Identify records with null character

[kasparov]

Some records in my database have got corrupted - when I extract the data to a text file I discover that some records have the null character (\0) in them. My question is how can I easily determine which records have the null char. What I've done so far is:

od -c <datafile>

This works but it's not easy to identify the record. Is there a way to use grep, sed, or anything else to identify these corrupted records? Everything I try seems to either treat every line or none of them as including the char.

TIA, Chris

 

[jamisar]

interesting question.
/bin/od is OK and if can live with it's
(stupid) output

od -c file | sed -ne 's/\\0/ERROR/p'

this works. following code gives you a better output

--------------------------------

/* jean-michel.sarbach@ubs.com */

/* to compile: cc -o exec-name nod.c */
/* usage: exec-name filename */
/* quiq && dirty script to reformat od's output, marking special characters */

#include <stdio.h>

#define MaxBuff 128 /* od's output is max 80 char */
#define ERRO &quot;***ERROR***&quot;

int main(int argc, char **argv)
{
FILE *in;
char *ppp, buff[MaxBuff+1];
int space, special, input;

if(!--argc) exit(1); ++argv;

/* we don't need to rewrite 'od' exec it and capture the output */

sprintf(buff,&quot;/bin/od -c %s&quot;,*argv); in = popen(buff,&quot;r&quot;

while(fgets(buff,MaxBuff,in)){

input = special = space = 0;

for(ppp = buff+7; *ppp && *ppp != '\n'; ++ppp){

++input;
switch(*ppp){
case ' ':
if(3 > space++) break;
space = 0;
printf(&quot; &quot;
break;
case '\\':
if(2 == space){ special = 1; break; }
special = space = 0;
printf(&quot;\\&quot;
break;
default :
space = 0;
if(!special){ printf(&quot;%c&quot;,*ppp); break; }
special = 0;
switch(*ppp){
case 'n': printf(&quot;\n&quot; break;
case 't': printf(&quot;\t&quot; break;
case '0': printf(ERRO); break;
default: break;
}
break;
}
}
}
pclose(in);
if(input) printf(&quot;\n&quot;
exit(0);
}