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extracting a line based on line number
i want to cut all the entries from the /etc/passwd file in which the uid is> 500
for this i was writing this ,m quiet new to all this.. scripting but on the 6th n 8th line ,, i hav to specify a line number .. to get the commnd working .. but i want to use variable i instead of that ,, which is not allowed .. my problem is that ,, i have a variable i , which is being incremented n it is containing an integer value as u can c , now i want to print that line which is satisfying the if condition . because i was not working so i wrote 56 instead of that , as in my passwd file it was havin pid >500 just to test .. #!/bin/bash nol=`wc -l /etc/passwd | cut -d " " -f 1` i=1 if [ $nol -ge $i ] then if [ `sed -n '56p' /etc/passwd | cut -d \: -f 3` -ge 500 ] then sed -n '56p' /etc/passwd >hii i=`expr $i + 1` fi fi exit |
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