我有一个日志文件，其中包含“随机”内容和json字符串。我想要的是，从json字符串中提取某些变量/值对，但是我不确定该怎么做。例如，我有类似的东西：
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并假设我想提取文件中每个json行的“牌照”和以下“信心”值。 如何在bash中做到最好？
答案 0 :(得分：1)
您可以使用以下grep命令：
grep -o '"\(licenseplate\|confidence\)":"[^"]*"' inputfile
打印：
"licenseplate":"49239BA"
"confidence":"77"
"confidence":"98"
"confidence":"94"
...
"licenseplate":"BLEBR1"
"confidence":"46"
"confidence":"94"
"confidence":"76"
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答案 1 :(得分：1)
您可以使用grep提取JSON数据并使用jq进行解析：
grep -o '{.*}' file | jq -r '.frames.frame.lps.lp | .licenseplate, .confidence'
BLEBR1
46
如果JSON数据为1行，这将起作用。由于正则表达式{.*}是贪婪的，因此grep将立即捕获整个JSON数据。
jq选择JSON结构frames.frame.lps.lp，从中提取值licenseplate和confidence。请注意-r以获取原始数据（不带双引号）。
如果您需要在同一行中同时使用这两个值（以进行进一步解析），请使用以下jq命令：
grep -o '{.*}' file | jq -r '.frames.frame.lps.lp | "\(.licenseplate) \(.confidence)"'
BLEBR1 46