1. There is a text test.log, content that is:
1 an A
2 b B
3 c C
2. Execute while to get normal output
cat test.log |while read item;do echo $item;done1 an A
2 b B
3 c C
3. But executing while+awk loses the first line
cat test.log |while read item;do awk "{print $1"A"}";done2A< hr > < H2 > question: why did you lose the first row of data? < / H2 > < H2 > second question: how do I get the following output? < / H2 >
3A
1A
2A
3A
@ Yujiaao has answered your second question, that is, how to get the correct output. Because awk can loop through all delimited lines (usually separated by \ n ), it is no longer necessary to use while to read.
therefore has the answer to your question, why is there a line missing:
you used the read command in while, which reads from standard input, so it reads the first line into the item variable. When awk does not follow the file parameters, it also reads the contents from the standard input, so awk reads and outputs the rest of the content.
in addition, if you use awk, there is no need to use cat to read the contents of the file, and then pipe it to awk; because awk can read the file directly:
awk ' {print $1"A"}' test.logawk
cat test.log |awk ' {print $1"A"}'
1A
2A
3A
Previous: In multiple SQL tables, how does the average result of one query relate to another table?
Next: Oracle trigger executes Times ORA-04091 and ORA-00060 errors
when using read, the following problems are encountered, which are not consistent with what you think. The code and results are as follows -sharp! bin bash IFS= read -d - var1 var2 <<< " 123 -" printf "%s n" "...
1, if you use the linux that comes with win10 to ssh 1: 2, use cmd to ssh other ssh software can display normally, the picture is not posted, wsl local adjustment normal response far end is centos no usr bin resize Type stty cols xx stty...