Can I grep a certain amount of lines before or after a string I want?

Question

Let's say I'm using lshw to get my system memory, and I want it printed out so that I only get the system memory option. Can you specify how many lines you want to print before or after a string in grep? A snippet of the lshw command is output below for reference:

 description: Computer
    width: 64 bits
  *-core
       description: Motherboard
       physical id: 0
     *-memory
          description: System memory
          physical id: 0
          size: 23GiB
     *-cpu
          product: Intel(R) Core(TM) i5-8350U CPU @ 1.70GHz
          vendor: Intel Corp.
          physical id: 1
          bus info: cpu@0
          capacity: 1896MHz
          width: 64 bits

I could use lshw | grep size | awk -F: '{print $2}' to get the 23 GiB portion, but I want to see if there's a way to get a block of text with grep to get the full memory section.

Are you looking for `-C`, e.g. `grep -C 5`? C is for context. You can also explicitly show lines `-B`efore and `-A`fter your matches. — L3viathan, Jul 30 '18 at 18:58
Use the -A (after) and -B (before) command-line options in grep. — FBergo, Jul 30 '18 at 18:58
@L3viathan yeah you're totally right. Thanks a lot, I changed my post and credited you. — AndreasKralj, Jul 30 '18 at 19:05

score 5 · Accepted Answer · answered Jul 30 '18 at 19:03

@L3viathan helped me figure it out. Grep has an -A flag that allows you to specify how many lines you want after the string. You can also use the -B flag to specify how many lines you want before the string.

After Example:

lshw | grep *-memory -A 3

Output:

     *-memory
      description: System memory
      physical id: 0
      size: 23GiB

Before Example:

lshw | grep size -B 3

Output:

     *-memory
      description: System memory
      physical id: 0
      size: 23GiB

Can I grep a certain amount of lines before or after a string I want?

1 Answers1