Escape a backup date in a file name












3















I have been trying to:



cp file.csv file.$(date +%D).csv


But it fails because the filenames is: file.03/27/19.csv with the slash of separate directories.



And I have been trying again to:



cp file.csv file.$(printf "%q" $(date +%D)).csv


But it still fails.






shell filenames date escape-characters slash







share|improve this question

























  • You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

    – tres.14159
    9 hours ago











  • the problem is your use of the date format using the / character. You said it yourself, the shell is seeing them as directory markers. Try one of the many other options available from date. You might be able to get the / escaped so the filename uses the character code (like putting a space in a filename), but that is often problematic.

    – 0xSheepdog
    9 hours ago


















3















I have been trying to:



cp file.csv file.$(date +%D).csv


But it fails because the filenames is: file.03/27/19.csv with the slash of separate directories.



And I have been trying again to:



cp file.csv file.$(printf "%q" $(date +%D)).csv


But it still fails.






shell filenames date escape-characters slash







share|improve this question

























  • You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

    – tres.14159
    9 hours ago











  • the problem is your use of the date format using the / character. You said it yourself, the shell is seeing them as directory markers. Try one of the many other options available from date. You might be able to get the / escaped so the filename uses the character code (like putting a space in a filename), but that is often problematic.

    – 0xSheepdog
    9 hours ago
















3












3








3








I have been trying to:



cp file.csv file.$(date +%D).csv


But it fails because the filenames is: file.03/27/19.csv with the slash of separate directories.



And I have been trying again to:



cp file.csv file.$(printf "%q" $(date +%D)).csv


But it still fails.






shell filenames date escape-characters slash







share|improve this question
















I have been trying to:



cp file.csv file.$(date +%D).csv


But it fails because the filenames is: file.03/27/19.csv with the slash of separate directories.



And I have been trying again to:



cp file.csv file.$(printf "%q" $(date +%D)).csv


But it still fails.






shell filenames date escape-characters slash




shell filenames date escape-characters slash






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago









Gilles

544k12811041621




544k12811041621










asked 9 hours ago









tres.14159tres.14159

2814




2814













  • You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

    – tres.14159
    9 hours ago











  • the problem is your use of the date format using the / character. You said it yourself, the shell is seeing them as directory markers. Try one of the many other options available from date. You might be able to get the / escaped so the filename uses the character code (like putting a space in a filename), but that is often problematic.

    – 0xSheepdog
    9 hours ago





















  • You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

    – tres.14159
    9 hours ago











  • the problem is your use of the date format using the / character. You said it yourself, the shell is seeing them as directory markers. Try one of the many other options available from date. You might be able to get the / escaped so the filename uses the character code (like putting a space in a filename), but that is often problematic.

    – 0xSheepdog
    9 hours ago



















You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

– tres.14159
9 hours ago





You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

– tres.14159
9 hours ago













the problem is your use of the date format using the / character. You said it yourself, the shell is seeing them as directory markers. Try one of the many other options available from date. You might be able to get the / escaped so the filename uses the character code (like putting a space in a filename), but that is often problematic.

– 0xSheepdog
9 hours ago







the problem is your use of the date format using the / character. You said it yourself, the shell is seeing them as directory markers. Try one of the many other options available from date. You might be able to get the / escaped so the filename uses the character code (like putting a space in a filename), but that is often problematic.

– 0xSheepdog
9 hours ago












1 Answer
1






active

oldest

votes


















12














You can't have / (byte 0x2F on ASCII-based systems) in a file name, period.



You can use characters that look like / like ⁄ (fraction slash), so you could do:



 cp file.csv "file.$(date +%D | sed 's|/|⁄|g').csv"


But you may run into problems like the file name being rendered differently in locales using a different charset.



My advice would be to use the standard non-ambiguous (for most people outside the US, 03/12/18 would be interpreted as the 3rd of December 2018 for instance) YYYY-mm-dd format instead (which also helps wrt sorting):



 cp file.csv "file.$(date +%Y-%m-%d).csv"


Which with many date implementations you can shorten to:



 cp file.csv "file.$(date +%F).csv"





share|improve this answer

























    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "106"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f509043%2fescape-a-backup-date-in-a-file-name%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    12














    You can't have / (byte 0x2F on ASCII-based systems) in a file name, period.



    You can use characters that look like / like ⁄ (fraction slash), so you could do:



     cp file.csv "file.$(date +%D | sed 's|/|⁄|g').csv"


    But you may run into problems like the file name being rendered differently in locales using a different charset.



    My advice would be to use the standard non-ambiguous (for most people outside the US, 03/12/18 would be interpreted as the 3rd of December 2018 for instance) YYYY-mm-dd format instead (which also helps wrt sorting):



     cp file.csv "file.$(date +%Y-%m-%d).csv"


    Which with many date implementations you can shorten to:



     cp file.csv "file.$(date +%F).csv"





    share|improve this answer






























      12














      You can't have / (byte 0x2F on ASCII-based systems) in a file name, period.



