bash script not executing mysql command with values passed as variables - php

Question

I am modifying a set of bash scripts that process video files and reports the processing steps to a mysql database (here is the original code in question).
The function that does the database reporting is called from the main processing script and looks like this in the original:
_report_to_db(){
if [ "${REPORT_TO_DB}" = "Y" ] ; then
echo "INSERT IGNORE INTO tableA (objectIdentifierValue,object_LastTouched) VALUES ('${MEDIA_ID}',NOW()) ON DUPLICATE KEY UPDATE object_LastTouched = NOW()" | mysql --login-path="${LOGIN_PROFILE}" "${DB_NAME}" 2> /dev/null
_db_error_check
fi
}
Since the scripts are meant to be run directly from the command line, when you run them that way it works fine. But I'm running them via php from a web interface and there's some shenanigans going on with the quoting/escaping of whitespace and/or variables.
For instance, the script breaks on the whitespace after ...| mysql and it thinks I'm trying to run mysql as root without a password and totally ignores the --login-path and the other stuff I'm piping to it.
When I call mysql from a variable like so:
_report_to_db(){
if [ "${REPORT_TO_DB}" = "Y" ] ; then
SQL_ARRAY=(INSERT IGNORE INTO tableA (columnA,lastTouched) VALUES ("${SOME_PASSED_VALUE}",NOW()) ON DUPLICATE KEY UPDATE object_LastTouched = NOW();)
MYSQL_COMMAND_ARRAY=(mysql --login-path="${LOGIN_PROFILE}" -e "${SQL_ARRAY[#]}" "${DB_NAME}")
echo "$(${MYSQL_COMMAND_ARRAY[#]})"
_db_error_check
fi
}
... I am able to log into mysql correctly but the SQL query is ignored (when it echos the result you get the standard MySQL --help output.
So far I have tried all kinds of variations on quoting, escaping, referencing the query as a separate string variable, as an array (as you see here).
What is also not helpful is that the original _db_error_check() function only checks the value of the pipe exit status. So if the pipe is ok, but there's a problem further down the path, it fails silently.
_db_error_check(){
if [ "$?" != "0" ] ; then
# reports an error if the pipe exit value ≠ 0
else
# everything is ok! even if there was a mysql error
fi
}
This is not a file or database permissions issue (already triple checked that). Are there quotes or some other stupid thing that I am missing?? Thanks! Oh, I am running OSX El Capitan.
UPDATE
Lol, I was going to post the PHP that calls the script and then I remembered that the PHP is actually calling a Pyhton script that does some other processing too, and that is what calls the bash script. Here it all is:
PHP
$command = escapeshellcmd("/usr/local/bin/python3 /Users/user/path/to/ingest.py " . $user . " 2>&1");
while (# ob_end_flush());
$proc = popen($command, 'r');
echo '<pre>';
while (!feof($proc))
{
echo fread($proc, 4096);
# flush();
}
echo '</pre>';
PYTHON
for item in os.listdir(ingestDir):
if not item.startswith("."):
filePath = os.path.abspath(ingestDir+"/"+item)
fileNameForMediaID = os.path.splitext(item)[0]
try:
ingest = subprocess.Popen(['/usr/local/bin/ingestfile','-e','-u',user,'-I',filePath,'-m',fileNameForMediaID])
ingest.wait()
os.remove(filePath)
except IOError as err:
print("OS error: {0}".format(err))
UPDATE 2
I think this might actually be a weird quirk of my installation (go figure). Using mysql --login-path=myDbUser [etc...] from a shell on my host machine I keep getting the error ERROR 1045 (28000): Access denied for user 'ADMIN'#'localhost' (using password: NO) where the client user is ADMIN and I am trying to login as myDbUser.
I actually uninstalled and reinstalled mysql (via Homebrew) and still have the same results. Using a different machine (running Sierra, but the same mysql version) I can run the above shell command successfully and log into mysql as the target user.
Also on the host machine, I can sudo -u _www zsh and run the command as the Apache user (which is the user running the whole show) without a problem. SO WHY IS IT NOT RUNNING CORRECTLY EITHER IN THE SCRIPT OR EVEN RUN FROM SHELL AS MY MAIN CLIENT USER???
Any ideas? $PATH is identical in all cases mentioned above. Same ~/.mylogin.cnf setups. Is there anything else stupid obvious I missed?

Answer 1

You need to use indirect expansion here:
echo "$(${MYSQL_COMMAND_ARRAY[#]})"
the man says:
If the first character of parameter is an exclamation point (!), and
parameter is not a nameref, it introduces a level of variable
indirection. Bash uses the value of the variable formed from the rest
of parameter as the name of the variable; this variable is then
expanded and that value is used in the rest of the substitution,
rather than the value of parameter itself. This is known as indirect
expansion. If parameter is a nameref, this expands to the name of the
variable referenced by parameter instead of performing the complete
indirect expansion. The exceptions to this are the expansions of
${!prefix*} and ${!name[#]} described below. The exclamation point
must immediately follow the left brace in order to introduce
indirection.
${!name[#]}
${!name[*]}
If name is an array variable, expands to the list of array indices
(keys) assigned in name. If name is not an array, expands to 0 if name
is set and null otherwise. When ‘#’ is used and the expansion appears
within double quotes, each key expands to a separate word.
PS: If I may put forward a piece of my personal opinion, having a chain of php -> python -> bash is the worst coding style one can ever met, you may want to rewrite it into single langue so it will be easier to track down further issues at least.