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Renaming files like 20141207_190822.jpg to "2014-12-07 19.08.22.jpg" in linux or MacOS X


Renaming files like 20141207_190822.jpg to "2014-12-07 19.08.22.jpg" in linux or MacOS X

By : user2955754
Date : November 22 2020, 10:48 AM
this one helps. How in Linux or MacOS X to rename a bunch of files with names 20141207_190822.jpg and 20141207_190823.mp4 to this format: , Try doing this :
code :
$ rename 's@^(\d{4})(\d{2})(\d{2})_(\d{2})(\d{2})(\d{2})@$1-$2-$3 $4.$5.$6@' *
$ file $(readlink -f $(type -p rename))
.../rename: Perl script, ASCII text executable


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How to make the linux/macOS command " .ls >> files.txt " run at php using exec()?

How to make the linux/macOS command " .ls >> files.txt " run at php using exec()?


By : user3310417
Date : March 29 2020, 07:55 AM
Hope this helps You can use 'opendir' and 'readdir' to do that, both native php commands...
Bash regex to match a "./" at the start of a line? OR renaming files with a leading "-"

Bash regex to match a "./" at the start of a line? OR renaming files with a leading "-"


By : J Reimer
Date : March 29 2020, 07:55 AM
To fix the issue you can do Use rename -n -- $regex $filenames.
The key here is --, which is a widely used method of saying "options end here, filenames start here". find itself doesn't support it, but rename does.
How to most efficiently convert a character string of "01 Jan 2014" to POSIXct i.e. "2014-01-01" yyy

How to most efficiently convert a character string of "01 Jan 2014" to POSIXct i.e. "2014-01-01" yyy


By : Anurag Singh
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further I already have a partial answer to the problem here, which I understand as far as it is explained: How to most efficiently restructure a character string for fasttime in data.table , What about using lubridate:
code :
x <- "01 Jan 2014"
x
[1] "01 Jan 2014"
library(lubridate)
dmy(x)
[1] "2014-01-01 UTC"
library(micro benchmark)
microbenchmark(dmy(x), times = 10000)
Unit: milliseconds
   expr      min      lq     mean   median      uq     max neval
 dmy(x) 1.992639 2.02567 2.142212 2.041514 2.07153 39.1384 10000

options(lubridate.fasttime = T)

microbenchmark(dmy(x), times = 10000)
Unit: milliseconds
   expr      min      lq     mean   median       uq      max neval
 dmy(x) 1.993326 2.02488 2.136748 2.039467 2.065326 163.2008 10000
How do I fix elasticsearch returning "2014-01" for date "2014-12-31" with format "yyyy-ww"

How do I fix elasticsearch returning "2014-01" for date "2014-12-31" with format "yyyy-ww"


By : Amy Lin
Date : March 29 2020, 07:55 AM
I hope this helps . You can use the format x-ww. I found the answer here: Leap year and aggregations per week in ElasticSearch -- which explains in detail how it works. Ultimately the key returned is the unix timestamp of the beginning of the week (which is in 2014), and the key_as_string is exactly that timestamp formatted according to your format. yyyy means the exact year of that timestamp, and x is the "weekyear". (joda time docs)
How I can pass this data format "27/04/2014 00:40" to this: "Sunday, 27 April, 2014"?

How I can pass this data format "27/04/2014 00:40" to this: "Sunday, 27 April, 2014"?


By : CocaFantaSprite
Date : March 29 2020, 07:55 AM
Does that help You want to first parse the date with $date = date_create_from_format('d/m/Y H:i:s', $f['date']) and then format according to your new requirement: date_format('l, d F, Y', $date).
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