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# 1  
Old 03-23-2011
Report missing files

Hello I need to check various directories for the existence of files. If a file or more are missing I need to display it:

e.g files in a dir:

data_file.11032300.Z
data_file.11032301.Z
data_file.11032302.Z
data_file.11032303.Z
data_file.11032304.Z
data_file.11032305.Z
data_file.11032306.Z
data_file.11032307.Z
data_file.11032308.Z
data_file.11032309.Z
data_file.11032310.Z
data_file.11032311.Z
data_file.11032312.Z
data_file.11032313.Z
data_file.11032314.Z
data_file.11032315.Z
data_file.11032316.Z
data_file.11032317.Z
data_file.11032318.Z
data_file.11032319.Z
data_file.11032320.Z
data_file.11032321.Z
data_file.11032322.Z
data_file.11032323.Z
(Where the numbers are YYMMDDHH)

So far:
Code:
#!/usr/bin/sh

DATA_DIR=/RawData/processed_files
DATA_FILE_NO=$(ll data_file.$DATE_YYMMDD*.Z | wc -l)

echo "How many days before to check?"
read THE_DAYS

#Gather YYMMDD from Oracle
DATE_YYMMDD=`${LIB}/dt_YYMMDD -${THE_DAYS}`

cd $DATA_DIR

if  ((DATA_FILE_NO != 24))
then
echo amr_date file is missing
fi

How can I display e.g.

Missing data_file:
--------------------
data_file.11032308
data_file.11032309

Instead of just:
date_file is missing

Thanks in advance.

Last edited by drbiloukos; 03-23-2011 at 07:46 AM..
# 2  
Old 03-23-2011
You can wrap it in a for loop:
Code:
for hr in `seq 0 23` ; do 
  fname=data_file.$DATE_YYMMDD${hr}.Z
  [[ ! -f "$fname" ]] && echo Missing $fname
done


Last edited by mirni; 03-23-2011 at 08:05 AM..
# 3  
Old 03-23-2011
./missing files.sh[20]: seq: not found.

HP-UX B.11.11 U 9000/800
# 4  
Old 03-23-2011
Code:
for i in {0..23}
do
  hr=$(printf "%02s" $i)
  fname=data_file.$DATE_YYMMDD${hr}.Z
  [[ ! -f "$fname" ]] && echo Missing $fname
done

or change the for loop to:

Code:
for (( i=0; i<=23; i++ ))


Last edited by rdcwayx; 03-23-2011 at 08:37 AM..
This User Gave Thanks to rdcwayx For This Post:
# 5  
Old 03-23-2011
how about
Code:
hr=0;
while [ $hr -lt 24 ]; do
  [[ $hr -lt 10 ]] && fname=data_file.$DATE_YYMMDD0${hr}.Z
  [[ $hr -ge 10 ]] && fname=data_file.$DATE_YYMMDD${hr}.Z
  [[ ! -f "$fname" ]] && echo Missing $fname
((hr++))
done

If that increment doesn't work (check with 'echo $hr' inside loop), try
hr=$(($hr+1))

---------- Post updated at 12:06 AM ---------- Previous update was at 12:05 AM ----------

@rdcwayx : the loop should have been from 0 to 23; I messed up in my orig reply...

Last edited by mirni; 03-23-2011 at 08:23 AM..
This User Gave Thanks to mirni For This Post:
# 6  
Old 03-23-2011
Mirni it works but i need to generate 00 01 02..24 not 0 1 2..24

I can break into to checks:
from 0-9
data_file.$DATE_YYMMDD$0{hr}.Z

and

from 10-24
data_file.$DATE_YYMMDD${hr}.Z

If something more efficient please suggest....

Last edited by drbiloukos; 03-23-2011 at 08:14 AM..
# 7  
Old 03-23-2011
Code:
FILE=`find ${DATA_DIR}`
COUNT="11032300"

for LINE in `echo ${FILE}`
do
     BASENAME_FILE=`basename ${LINE}`
     DATE_FILE=`echo "$BASENAME_FILE" | awk -F "." '{ print $2 }'`
     if [ "${COUNT}" != "${DATE_FILE}" ]
          do
                 echo "..."  ----> whatever you like
          done
     COUNT=`expr ${COUNT} + 1`
done

This User Gave Thanks to acmvillareal For This Post:

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