[英]How to get the first line of a file in a bash script?
I have to put in a bash variable the first line of a file. 我必须在文件的第一行放入一个bash变量。 I guess it is with the grep command, but it is any way to restrict the number of lines? 我想这是用grep命令,但它是否有任何限制行数的方法?
head
需要从文件中第一线,以及-n
参数可用于指定多少线应该提取:
line=$(head -n 1 filename)
to read first line using bash, use read
statement. 使用bash读取第一行,使用read
语句。 eg 例如
read -r firstline<file
firstline
will be your variable (No need to assign to another) firstline
将是你的变量(不需要分配给另一个)
This suffices and stores the first line of filename
in the variable $line
: 这足以将第一行filename
存储在变量$line
:
read -r line < filename
I also like awk
for this: 我也喜欢awk
:
awk 'NR==1 {print; exit}' file
To store the line itself, use the var=$(command)
syntax. 要存储行本身,请使用var=$(command)
语法。 In this case, line=$(awk 'NR==1 {print; exit}' file)
. 在这种情况下, line=$(awk 'NR==1 {print; exit}' file)
。
Or even sed
: 甚至是sed
:
sed -n '1p' file
With the equivalent line=$(sed -n '1p' file)
. 使用等效的line=$(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: 当我们用seq 10
提供read
时,查看样本,即1到10的数字序列:
$ read -r line < <(seq 10)
$ echo "$line"
1
$ line=$(awk 'NR==1 {print; exit}' <(seq 10))
$ echo "$line"
1
line=$(head -1 file)
Will work fine. 会工作得很好。 (As previous answer). (如前所述)。 But 但
line=$(read -r FIRSTLINE < filename)
will be marginally faster as read
is a built-in bash command. 因为read
是内置的bash命令,所以会略快一些。
只需将源文件的第一个列表echo
显到目标文件中即可。
echo $(head -n 1 source.txt) > target.txt
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. 问题并没有问哪个是最快的,但是为了添加到sed答案,-n'1p'表现不佳,因为模式空间仍在大文件上扫描。 Out of curiosity I found that 'head' wins over sed narrowly: 出于好奇,我发现“头部”以微弱优势赢得了胜利:
# best:
head -n1 $bigfile >/dev/null
# a bit slower than head (I saw about 10% difference):
sed '1q' $bigfile >/dev/null
# VERY slow:
sed -n '1p' $bigfile >/dev/null
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