检查文件中所有行的格式是否相同

Question

I would like to wrote a little shell script that permit to check if all lines on a file has the same number of ;

I have a file containing the following format :

$ cat filename.txt

34567890;098765456789;098765567;9876;9876;EXTG;687J;
4567800987987;09876789;9667876YH;9876;098765;098765;09876;
SLKL987H;09876LKJ;POIUYT;PÖIUYT;88765K;POIUYTY;LKJHGFDF;
TYUIO;09876LKJ;POIUYT;LKJHG;88765K;POIUYTY;OIUYT;
...
...
...
SDFGHJK;RTYUIO9876;4567890LKJHGFD;POIUYTRF56789;POIUY;POIUYT;9876;

I use the following command for determine of the number of ; of each line :

awk -F';' 'NF{print (NF-1)}' filename.txt

I have the following output :

7
7
7
7
...
...
...
7

Because number of ; on each line of this file is 7.

Now, I want to wrote a script that permit me to verify if all the lines in the file have 7 commas. If it's OK, it tells me that the file is correct. Otherwise, if there is a single line containing more than 7 commas, it tells me that the file is not correct.

Answer 1

Rather than printing output, return a value. eg

awk -F',' 'NR==1{count = NF} NF!=count{status=1}END{exit status}' filename.txt

If there are no lines or if all lines contain the same number of commas, this will return 0. Otherwise, it returns 1 to indicate failure.

Answer 2

Count the number of unique lines and verify that the count is 1.

if (($(awk -F';' 'NF{print (NF-1)}' filename.txt | uniq | wc -l) == 1)); then
    echo good
else
    echo bad
fi

Answer 3

Just pipe the result through sort -u | wc -lsort -u | wc -l . If all lines have the same number of fields, this will produce one line of output.

Alternatively, just look for a line in awk that doesn't have the same number of fields as the first line.

awk -F';' 'NR==1 {linecount=NF}
           linecount != NF { print "Bad line " $0; exit 1}
          ' filename.txt && echo "Good file"

You can also adapt the old trick used to output only the first of duplicate lines.

awk -F';' '{a[NF]=1}; length(a) > 1 {exit 1}' filename.txt

Each line updates the count of lines with that number of fields. Exit with status 1 as soon as a has more than one entry. Basically, a acts as a set of all field counts seen so far.

Answer 4

Based on all the information you have given me, I ended up doing the following. And it works for me.

nbCol=`awk -F';' '(NR==1){print NF;}' $1`
val=7
awk -F';' 'NR==1{count = NF} NF != count { exit 1}' $1
result=`echo $?`

if [ $result -eq 0  ] && [ $nbCol -eq $val ];then
echo "Good Format"
else
echo "Bad Format"
fi