第二个join(),只需将结果包含在数组中[...].

请注意,第一个键/值未交换,如示例所示(即Autoscaling值是键而不是键).