detecting the part of a filename

Shell Programming and Scripting
1

I like to have the date in the 2008-09-01 format at the beginning of my filenames. I then hyphenate after that and then have my filename.

I have a script that creates this for me. However, I may be working on files that already have the date format already in there and so I don't want to have a date twice. Is there a way to detect number or something at the beginning of the filename and if so then cut out the date?

2 
filename=2008-11-10-myfilename
newfilename=${filename##*-}
echo $newfilename
3 
filename=2008-11-17-qwerty-uiop
dpat=[12][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]
newfilename=${filename#$dpat-}
echo $newfilename

Not ${filename##*-}, as that fails if there is a hyphen in the balance of the filename, or if there is no date but there is a hyphen in the name.

4

That was in lieu of this, dates are messy

echo $filename | \
 sed  's/\(^[1-2][0-9]\{3\}\)-\([0-1][0-9]\)-\([0-3][0-9]\)-\([A-Z_0-9a-z]*\)/\4/'  | read newfilename
if [[ ! -z newfilename ]] ; then
   filename=$newfilename
fi

Basically I would suggest that naming every file with a leading date has problems. try
ls -l instead. Chris was technically correct -no doubt - but the usefulness of the above is limited. In part this is due to the way dates are constructed - and the above regex is not perfect. It will think 2008-19-19 is a date - which it is not. Most correct date regexes are several hundred character long. The only gold standard is to refer to the output of cal.

We use part of this date check algorithm below (simplified greatly) for validating old dates. This is more nearly complete && correct than most straight shell implementations using regex are likely to be. I deliberately excluded

/* chpdt.c 
usage chpdt filename 
      echo "$filname" | chpdt
*/
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <stdio.h>
#define isleapyr(y) (((y) % 4) == 0 && ((y) % 100) != 0 || ((y) % 400) == 0)
int leap[12]= {31,29,31,30,31,30,31,31,30,31,30,31};
int nleap[12]={31,28,31,30,31,30,31,31,30,31,30,31};


char  *date_ok(char *result, char *src)
{
	int value=atoi(src);
	int month=0;
	char *p=src;
	int *v=nleap;

    if( value >= 1900 && value < 2038 && (p=strchr(p, '-')) != NULL )
    {
        if(isleapyr(value))        
        	v=leap;
		value=atoi(++p);
		if(value >= 1 && value < 13 && (p=strchr(p, '-')) != NULL)
		{
			month=value;
			value=atoi(++p);
			if( value > 0 && value < v[ month - 1 ] )
				if( (p=strchr(p, '-')) != NULL)
					result=++p;
		}
    }
    return result;
}

char  *name_ok(char *p, char *src)
{
	if(src[4]=='-' && src[7]=='-' && src[10]=='-')
		p=date_ok(p, src);
	return p;
}

int main(int argc, char **argv)
{
	int retval=0;
	char tmp[PATH_MAX + 2]={0x0};
	char *p=NULL;
	if (argc < 2)
	{
		if(fgets(tmp, sizeof(tmp), stdin) ==NULL)
		{
			perror("I/O error");
			retval=1;
		};
		p=strchr(tmp, '\n');
		if(p!=NULL)
			*p=0x0;
	}
	else
		(void)strncpy(tmp, argv[1], PATH_MAX);
	p=tmp;
	if(strlen(tmp) > 10 && !retval) /* will return zero length if string is "date-"*/
		p=name_ok(p, tmp);
    if( fprintf(stdout, "%s", p) < 0)
    {
    	perror("I/O error");
    	retval=1;	
    }
    return retval;
}

Good luck with this.

5

Thanks guys. I went with Chris's solution. It is limited like you said Jim (you need to have exact syntax for it to work) but Jim your solution is way over my head for a beginner! For what I am doing Chris's will do just fine. Thanks to all of you.