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How to extract and format specific values from a string

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1

We have files uploaded to our server in a specific format that indicates a version and build date, ex: 5_5_61__20180828.dav

The first group of numbers represents the version 5_5_61, and the last group of numbers represents the build date 20180828 and then .dav is just the file extension.

How would we be able to extract and format this file name to appear like this:

Version: 5.5.61 Build Date: 08/28/2018

I've been trying different variations of this snippet of code but can't seem to figure out the proper method:

$file = '5_5_61__20180828.dav';
echo substr($file, strpos($file , "_") + 1);

I should note that I am willing to change the file names if there's an easier way to achieve the final result using a different naming convention for the files.

php

share|improve this question

edited Mar 28 at 19:31

Qirel

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asked Mar 28 at 19:22

AJ47AJ47

26655 silver badges2121 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you're sure the filename will always be in the same format, i would use a regular expression. /(d+)_(d+)_(d+)__(d4)(d2)(d2)/ php.net/manual/en/function.preg-match.php
  
  – bassxzero
  Mar 28 at 19:25
  
  

  
  
  
  

add a comment
 | 

1

We have files uploaded to our server in a specific format that indicates a version and build date, ex: 5_5_61__20180828.dav

The first group of numbers represents the version 5_5_61, and the last group of numbers represents the build date 20180828 and then .dav is just the file extension.

How would we be able to extract and format this file name to appear like this:

Version: 5.5.61 Build Date: 08/28/2018

I've been trying different variations of this snippet of code but can't seem to figure out the proper method:

$file = '5_5_61__20180828.dav';
echo substr($file, strpos($file , "_") + 1);

I should note that I am willing to change the file names if there's an easier way to achieve the final result using a different naming convention for the files.

php

share|improve this question

edited Mar 28 at 19:31

Qirel

17.4k77 gold badges2828 silver badges4646 bronze badges

asked Mar 28 at 19:22

AJ47AJ47

26655 silver badges2121 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you're sure the filename will always be in the same format, i would use a regular expression. /(d+)_(d+)_(d+)__(d4)(d2)(d2)/ php.net/manual/en/function.preg-match.php
  
  – bassxzero
  Mar 28 at 19:25
  
  

  
  
  
  

add a comment
 | 

We have files uploaded to our server in a specific format that indicates a version and build date, ex: 5_5_61__20180828.dav

The first group of numbers represents the version 5_5_61, and the last group of numbers represents the build date 20180828 and then .dav is just the file extension.

How would we be able to extract and format this file name to appear like this:

Version: 5.5.61 Build Date: 08/28/2018

I've been trying different variations of this snippet of code but can't seem to figure out the proper method:

$file = '5_5_61__20180828.dav';
echo substr($file, strpos($file , "_") + 1);

I should note that I am willing to change the file names if there's an easier way to achieve the final result using a different naming convention for the files.

php

share|improve this question

edited Mar 28 at 19:31

Qirel

17.4k77 gold badges2828 silver badges4646 bronze badges

asked Mar 28 at 19:22

AJ47AJ47

26655 silver badges2121 bronze badges

Version: 5.5.61 Build Date: 08/28/2018

$file = '5_5_61__20180828.dav';
echo substr($file, strpos($file , "_") + 1);

share|improve this question

edited Mar 28 at 19:31

Qirel

17.4k77 gold badges2828 silver badges4646 bronze badges

asked Mar 28 at 19:22

AJ47AJ47

26655 silver badges2121 bronze badges

share|improve this question

edited Mar 28 at 19:31

Qirel

17.4k77 gold badges2828 silver badges4646 bronze badges

asked Mar 28 at 19:22

AJ47AJ47

26655 silver badges2121 bronze badges

edited Mar 28 at 19:31

Qirel

17.4k77 gold badges2828 silver badges4646 bronze badges

edited Mar 28 at 19:31

Qirel

17.4k77 gold badges2828 silver badges4646 bronze badges

asked Mar 28 at 19:22

AJ47AJ47

26655 silver badges2121 bronze badges

asked Mar 28 at 19:22

AJ47AJ47

26655 silver badges2121 bronze badges

2 Answers
2

active

oldest

votes

3

If you know that the versions and dates are separated by __ (two underscores), it's quite straight forward.

First split the string on __ (two underscores) - that will give you the version, and the date with extension. Explode the last part again to separate date from extension. Then replace _ (one underscore) with . on your version, and print it all.

Since 20180828 isn't a standard date-format for strtotime() functions, use DateTime::createFromFormat() to create a date-object.

