以cron形式运行的C脚本，授予权限被拒绝错误

Question

I have a .c file compiled and would like to run via a cron job but I end up getting this error:

/bin/sh: /usr/local/bin/get1Receive.c: Permission denied. 

What is causing this error and how do I fix it?

Should I be running the .c file in cron or a different compiled file?

Results from /tmp/myvars

GROUPS=()
HOME=/root
HOSTNAME=capture
HOSTTYPE=x86_64
IFS='
'
LOGNAME=root
MACHTYPE=x86_64-redhat-linux-gnu
OPTERR=1
OPTIND=1
OSTYPE=linux-gnu
PATH=/usr/bin:/bin
POSIXLY_CORRECT=y
PPID=11086
PS4='+ '
PWD=/root
SHELL=/bin/sh
SHELLOPTS=braceexpand:hashall:interactive-comments:posix
SHLVL=1
TERM=dumb
UID=0
USER=root
_=/bin/sh

Results from file get1Receive.c

file get1Receive.c
get1Receive.c: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.18, not stripped

Snippet of codes.

sprintf(queryBuf1,"SELECT ipDest, macDest,portDest, sum(totalBits) FROM dataReceive WHERE timeStampID between '%s' And '%s'  GROUP BY ipDest, macDest, portDest ",buff1,buff2);
                printf("\nQuery receive %s",queryBuf1);


                if(mysql_query(localConn, queryBuf1))
                {
                    //fprintf(stderr, "%s\n", mysql_error(localConn));
                    printf("Error in first query of select %s\n",mysql_error(localConn));
                    exit(1);
                }

                localRes1 = mysql_store_result(localConn);
                int num_fields = mysql_num_fields(localRes1);

                printf("\nNumf of fields : %d",num_fields);
                printf("\nNof of row : %lu",mysql_num_rows(localRes1));

Answer 1

If the output of this command:

file  get1Receive1.c

shows that file name to be a valid executable that part is very unusual, but okay.

Assuming you are using biz14 (or your real username's ) crontab try this:

use the command crontab -e to create this line in your crontab:

* * * * *  set > /tmp/myvars

Wait a few minutes, go back into crontab -e and delete that entry.

Use the set command from the command line to see what variables and aliases exist. Compare that with that you see in /tmp/myvars You have to change how your C code executes by changing the variables and aliases the cron job runs with.

If you are running the cron job in someone else's crontab, then you have a bigger problem. Check file permissions on get1Receive1.c. and the directory it lives in. That other user (the one who wons the crontab) has to have permissions set on your directory and get1Receive1.c so the job can run.

Example crontab entry:

0 10 * * 1-5 /path/to/get1Receive1.c > /tmp/outputfile

Read /tmp/outputfile to see what you got. You are using printf in your code. printf only writes to the controlling terminal. There is no controlling terminal, so redirect the printf stuff to a file.

Last effort on this problem: Check return codes on EVERYTHING. All C functions like fread(), any db function, etc. If a return code gives a fail response ( these are different for different function calls) then report the error number the line number and function - gcc provides LINE and func . Example:

printf("error on line %d in my code %s, error message =%s\n", __LINE__, __func__, [string of error message]);

If you do not check return codes you are writing very poor C code.

CHECK return codes, please, now!

Answer 2

Permission wise you could have two issues. 1. The 'c' file's permissions don't allow who you are running it as to run it. 2. You are running the cron with a script which doesn't have permissions.

Here's a helpful post: How to give permission for the cron job file?

The fact that you are running a 'c' file and referring to it as a script makes me think you're using C shell and not writing it as a C language program which would need to be compiled and have the generated executable run by the cron. If you're not using gcc or have never called gcc on your 'C' script then it's not C and call it C shell to avoid confusion.