      You can use characters that look like / like ⁄ (fraction slash), so you could do:



       cp file.csv "file.$(date +%D | sed 's|/|⁄|g').csv"


      But you may run into problems like the file name being rendered differently in locales using a different charset.



      My advice would be to use the standard non-ambiguous (for most people outside the US, 03/12/18 would be interpreted as the 3rd of December 2018 for instance) YYYY-mm-dd format instead (which also helps wrt sorting):



       cp file.csv "file.$(date +%Y-%m-%d).csv"


      Which with many date implementations you can shorten to:



       cp file.csv "file.$(date +%F).csv"





      share|improve this answer




























        12












        12








        12







        You can't have / (byte 0x2F on ASCII-based systems) in a file name, period.



        You can use characters that look like / like ⁄ (fraction slash), so you could do:



         cp file.csv "file.$(date +%D | sed 's|/|⁄|g').csv"


        But you may run into problems like the file name being rendered differently in locales using a different charset.



        My advice would be to use the standard non-ambiguous (for most people outside the US, 03/12/18 would be interpreted as the 3rd of December 2018 for instance) YYYY-mm-dd format instead (which also helps wrt sorting):



         cp file.csv "file.$(date +%Y-%m-%d).csv"


        Which with many date implementations you can shorten to:



         cp file.csv "file.$(date +%F).csv"





        share|improve this answer















        You can't have / (byte 0x2F on ASCII-based systems) in a file name, period.



        You can use characters that look like / like ⁄ (fraction slash), so you could do:



         cp file.csv "file.$(date +%D | sed 's|/|⁄|g').csv"


        But you may run into problems like the file name being rendered differently in locales using a different charset.



        My advice would be to use the standard non-ambiguous (for most people outside the US, 03/12/18 would be interpreted as the 3rd of December 2018 for instance) YYYY-mm-dd format instead (which also helps wrt sorting):



         cp file.csv "file.$(date +%Y-%m-%d).csv"


        Which with many date implementations you can shorten to:



         cp file.csv "file.$(date +%F).csv"






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 9 hours ago

























        answered 9 hours ago









        Stéphane ChazelasStéphane Chazelas

        311k57589946




        311k57589946






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Unix & Linux Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f509043%2fescape-a-backup-date-in-a-file-name%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            How to make a Squid Proxy server?

            Image
            1 I want to set up squid proxy server for monitoring user activity in a network and Internet Web URL filtering for user. I also want to configure the user login, user log, user access group, and block unwanted web sites in my network. Please help me and guide me the full process to do this. server proxy squid share | improve this question edited Jul 31 '12 at 14:45 Jorge Castro 37.1k 107 422 617 asked Jul 31 '12 at 5:16 Goivind Tiwari Goivind Tiwari ...
            Read more

            Is this a new Fibonacci Identity?

            Image
            6 1 $begingroup$ I have found the following Fibonacci Identity (and proved it). If $F_n$ denotes the nth Fibonacci Number, we have the following identity begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} end{equation} where $F_1 = F_2 = 1$ , $r leq n$ , $h leq g$ , and $n, g, k in mathbb{N}$ . It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few. I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see w...
            Read more

            19世紀

            Image
            この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2012年3月 ） 千年紀: 2千年紀 世紀: 18世紀 - 19世紀 - 20世紀 十年紀: 1800年代 1810年代 1820年代 1830年代 1840年代 1850年代 1860年代 1870年代 1880年代 1890年代 19世紀に君臨した大英帝国。 19世紀 （じゅうきゅうせいき）は、西暦1801年から西暦1900年までの100年間を指す世紀。 目次 1 19世紀の歴史 1.1 国民国家の成立 1.2 帝国主義の興隆 1.3 列強の植民地争奪戦 2 できごと 2.1 1800年代 2.2 1810年代 2.3 1820年代 2.4 1830年代 2.5 1840年代 2.6 1850年代 2.7 1860年代 2.8 1870年代 2.9 1880年代 2.10 1890年代 2.11 1900年代 3 文化 3.1 文学 3.2 音楽 3.3 思想 3.4 科学 3.5 技術 4 人物 4.1 ヨーロッパ 4.1.1 政治家・王族 4.1.1.1 フランス 4.1.1.2 オーストリア＝ハンガリー 4.1.1.3 ロシア 4.1.1.4 イギリス 4.1.1.5 ドイツ（プロイセンほかドイツ領邦を含む） 4.1.1.6 北欧 4.1.1.7 イタリア 4.1.1.8 スペイン 4.1.1.9 ベルギー 4.1.1.10 ギリシア 4.1.2 軍人 4.1.3 実業家 4.1.4 科学と技術 4.1.5 思想と哲学・人文諸学 4.1.6 宗教 4.1.7 文学 4.1.8 美術 4.1.9 音楽 4.1.10 社会事業家 4.1.11 探検家・旅行家 4.1.12 料...
            Read more