How to extract elements in a field using a number

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# 8  
Old 12-12-2007
Hi porter,

Still has error of:-



(standard_in) 1: parse error

My sample code is:-

cat a.txt|while read LINE

char=`echo "${LINE}"| awk '{print $2}'`

#echo $char
i=`grep -n "^$char" b.txt|awk '{print $2}'`
j=`grep -n "^$char" b.txt|awk '{print $3}'`
k=`grep -n "^$char" c.txt|awk '{print $1}'`
q=`echo \( $j / $i \) + $j | bc`

echo $i
echo $j
echo $k
echo $q


My sample input file (let say for illustration purpose here) is

aa 108
bb 109
cc 110
ddd 100

108 5 6
110 2 2
109 2 2


2 0.959341 109
3 0.959341 108
4 0.797717 367
5 0.788534 83

I dont see any problem in here though...please advise. Thanks.
# 9  
Old 12-12-2007
As always I recommend debugging scripts using

(a) strategically placed statements that print the value of variable

(b) using "#!/bin/sh -x" at the start of the script

(c) the equivalent of asserts, where you exit if some condition is not met, see (a)
# 10  
Old 12-12-2007
Hi Porter,

I debug the script and found out that:-

i) ++ echo '6/(' 5 '+6)'
++ bc
+ q=0

Does it means bc will round down the value? How can I get the exact value of division in this case?

ii) As I raised before, given the file has quite similar number.It will still print out all the possible number that found.

++ awk '{print $1}'
+ k='2:

This basically print the positions of file in c.txt(below) when 108 is grepped. Which means k=`grep -n "$char" c.txt|awk '{print $1}'` is not working to grab exact number.

2 0.959341 109
3 0.959341 108
4 0.797108 367
5 0.788534 83

iii)the (standard_in) 1: parse error comes in when the bc tries to calculate non value from (non match entry). Do you have any idea how to prevent the q to be calculate when the grep returns empty?
I tried to insert while and if inside but it returns me
./grep2.bash: line 49: syntax error near unexpected token `done'
./grep2.bash: line 49: `done '

Please help. Thanks.

# 11  
Old 12-13-2007
select b.col2 i, b.col3 j, c.col1 k, (b.col3/b.col2)+b.col3 q
from a.txt a, b.txt b, c.txt c
where a.col2 = b.col1
and a.col2 = c.col3

See how simple & clean? Now you just need to upload these three files into three database tables Smilie

On a serious note, use FILENAME variable in awk to differentiate between multiple files. Using awk is much better than wrestling with shell constructs/commands - particularly if your files are big. The performance gains are massive.

# 12  
Old 12-13-2007
Originally Posted by ahjiefreak
Do you have any idea how to prevent the q to be calculate when the grep returns empty?
"if/then/fi" springs to mind.

Regarding the calculations appearing like integer, try changing the formula to

j=`echo \( $j.00000 / $i.00000 \) + $j.00000 | bc`

# 13  
Old 12-13-2007

Do you mean have this code in the bash itself?
I have error of

./grep4: line 3: syntax error near unexpected token `i,'
./grep4: line 3: `select b.col2 i, b.col3 j, c.col1 k, (b.col3/b.col2)+b.col3 q'

# 14  
Old 12-13-2007

Sorry Jason - that was my lame try at humor - do not execute that Smilie

Try porter's solution & post the results.
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