I have the following file:
/home/user/public_html/web-site.org/download/audio/1999/01-03 Good News.mp3
The problem is that the error is thrown and the file is exited when it exists. I'm checking for file_exists because I'm doing a copy() later and it's failing cause the file doesn't existing. I'm hosted on hostGator and my permissions on these mp3s are 755.
if (!file_exists('/home/user/public_html/web-site.org/download/audio/'.$sermon['year'].'/'.$sermon['file'])) {
echo "Oops, file doesn't exist: <b>/home/user/public_html/web-site.org/download/audio/".$sermon['year']."/".$sermon['file']."</b>";
exit;
}
Could it be the space in the filename? I have about 1,100 audio files to process, so changing the filename manually would be a nightmare...
update
urlencode() and trim() on the filename has no effect
str_replace(' ', '-', $sermon['file']) on the filename doesn't work (after changing spaces to dashes in the filename itself)
I checked the permissions on the folder the file is in, it WAS 700 and I changed it to 755 and it had no effect.
update
I read in another thread that having a . in the path anywhere could cause a problem... so I changed .org to _org and the problem persisted, so I've reverted those changes.
Try escaping the spaces in the filename.
/home/user/public_html/web-site.org/download/audio/1999/01-03\ Good\ News.mp3
wow.... fail on my part.... epic fail.... my user path was wrong
Related
Upon looping a directory to delete txt files ONLY - a message is returned indicating The System cannot find the file specified: 'File.txt'.
I've made sure the txt files that I'm attempting to delete exist in the directory I'm looping. I've also checked my code and to make sure it can see my files by printing them in a list with the print command.
import os
fileLoc = 'c:\\temp\\files'
for files in os.listdir(fileLoc):
if files.endswith('.txt'):
os.unlink(files)
Upon initial execution, I expected to see all txt files deleted except for other non-txt files. The actual result was an error message "FileNotFoundError: [WinError 2] The system cannot find the file specified: 'File.txt'.
Not sure what I'm doing wrong, any help would be appreciated.
It isn't found because the the path you intended to unlink is relative to fileLoc. In fact with your code, the effect is to unlink the file relative to the current working directory. If there were *.txt files
in the cwd then the code would have unfortunate side-effects.
Another way to look at it:
Essentially, by analogy, in the shell what you're trying to do is equivalent to this:
# first the setup
$ mkdir foo
$ touch foo/a.txt
# now your code is equvalent to:
$ rm *.txt
# won't work as intended because it removes the *.txt files in the
# current directory. In fact the bug is also that your code would unlink
# any *.txt files in the current working directory unintentionally.
# what you intended was:
$ rm foo/*.txt
The missing piece was the path to the file in question.
I'll add some editorial: The Old Bard taught us to "when in doubt, print variables". In other words, debug it. I don't see from the OP an attempt to do that. Just a thing to keep in mind.
Anyway the new code:
Revised:
import os
fileLoc = 'c:\\temp\\files'
for file in os.listdir(fileLoc):
if file.endswith('.txt'):
os.unlink(os.path.join(fileLoc,file))
The fix: os.path.join() builds a path for you from parts. One part is the directory (path) where the file exists, aka: fileLoc. The other part is the filename, aka file.
os.path.join() makes a whole valid path from them using whatever OS directory separator is appropriate for your platform.
Also, might want to glance through:
https://docs.python.org/2/library/os.path.html
I am trying to zip a file using shell script command. I am using following command:
zip ./test/step1.zip $FILES
where $FILES contain all the input files. But I am getting a warning as follows
zip warning: name not matched: myfile.dat
and one more thing I observed that the file which is at last in the list of files in a folder has the above warning and that file is not getting zipped.
Can anyone explain me why this is happening? I am new to shell script world.
zip warning: name not matched: myfile.dat
This means the file myfile.dat does not exist.
You will get the same error if the file is a symlink pointing to a non-existent file.
As you say, whatever is the last file at the of $FILES, it will not be added to the zip along with the warning. So I think something's wrong with the way you create $FILES. Chances are there is a newline, carriage return, space, tab, or other invisible character at the end of the last filename, resulting in something that doesn't exist. Try this for example:
for f in $FILES; do echo :$f:; done
I bet the last line will be incorrect, for example:
:myfile.dat :
...or something like that instead of :myfile.dat: with no characters before the last :
UPDATE
If you say the script started working after running dos2unix on it, that confirms what everybody suspected already, that somehow there was a carriage-return at the end of your $FILES list.
od -c shows the \r carriage-return. Try echo $FILES | od -c
Another possible cause that can generate a zip warning: name not matched: error is having any of zip's environment variables set incorrectly.
From the man page:
ENVIRONMENT
The following environment variables are read and used by zip as described.
ZIPOPT
contains default options that will be used when running zip. The contents of this environment variable will get added to the command line just after the zip command.
ZIP
[Not on RISC OS and VMS] see ZIPOPT
Zip$Options
[RISC OS] see ZIPOPT
Zip$Exts
[RISC OS] contains extensions separated by a : that will cause native filenames with one of the specified extensions to be added to the zip file with basename and extension swapped.
ZIP_OPTS
[VMS] see ZIPOPT
In my case, I was using zip in a script and had the binary location in an environment variable ZIP so that we could change to a different zip binary easily without making tonnes of changes in the script.
Example:
ZIP=/usr/bin/zip
...
