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Check that Field Exists with MongoDB
So I'm attempting to find all records who have a field set and isn't null.
4 Answers
4...
Loop through all the files with a specific extension
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It will if it actually matches any files. You need to use shopt -s nullglob so that a non-matching pattern expands to the empty sequence rather than be treated literally.
– chepner
Aug 1 '15 at 23:08
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MySQL Workbench Dark Theme
...ave full of excitement bringing up my first question. My first question is all about changing the color appearance of MySQL Workbench from the default of white background to its negative value of black.
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Error 908: Permission Receive SMS - #5 by Taifun - MIT App Inventor Help - MIT App Inventor Community
...in points from that thread here
Google Play Store policy requires that all apps declaring the ability to send text and make phone calls directly without user intervention, or to receive texts and phone calls, require a manual review by Google staff. The MIT App Inventor Companion was one such ...
How do you convert an entire directory with ffmpeg?
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Previous answer will only create 1 output file called out.mov. To make a separate output file for each old movie, try this.
for i in *.avi;
do name=`echo "$i" | cut -d'.' -f1`
echo "$name"
ffmpeg -i "$i" "${name}.mov"
done
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Bash foreach loop
... input runs the cat program with the input lines as argument(s).
If you really want to do this in a loop, you can:
for fn in `cat filenames.txt`; do
echo "the next file is $fn"
cat $fn
done
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Bash set +x without it being printed
Does anyone know if we can say set +x in bash without it being printed:
5 Answers
5
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Extract file name from path, no matter what the os/path format
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Using os.path.split or os.path.basename as others suggest won't work in all cases: if you're running the script on Linux and attempt to process a classic windows-style path, it will fail.
Windows paths can use either backslash or forward slash as path separator. Therefore, the ntpath module (whi...
Split list into multiple lists with fixed number of elements
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I think you're looking for grouped. It returns an iterator, but you can convert the result to a list,
scala> List(1,2,3,4,5,6,"seven").grouped(4).toList
res0: List[List[Any]] = List(List(1, 2, 3, 4), List(5, 6, seven))
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What is your naming convention for stored procedures? [closed]
...ow the database is at 700 procedures plus, it becomes a lot harder to find all procedures on a specific object. For example i now have to search 50 odd Add procedures for the Product add, and 50 odd for the Get etc.
Because of this in my new application I'm planning on grouping procedure names by o...