Question

我有下面的json文件：

{
"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，并在单行上输出为：

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

我尝试过：

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一个新条目，我希望它位于具有不同分隔符（＃）的单行中

谢谢。

Answer 1

您可以尝试以下方法：

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()，只需将结果封装在数组[...]中。

请注意，第一个键/值不会像您的示例所示那样交换（即Autoscaling是值而不是键）。

用jq连接json数组对象

1 个答案: