13 lines
2.2 KiB
JavaScript
13 lines
2.2 KiB
JavaScript
OC.L10N.register(
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"files_versions",
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{
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"Versions" : "Stumm",
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"This application automatically maintains older versions of files that are changed." : "Ar meziant-man a laosk en un doare otomatek stumm ar resrtoù kozh a vez cheñchet",
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"This application automatically maintains older versions of files that are changed. When enabled, a hidden versions folder is provisioned in every user’s directory and is used to store old file versions. A user can revert to an older version through the web interface at any time, with the replaced file becoming a version. The app automatically manages the versions folder to ensure the user doesn’t run out of Quota because of versions.\n\t\tIn addition to the expiry of versions, the versions app makes certain never to use more than 50% of the user’s currently available free space. If stored versions exceed this limit, the app will delete the oldest versions first until it meets this limit. More information is available in the Versions documentation." : "Ar meziant a dalc'h en un doare otomatek stummoù kozh ar restroù cheñchet. P'az eo aotreet, un teuliad koachet a vez roet e pep teuliad an implijour evit gwarn stummoù kozh ar restr. An implijer a c'hel adkemer ur stumm kozh n'eus forz peseurt amzer dre an etrefas web. ar meziant a gra war dro teuliad ar stummoù evit vbezhaé sur e eus traouac'h a cQuota gant an implijer.\n\t\tOuzhpenn ar stumm diskarzed, stumm ar meziant a ra ma ne vefe ket implijet muioc'h eget 50% eus a plass vak an implijer. M'a vez dibaset e vo lemmet gant ar meziant ar stumm kozhañ da gentañ. Muioc'h titouriù en dielvadur Versions. ",
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"Failed to revert {file} to revision {timestamp}." : "C'hwitet m ar mon en dro {file} evit adober {timestamp}.",
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"_%n byte_::_%n bytes_" : ["%n okted","%n okted","%n okted","%n okted","%n octed"],
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"Restore" : "Adkrouiñ",
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"No other versions available" : "Stumm all ebet kavet"
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},
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"nplurals=5; plural=((n%10 == 1) && (n%100 != 11) && (n%100 !=71) && (n%100 !=91) ? 0 :(n%10 == 2) && (n%100 != 12) && (n%100 !=72) && (n%100 !=92) ? 1 :(n%10 ==3 || n%10==4 || n%10==9) && (n%100 < 10 || n% 100 > 19) && (n%100 < 70 || n%100 > 79) && (n%100 < 90 || n%100 > 99) ? 2 :(n != 0 && n % 1000000 == 0) ? 3 : 4);");
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