${ZIP} -r folder.zip folder
This is then processed as:
/usr/bin/zip /usr/bin/zip -r folder.zip folder
And generates the errors:
zip warning: name not matched: folder.zip
zip I/O error: Operation not permitted
zip error: Could not create output file (/usr/bin/zip.zip)
The first because it's now trying to add folder.zip to the archive instead of using it as the archive. The second and third because it's trying to use the file /usr/bin/zip.zip as the archive which is (fortunately) not writable by a normal user.
Note: This is a really old question, but I didn't find this answer anywhere, so I'm posting it to help future searchers (my future self included).
eebbesen hit the nail in his comment for my case (but i cannot vote for comment).
Another possible reason missed in the other comments is file exceeding the file size limit (4GB).
I converted my script for unix environment using dos2unix command and executed my script as ./myscript.sh instead bash myscript.sh.
I just discovered another potential cause for this. If the permissions of the directory/subdirectory don't allow the zip to find the file, it will report this error. Actually, if you run a chmod -R 444 on the directory, and then try to zip it, you will reproduce this error, and also have a "stored 0%" report, like this:
zip warning: name not matched: borrar/enviar
adding: borrar/ (stored 0%)
Hence, try changing the permissions of the file. If you are trying to send them through email, and those email filters (like Gmail's) invent silly filters of not sending executables, don't forget that making permissions very strict when making zip compression can be the cause of the error you are reporting, of "name not matched".
spaces are not allowed:
it would fail if there are more than one files(s) in $FILES unless you put them in loop
I also encountered this issue. In my case, the line separate is CRLF in my zip shell script which causes the problem. Using LF fixed it.
I have a textfile (qrs.txt) which contains dir names (one per line) and on my server in the same directory as the script I have those folders with corresponding names from the text file.
This is my script:
#!/bin/bash
while read p; do
if [ ! -d "$p" ];
then
echo "ERROR $p" >> log.txt
else
echo "GOOD" >> log.txt
fi
done < qrs.txt
qrs.txt:
1992300000183805
1992300001176204
1992300002145500
1992300003104507
1992300004104902
1992300005133703
1992300006117802
1992300007144501
1992300008172803
1992300009189005
1992300010146307
1992300011151700
1992300012190007
1992300013126802
1992300014111508
1992300015193908
When that if statement is inside the loop it always returns error which is incorrect because I can see the folders exist. When I take it out of the loop and check for just 1, it works fine... When I echo $p on the same line as error, I can see the file name its checking is indeed correct.
What am I missing here..?
EDIT:
Screenshot of qrs.txt in hex mode:
http://i.snag.gy/25mqJ.jpg
RESOLVED!
My qrs.txt was in [dos] format originally but once converted to unix format using ":set ff=unix" the script worked like a charm!
Your script works fine.
I copied your script to my local machine. When I put blh blah in the qrs.txt file, I got ERROR for each time I ran your script. I ran it four times. I changed the blh blah to a valid path and I received GOOD.
The directory 1992300000183805 for instance, may be not be a valid path. You need the fully qualified path name! For example, /home/user/1992300000183805.
ERROR blh blah
ERROR blh blah
GOOD
GOOD
EDIT
Looking at #chepner comments, I recreated your problem:
Open your qrs.txt file in vi or vim. You should see ^M at the end of your lines. To remove the ^M characters at the end of all lines in vi, use:
:%s/^M//g
This should fix your problem. If not, in vim type this:
:set ff=unix
save the file.
Re-open qrs.txt in vim, then run the regex above again, or manually delete the ^M.
Or you can use perl:
perl -pi -e "s/\r/\n/g;" <file>
OK so looking at your provided file it seems those are relative directory names -- as such current directory is very important when you execute the script. Do you execute the script from its own directory or from the parent directory to all the (sub)directories shown in your example?
In other words have you tried:
cd <parent directory>
/path/to/yourscript.sh
?
Not to mention the location of qrs.txt seems to be specified relative rather than absolute path. So if there's no qrs.txt in the current directory I don't think your script would work.
I am creating a script which will go into into a particular column and row in a CSV file and use this data as part of a file subdirectory. So, I have a directory = $d and a sub-directory = $s
My scripts works really well, but this is not the important part. When I try to output some information in the script like..
echo "file $d/$s was unable to be replicated"
I get this output on the screen...
file /home/jsigel//filename was unable to be replicated
What do I need to do to get rid of this extra slash? I've tried a thousand things and nothing seems to work.
Super simple:
echo "file $d$s was unable to be replicated"
just as Kevin said a few seconds faster than me...
Your data contains the '/' so it's not necessary in your output specifier.
The extra slash is harmless, but you can remove it from your output if you like with
echo "file ${d%/}/$s was unable to be replicated"
If $d does have a trailing slash, it will be removed and an explicit / output in its place. If $d does not have a trailing slash, no change is made, and again the explicit / will be output.
I'm writing a script that needs to find a file in a directory based on the user input. That file contains a filepath, and I need to use that filepath as a variable so I can use it later in a mv command. So far :-
read x
path = `cat ~/filepaths/$x`
Later it needs to move a file from trash using the filepath read from this file
mv ~/trash/$x $path
Currently, it doesn't appear to work, and hangs when it runs. Is there something stupid I've missed here?
EDIT: Solved, was a stupid syntax mistake. Thanks for your help!
Remove the spaces around the assignment:
path=`cat ~/filepaths/$x`
or:
path=$(< ~/filepaths/$x)