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How to extract and format specific values from a string


Convert from MySQL datetime to another format with PHPHow do I get (extract) a file extension in PHP?How do I get a YouTube video thumbnail from the YouTube API?How to Sort Multi-dimensional Array by Value?How to check if a string starts with a specified string?How do I check if a string contains a specific word?Remove the last character from stringExtract numbers from a stringQuery on a piece of codeHow to upload file and rename if file is exist in folder while we keep original file name?






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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.










share|improve this question


























  • 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


















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.










share|improve this question


























  • 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














1












1








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.










share|improve this question
















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















share|improve this question













share|improve this question




share|improve this question








edited Mar 28 at 19:31









Qirel

17.4k7 gold badges28 silver badges46 bronze badges




17.4k7 gold badges28 silver badges46 bronze badges










asked Mar 28 at 19:22









AJ47AJ47

2665 silver badges21 bronze badges




2665 silver badges21 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


















  • 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

















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






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













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




















  • 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


















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



























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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




















    • 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















    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




















    • 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













    3














    3










    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













    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












    share|improve this answer



    share|improve this answer










    answered Mar 28 at 19:28









    QirelQirel

    17.4k7 gold badges28 silver badges46 bronze badges




    17.4k7 gold badges28 silver badges46 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












    • 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







    1




    1





    The second explode (.) can be replaced with a substr since the date is fixed in length

    – Andreas
    Mar 28 at 19:33





    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





    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





    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





    Thank you, exactly what I was looking for!

    – AJ47
    Mar 28 at 19:48




    1




    1





    Or just $parts = explode("__", pathinfo($filename, PATHINFO_FILENAME));

    – AbraCadaver
    Mar 28 at 20:22





    Or just $parts = explode("__", pathinfo($filename, PATHINFO_FILENAME));

    – AbraCadaver
    Mar 28 at 20:22













    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





























      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



























        1














        1










        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













        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












        share|improve this answer



        share|improve this answer










        answered Mar 28 at 19:42









        AlexAlex

        15.5k1 gold badge20 silver badges39 bronze badges




        15.5k1 gold badge20 silver badges39 bronze badges































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