I have a log that contains records like:

LineFFF 1234567 0002 @2018-06-12T17:02:50 (len:0)  99 OK  : jAFoSSYut2iJLmrk6fdQPL4OMRgd1Ybzb6U9MG1GeWKWqXlYPyfzyMVb9zn3
LineFFF 1234568 0044 @2018-06-12T17:02:57 (len:0)  84 OK  : jAFoDibWjaH4G0AEuA2dLRilreLDVAV0krEzk58ksAsp8yf38mqBzbMMrMOB
LineFFF 1234569 0053 @2018-06-12T17:02:58 (len:0)  11 OK  : jAFoFyaQtjxC3U0Q6CH+EFd0sC5QoJvo9XGLj87vfeapTrb2qZdNhcYUSGr1

Occasionally, the part of the string to the right of the ':' contains some spaces, and I want to grep/sed/awk the log and only return those lines.. I.e. 'grep for a space character from position 69 and return the full line if present'.

Any ideas? Thanks..


1 Answer 1


You can grep for a space anywhere after the 68th character from the beginning simply using

grep '^.\{68,\} ' file
  • Too easy - thank you.
    – rack201
    Sep 13, 2019 at 0:42

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