linux有什么命令可以提取指定的字符串？ 如何操作

Question

有个日志文件如下：

// demo.log
"time":"2017","id":"AA2342ds34234","content":"asdfasdfafwerfasdf"
"time":"2017","content":"asdfasdfafwerfasdf","id":"AAd234s234234"
"time":"2017","id":"AA23as423d4234","content":"asdfasdfafwerfasdf"
"id":"BB234234234","content":"asdfasdfafwerfasdf","time":"2017"

我想只输出id的值：
例如这样

AA2342ds34234
AAd234s234234
AA23as423d4234
BB234234234

有类似功能的的命令有awk, gawk，sed, grep等，看了半天也没找到我想要的这种功能的示例。
请问有什么命令可以提取，麻烦给个例子，谢谢

Answer 1

grep -Po '(?<="id":")[^"]+' demo.log

awk -F'[,:]' '{gsub(/"/,"");for(n=0;n++<NF;)if($n=="id")print $++n}' demo.log

sed 's/.*"id":"\([^"]\+\)".*/\1/' demo.log

Answer 2

cat demo.log  | sed 's/.*"id":"\([0-9]*\)".*/\1/'

Answer 3

javascript@linux:~/test$ cat demo.log
"time":"2017","id":"234234234","content":"asdfasdfafwerfasdf"
"time":"2017","content":"asdfasdfafwerfasdf","id":"234234234"
"time":"2017","id":"234234234","content":"asdfasdfafwerfasdf"
"id":"234234234","content":"asdfasdfafwerfasdf","time":"2017"
javascript@linux:~/test$ sed -r 's/.*"id":"([0-9]+)".*/\1/g' demo.log
234234234
234234234
234234234
234234234

Answer 4

awk是个比较经典的文本分析工具，支持二次编程，我做数据采集是经常使用它！
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