$filename = '5_5_61__20180828.dav';
$parts = explode("__", $filename);
$parts2 = explode(".", $parts[1]);

$version = str_replce("_", ".", $parts[0]);
$date = $parts2[0];
$extension = $parts2[1];

$date = DateTime::createFromFormat("Ymd", $date)->format("m/d/Y");

echo "Version: $version. Build Date: $date";

• Live demo at https://3v4l.org/4Lk2T
  

share|improve this answer

answered Mar 28 at 19:28

QirelQirel

17.4k77 gold badges2828 silver badges4646 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The second explode (.) can be replaced with a substr since the date is fixed in length
  
  – Andreas
  Mar 28 at 19:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Fair point, that'll be a little more efficient too. Can probably reduce this whole split ordeal into a regex as well.
  
  – Qirel
  Mar 28 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, regex is possible, but not sure it will be faster. (But then again, I don't think this code will run thousands of times). But I just commented since it may make the code slightly cleaner to add the substr in the DateTime line than a separate line. All in all, +1
  
  – Andreas
  Mar 28 at 19:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, exactly what I was looking for!
  
  – AJ47
  Mar 28 at 19:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Or just $parts = explode("__", pathinfo($filename, PATHINFO_FILENAME));
  
  – AbraCadaver
  Mar 28 at 20:22
  

  
  
  
  

add a comment
 | 

1

https://ideone.com/a5Vudv

$file = '5_5_61__20180828.dav';
$version = str_replace("_", ".", substr($file, 0, -14));
$d = DateTime::createFromFormat("Ymd", substr($file, -12, 8))->format("m/d/Y");
echo 'Version: ' . $version . ' Build Date: '. $d;

share|improve this answer

answered Mar 28 at 19:42

AlexAlex

15.5k11 gold badge2020 silver badges3939 bronze badges

add a comment
 | 

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3

If you know that the versions and dates are separated by __ (two underscores), it's quite straight forward.

First split the string on __ (two underscores) - that will give you the version, and the date with extension. Explode the last part again to separate date from extension. Then replace _ (one underscore) with . on your version, and print it all.

Since 20180828 isn't a standard date-format for strtotime() functions, use DateTime::createFromFormat() to create a date-object.

$filename = '5_5_61__20180828.dav';
$parts = explode("__", $filename);
$parts2 = explode(".", $parts[1]);

$version = str_replce("_", ".", $parts[0]);
$date = $parts2[0];
$extension = $parts2[1];

$date = DateTime::createFromFormat("Ymd", $date)->format("m/d/Y");

echo "Version: $version. Build Date: $date";

• Live demo at https://3v4l.org/4Lk2T
  

share|improve this answer

answered Mar 28 at 19:28

QirelQirel

17.4k77 gold badges2828 silver badges4646 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The second explode (.) can be replaced with a substr since the date is fixed in length
  
  – Andreas
  Mar 28 at 19:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Fair point, that'll be a little more efficient too. Can probably reduce this whole split ordeal into a regex as well.
  
  – Qirel
  Mar 28 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, regex is possible, but not sure it will be faster. (But then again, I don't think this code will run thousands of times). But I just commented since it may make the code slightly cleaner to add the substr in the DateTime line than a separate line. All in all, +1
  
  – Andreas
  Mar 28 at 19:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, exactly what I was looking for!
  
  – AJ47
  Mar 28 at 19:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Or just $parts = explode("__", pathinfo($filename, PATHINFO_FILENAME));
  
  – AbraCadaver
  Mar 28 at 20:22
  

  
  
  
  

add a comment
 | 

3

If you know that the versions and dates are separated by __ (two underscores), it's quite straight forward.

First split the string on __ (two underscores) - that will give you the version, and the date with extension. Explode the last part again to separate date from extension. Then replace _ (one underscore) with . on your version, and print it all.

Since 20180828 isn't a standard date-format for strtotime() functions, use DateTime::createFromFormat() to create a date-object.

$filename = '5_5_61__20180828.dav';
$parts = explode("__", $filename);
$parts2 = explode(".", $parts[1]);

$version = str_replce("_", ".", $parts[0]);
$date = $parts2[0];
$extension = $parts2[1];

$date = DateTime::createFromFormat("Ymd", $date)->format("m/d/Y");

echo "Version: $version. Build Date: $date";

• Live demo at https://3v4l.org/4Lk2T
  

share|improve this answer

answered Mar 28 at 19:28

QirelQirel

17.4k77 gold badges2828 silver badges4646 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  The second explode (.) can be replaced with a substr since the date is fixed in length
  
  – Andreas
  Mar 28 at 19:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Fair point, that'll be a little more efficient too. Can probably reduce this whole split ordeal into a regex as well.
  
