sed or awk grep, that will only get the line with more characters.


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# 1  
sed or awk grep, that will only get the line with more characters.

Is there a command for sed and awk that will only sort the line with more characters?

Code:
#cat file
123
12345
12
asdgjljhhho
bac
ss

Output:
asdgjljhhho


#cat file2
11.2
12345.00
21.222
12345678.10

Output:
12345678.10

--- Post updated at 12:40 AM ---

Closing this now. I figured it out.
Thanks
# 2  
Print the longest line in the file?
awk:
Code:
awk '{ x=length } x>max { max=x; save=$0 } END { print save }' file

shell built-ins:
Code:
max=0; while read line; do x=${#line}; if [ $x -gt $max ]; then max=$x; save=$line; fi; done < file; echo "$save"

sed: would be (too) complicated.
# 3  
Code:
sed -n "/.\\{$(wc -L < file)\\}/{p;q}" file
grep -m 1 ".\\{$(wc -L < file)\\}" file
awk 'length==l {print;exit}' l=$(wc -L < file) file


Last edited by rdrtx1; 08-20-2019 at 02:54 PM..
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