Bash linux，JPG图像的小时，分​​钟和秒

Question

I am trying to get the complete date of an image in JPG format. The format I want to get is 14:25:38 (Hour:Minutes:Seconds).

I tried with the commands:

$ stat -c %y DSC_0002.JPG | sed 's/^\([0-9\-]*\).*/\1/'
=> 2017-05-19  -Not that way

$ file DSC_0002.JPG
=> DSC_0002.JPG: JPEG image data, Exif standard: [TIFF image data, little-endian, direntries=11, manufacturer=NIKON CORPORATION, model=NIKON D5200, orientation=upper-left, xresolution=180, yresolution=188, resolutionunit=2, software=Ver.1.01 , datetime=2017:05:19 13:30:34, GPS-Data], baseline, precision 8, 6000x4000, frames 3

This last ( file DSC_0002.JPG ) command displays datetime=2017:05:19 13:30:34 , but I need to get only 13:30:34

Preferably without using add-ons or programs external to Linux bash.

Thank you very much for any help.

Answer 1

My 2 cents...

As exif is a big part of my job and because I already spent a lot of time building my own scripts, time is comming to make a little bench.

Getting result of jhead using bash :

This use jhead because it's my prefered. See further to know why...

DateTime=$(jhead -nofinfo DSC_0002.JPG)
DateTime=${DateTime#*Date/Time*: }
DateTime=${DateTime%%$'\n'*}

This is the quickest way, (a lot quicker than using bash regex)!

echo $DateTime 
2011:02:27 14:53:32

Different ways

Inquiring a jpeg file for knowing his date is possible by using at least 4 different tools:

1. file The magic file recognition command make some light inquiring on any file for determining is nature and print some more informations.

    file DSC_0002.JPG DSC_0002.JPG: JPEG image data, Exif standard: [TIFF image data, big-endian, direntries=10, manufacturer=CANIKON, model=CANIKON AB12, orientation=upper- left, xresolution=168, yresolution=176, resolutionunit=2, software=Ver.1.00 , datetime=2011:02:27 14:53:32], baseline, precision 8, 3872x2592, frames 3 

   file print size, resolution and dateTime from a jpeg file.

2. jhead is a jpeg's header dedicated tool:

    jhead DSC_0002.JPG File name : DSC_0002.JPG File size : 4940925 bytes File date : 2011:02:27 14:53:32 Camera make : CANIKON Camera model : CANIKON AB12 Date/Time : 2011:02:27 14:53:32 Resolution : 3872 x 2592 Flash used : No Focal length : 55.0mm (35mm equivalent: 82mm) Exposure time: 0.0080 s (1/125) Aperture : f/5.6 ISO equiv. : 180 Whitebalance : Auto Metering Mode: pattern JPEG Quality : 97 

3. identify is part of ImageMagick package wich is a kind of all purpose tool for bitmap images... (Due to some security bugs in past and overall perfomance, I'ts not my personal choice).

    identify DSC_0002.JPG DSC_0002.JPG JPEG 3872x2592 3872x2592+0+0 8-bit sRGB 4.941MB 0.010u 0:00.020 identify -format "%[EXIF:DateTime]\\n" DSC_0002.JPG 2011:02:27 14:53:32 

4. exiftool is a dedicated tool using libimage .

    exiftool DSC_0002.JPG ExifTool Version Number : 9.74 File Name : DSC_0002.JPG Directory : . File Size : 4.7 MB File Modification Date/Time : 2011:02:27 14:53:32+01:00 File Access Date/Time : 2017:06:05 08:40:26+02:00 File Inode Change Date/Time : 2017:06:05 08:40:04+02:00 File Permissions : rw-r--r-- File Type : JPEG MIME Type : image/jpeg Exif Byte Order : Big-endian (Motorola, MM) ... Modify Date : 2011:02:27 14:53:32 ... Thumbnail Image : (Binary data 8965 bytes, use -b option to extract) Circle Of Confusion : 0.020 mm Depth Of Field : 15.07 m (8.58 - 23.65) Field Of View : 24.7 deg (5.50 m) Focal Length : 55.0 mm (35 mm equivalent: 82.0 mm) Hyperfocal Distance : 26.80 m Light Value : 11.1 

   The default output of this is 185 lines length, I've dropped a lot.

5. pure bash As requested and because Toby Speight recommend to not doing this :

    FDate=;while IFS= read -d $'\\0' -n 1024 raw;do [ "$raw" ] && \\ [ -z "${raw#[0-9][0-9][0-9][0-9]:[0-9][0-9]:[0-9][0-9] [0-9][0-9]:[0-9][0-9]:[0-9][0-9]}" ] && FDate=$raw && break done <DSC_0002.JPG echo $FDate 2011:02:27 14:53:32 

   Ok, this a far away a perfect function: first date found in file is considered, regardless of fieldname.

Little bench

As the goal is to extract date time part of exif infos, the bench would only do this.

time for i in {1..100};do ... done >/dev/null
export TIMEFORMAT="R:%4R u:%4U s:%4S"

Like this (from quicker to slower):

# quicker: jhead
time for i in {1..100};do jhead DSC_0002.JPG ;done >/dev/null
R:0.115  u:0.000  s:0.028

# 2nd: file
time for i in {1..100};do file DSC_0002.JPG; done >/dev/null
R:0.226  u:0.000  s:0.044

# 3nd: pure bash
time for i in {1..100};do 
    while IFS= read -d  $'\0' -n 1024 raw ;do
        [ "$raw" ] && 
        [ -z "${raw#[0-9][0-9][0-9][0-9]:[0-9][0-9]:[0-9][0-9] [0-9][0-9]:[0-9][0-9]:[0-9][0-9]}" ] &&
         ftim=$raw && break
      done <DSC_0002.JPG
   done >/dev/null
R:0.393  u:0.380  s:0.012

# 4nd: best dedicated: exiftool
time for i in {1..100};do exiftool -time:CreateDate DSC_0002.JPG ;done >/dev/null
R:14.921  u:13.064  s:0.956

# slower: imagemagick's identify
time for i in {1..100};do identify -format "%[EXIF:DateTime]\n" DSC_0002.JPG ;done >/dev/null
R:21.609  u:15.712  s:5.060

Sorry, I agree with Toby Speight : Doing this in pure bash is a not so good idea!

Setting file date

Doing date manipulation is very easy under pure bash, but working on file date. For this, I asked my system to set all photos files datetime to creation date found in exif.

For this, exiftool offer a specific syntax:

exiftool '-FileModifyDate<DateTimeOriginal /path/to/myphotos

and jhead too:

jhead -q -ft /path/to/myphotos/*.JPG

This will set all creation file date based on creation date in exif infos. Once done, you could use standard tools for filesystem inquiry:

ls -l DSC_0002.JPG
stat DSC_0002.JPG

find /path/to/myphotos -type f -name 'DSC*JPG' -mtime +400 -mtime -410 -ls

And so on...

Answer 2

You can use cut command to parse output.

In your case stat gives output

2017-05-05 06:12:37.228033281 -0500

So to get desire output you can use stat -c %y popen.c | cut -f2 -d' ' | cut -f1 -d'.' stat -c %y popen.c | cut -f2 -d' ' | cut -f1 -d'.'

Refer: man cut

Answer 3

Not all images have metadata, but those who have them can get them in the following ways:

identify -format "%[EXIF:DateTime]\n" image.jpg | awk '{print $2}

As said Toby Speight:

exiftool -time:CreateDate -a -G0:1 -s image.jpg | awk '{print $5}'

and

jhead image.jpg | awk '/^Date\/Time/{print $4}'

I'm sure there are other options, but I did not practice them