问 (Linux)具体使用Grep提取JSON值(没有json解析器)

EN

由于OP清楚地提到了json解析器不能被使用，所以在GNU grep 中不能这样回答。用GNU grep 编写和测试，仅显示示例。而且，专家们总是建议使用json解析器，所以如果您有兴趣的话。

echo "$Response" | grep -ozP '(^|\n){\n"count":\s+[0-9]+,\n"items":\s+\[{\n\s+"oid":\s+"\K[^"]*'

输出如下： xyzxyzxyzxyzxyzxyzxyz

regex的 解释： 添加了对上述regex使用的详细说明，它仅用于解释目的，如需使用，请参阅上面的GNU grep 命令。

(^|\n){\n    ##Matching new line OR starting here followed by { and new line.
"count":\s+  ##Matching "count": followed by 1 or more spaces.
[0-9]+,\n    ##Matching 1 or more digits followed by comma followed by new line.
"items":\s+  ##Matching "items": followed by 1 or more spaces.
\[{\n\s+     ##matching literal [ followed by { new line and spaces.
"oid":\s+"   ##Matching "oid": followed by 1 or more spaces followed by " here.
\K           ##Here is GNU grep's GREAT option \K which helps us to forget previous match.
             ##Basically match everything but forget its value so that we can get only required values.
[^"]*        ##Match everything just before next occurrence of " here.

试试这个：

Response='
"count": 1,
"items": [{
        "oid": "xyzxyzxyzxyzxyzxyzxyz",
        "creationDate": "2019-02-05T02:21:08.662+0000"