How to get the first line of a file in a bash script

Question

I have to put in a bash variable the first line of a file  I guess it is with the grep command  but it is any way to restrict the number of lines

User · Answer

The question didn t ask which is fastest  but to add to the sed answer  -n  1p  is badly performing as the pattern space is still scanned on large files  Out of curiosity I found that  head  wins over sed narrowly     best  head -n1  bigfile  gt  dev null    a bit slower than head  I saw about 10  difference   sed  1q   bigfile  gt  dev null    VERY slow  sed -n  1p   bigfile  gt  dev null

User · Answer

head takes the first lines from a file  and the -n parameter can be used to specify how many lines should be extracted   line   head -n 1 filename

User · Answer

This suffices and stores the first line of filename in the variable  line   read -r line  lt  filename   I also like awk for this   awk  NR  1  print  exit   file   To store the line itself  use the var   command  syntax  In this case  line   awk  NR  1  print  exit   file    Or even sed   sed -n  1p  file   With the equivalent line   sed -n  1p  file      See a sample when we feed the read with seq 10  that is  a sequence of numbers from 1 to 10     read -r line  lt   lt  seq 10     echo   line  1    line   awk  NR  1  print  exit    lt  seq 10     echo   line  1

User · Answer

to read first line using bash  use read statement  eg   read -r firstline lt file   firstline will be your variable  No need to assign to another

User · Answer

Just echo the first list of your source file into your target file   echo   head -n 1 source txt   gt  target txt

User · Answer

line   head -1 file    Will work fine   As previous answer   But  line   read -r FIRSTLINE  lt  filename    will be marginally faster as read is a built-in bash command

[bash] How to get the first line of a file in a bash script?

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