  – Qirel
  Mar 28 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, regex is possible, but not sure it will be faster. (But then again, I don't think this code will run thousands of times). But I just commented since it may make the code slightly cleaner to add the substr in the DateTime line than a separate line. All in all, +1
  
  – Andreas
  Mar 28 at 19:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you, exactly what I was looking for!
  
  – AJ47
  Mar 28 at 19:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Or just $parts = explode("__", pathinfo($filename, PATHINFO_FILENAME));
  
  – AbraCadaver
  Mar 28 at 20:22
  

  
  
  
  

add a comment
 | 

If you know that the versions and dates are separated by __ (two underscores), it's quite straight forward.

First split the string on __ (two underscores) - that will give you the version, and the date with extension. Explode the last part again to separate date from extension. Then replace _ (one underscore) with . on your version, and print it all.

Since 20180828 isn't a standard date-format for strtotime() functions, use DateTime::createFromFormat() to create a date-object.

$filename = '5_5_61__20180828.dav';
$parts = explode("__", $filename);
$parts2 = explode(".", $parts[1]);

$version = str_replce("_", ".", $parts[0]);
$date = $parts2[0];
$extension = $parts2[1];

$date = DateTime::createFromFormat("Ymd", $date)->format("m/d/Y");

echo "Version: $version. Build Date: $date";

• Live demo at https://3v4l.org/4Lk2T
  

share|improve this answer

answered Mar 28 at 19:28

QirelQirel

17.4k77 gold badges2828 silver badges4646 bronze badges

$filename = '5_5_61__20180828.dav';
$parts = explode("__", $filename);
$parts2 = explode(".", $parts[1]);

$version = str_replce("_", ".", $parts[0]);
$date = $parts2[0];
$extension = $parts2[1];

$date = DateTime::createFromFormat("Ymd", $date)->format("m/d/Y");

echo "Version: $version. Build Date: $date";

share|improve this answer

answered Mar 28 at 19:28

QirelQirel

17.4k77 gold badges2828 silver badges4646 bronze badges

answered Mar 28 at 19:28

QirelQirel

17.4k77 gold badges2828 silver badges4646 bronze badges

answered Mar 28 at 19:28

QirelQirel

17.4k77 gold badges2828 silver badges4646 bronze badges

1

https://ideone.com/a5Vudv

$file = '5_5_61__20180828.dav';
$version = str_replace("_", ".", substr($file, 0, -14));
$d = DateTime::createFromFormat("Ymd", substr($file, -12, 8))->format("m/d/Y");
echo 'Version: ' . $version . ' Build Date: '. $d;

share|improve this answer

answered Mar 28 at 19:42

AlexAlex

15.5k11 gold badge2020 silver badges3939 bronze badges

add a comment
 | 

1

https://ideone.com/a5Vudv

$file = '5_5_61__20180828.dav';
$version = str_replace("_", ".", substr($file, 0, -14));
$d = DateTime::createFromFormat("Ymd", substr($file, -12, 8))->format("m/d/Y");
echo 'Version: ' . $version . ' Build Date: '. $d;

share|improve this answer

answered Mar 28 at 19:42

AlexAlex

15.5k11 gold badge2020 silver badges3939 bronze badges

add a comment
 | 

https://ideone.com/a5Vudv

$file = '5_5_61__20180828.dav';
$version = str_replace("_", ".", substr($file, 0, -14));
$d = DateTime::createFromFormat("Ymd", substr($file, -12, 8))->format("m/d/Y");
echo 'Version: ' . $version . ' Build Date: '. $d;

share|improve this answer

answered Mar 28 at 19:42

AlexAlex

15.5k11 gold badge2020 silver badges3939 bronze badges

$file = '5_5_61__20180828.dav';
$version = str_replace("_", ".", substr($file, 0, -14));
$d = DateTime::createFromFormat("Ymd", substr($file, -12, 8))->format("m/d/Y");
echo 'Version: ' . $version . ' Build Date: '. $d;

share|improve this answer

answered Mar 28 at 19:42

AlexAlex

15.5k11 gold badge2020 silver badges3939 bronze badges

answered Mar 28 at 19:42

AlexAlex

15.5k11 gold badge2020 silver badges3939 bronze badges

answered Mar 28 at 19:42

AlexAlex

15.5k11 gold badge2020 silver badges3939 bronze badges