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  • A download manager for Linux which saves downloaded files in directories by date like 2012_06_29

    - by Gart
    I've been using Download Master on Windows for years and what I liked most about it is that this program can automatically put downloaded files into directories by download date: /Downloads | |--/2012_06_28 | | | |--a.zip | |--b.pdf | ... | |--/2012_06_29 | | | |--c.txt | ... ... I'm looking for something similar for Linux. Is there any free download manager that can do this? I have tried KGet and uGet but they both seem to lack this feature. If there is a way to configure them to do that, I'll be happy to know about it. Thank you.

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  • Internet Time tab has disappeared from the Date and Time applet of the Control Panel

    - by Robert Thornton
    Previously, there was an Internet Time tab on the Date and Time applet of the Control Panel, wherein one could force a query of an internet time server and also type in a different server from the ones supplied. However, this tab has now disappeared, and I need to have it back. I should mention that this machine has never been part of a domain, since it seems that machines that are such do not have such a tab. I should be obliged to anyone who can help me restore the missing tab. Windows 7 Home Premium Service Pack 1

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  • How can I change the default date format?

    - by RossFabricant
    When I paste a date into Excel 2010 in the format "2012-12-07 00:00:00.000" I'd like it to be displayed by default as "2012-12-07", but it is actually displayed as "00:00.0". I know I can work around this by prepending a ' or changing the cell format, but I'm interested in changing the default format. An approach that almost works is going to Control Panel-Region and Language-Long Time format and changing it to something like "h". This results in Excel displaying the dates I paste in as "2012-12-07 00:00:00.000", but screws up times displayed outside of Excel. The dates I'm pasting in are from SQL Server.

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  • bash: listing files in date order, with spaces in filenames

    - by Jason Judge
    I am starting with a file containing a list of hundreds of files (full paths) in a random order. I would like to list the details of the ten latest files in that list. This is my naive attempt: ls -las -t `cat list-of-files.txt` | head -10 That works, so long as none of the files have spaces in, but fails if they do as those files are split up at the spaces and treated as separate files. I have tried quoting the files in the original list-of-files file, but the here-document still splits the files up at the spaces in the filenames. The only way I can think of doing this, is to ls each file individually (using xargs perhaps) and create an intermediate file with the file listings and the date in a sortable order as the first field in each line, then sort that intermediate file. However, that feels a bit cumbersome and inefficient (hundreds of ls commands rather than one or two). But that may be the only way to do it? Is there any way to pass "ls" a list of files to process, where those files could contain spaces - it seems like it should be simple, but I'm stumped.

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  • How do I ask screen to behave like a standard bash shell?

    - by thornomad
    Just learned about the screen command on linux - it is genius. I love it. However, the actual terminal/prompt in screen looks and behaves differently than my standard bash prompt. That is, the colors aren't the same, tab completion doesn't seem to work, etc. Is there a way I can tell screen to behave just like a normal (at least, normal as in what I am used to) bash prompt ?

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  • How could I make a bash script to execute apt-get?

    - by poz2k4444
    I'm trying to automatize some configurations I have with bash script, I've never done this before so I tried with something easy like a Hello World! and everything works just fine, but then I tried something like this: #!/bin/bash sudo su apt-get purge postfix and it doesn't do anything, I check and postfix is still installed, and at any time it asks for any entry of mine, I just tried with apt-get but I'll do things like ssh-keygen or even write files I guess with cat or something, how could I do the script working and also seeing what's going on?

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  • The Date type and the Android database

    - by cdonner
    I understand that SQLite stores dates as long integers. When I read rows into a cursor using the standard method (i.e. using the query() method that reads data into a cursor), the result is a date string that includes the time. If I want a different format, I have to parse the string back into a date - possible, but a bit backwards. By using a ViewBinder (as suggested in this question), I can pretty much do anything I want, but the date is already a string at the time the method executes. The accepted answer to the above question also suggests that storing dates as longs would help avoid this problem. I don't want to do that, just in case I want to interpret my data with something else than this application. Maybe I want to expose it via a provider. Is there a way to handle this in the database adapter instead, i.e. controlling the date format that the cursor contains, so that I don't have to change the schema, and don't have to parse the default output back into a Date type?

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  • Formatting the date in unix to include suffix on day (st, nd, rd and th)

    - by skymook
    How can I add the suffix on the day number of a unix date? I'll explain. I have a TextMate bundle snippit that writes out today's date. It uses unix date and formatting. Here is the code: `date +%A` `date +%d` `date +%B` `date +%Y` It outputs: Monday 22 March 2010 I would like to add the suffix to the day (st, nd, rd and th) like so: Monday 22nd March 2010 As far as I can see, there is no native function in the unix date formatting, like there is in PHP (j). How would I achieve this in unix? A complicated regex on the day number?

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  • Convert String to java.util.Date

    - by Vinayak.B
    Hi Folks, I storing the date to SQLite database in the format d-MMM-yyyy,HH:mm:ss aaa And again retrieving it with the same format, the problem now is, I am gettin every thing fine exepth the Hour. Hour I am geting 00 every time, Here the print statement String date--->29-Apr-2010,13:00:14 PM After convrting Date--->1272479414000--Thu Apr 29 00:00:14 GMT+05:30 2010 Please where I am doing wrong. Cheers, Vinayak

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  • Checking if date parsing is correct

    - by Javi
    Hello, I have this code for checking whether the Date is OK or not, but it's not ckecking all the cases. For example when text="03/13/2009" as this date doesn't exist in the format "dd/MM/yyyy" it parses the date as 03/01/2010. Is there any way to change this behaviour and getting an exception when I try to parse a Date which is not correct? What's the best way to do this validation? public static final String DATE_PATTERN = "dd/MM/yyyy"; public static boolean isDate(String text){ SimpleDateFormat formatter = new SimpleDateFormat(DATE_PATTERN); ParsePosition position = new ParsePosition(0); formatter.parse(text, position); if(position.getIndex() != text.length()){ return false; }else{ return true; } } Thanks.

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  • How to set default date in date_select helper in Rails

    - by brad
    I'm trying to set up a date of birth helper in my Rails app (2.3.5). At present it is like so. <%= f.date_select :date_of_birth, :start_year => Time.now.year - 110, :end_year => Time.now.year %> This generates a perfectly functional set of date fields that work just fine but.... They default to today's date which is not ideal for a date of birth field (I'm not sure what is but unless you're running a neonatal unit today's date seems less than ideal). I want it to read Jan 1 2010 instead (or 2011 or whatever year it happens to be). Using the :default option has proven unsuccessful. I've tried many possibilities including; <%= f.date_select :date_of_birth, :default => {:year => Time.now.year, :month => 'Jan', :day => 1}, :start_year => Time.now.year - 110, :end_year => Time.now.year %> and <%= f.date_select :date_of_birth, :default => Time.local(2010,'Jan',1), :start_year => Time.now.year - 110, :end_year => Time.now.year %> None of this changes the behaviour of the first example. Does the default option actually work as described? It seems that this should be a fairly straightforward thing to do. Ta.

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  • Java Date vs Calendar

    - by Marty Pitt
    Could someone please advise the current "best practice" around Date and Calendar types. When writing new code, is it best to always favour Calendar over Date, or are there circumstances where Date is the more appropriate datatype?

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  • Human readable and parsable date format in Java.

    - by Savvas Dalkitsis
    I want to save a Date object to a readable string (for example 22/10/2009 21:13:14) that is also parsable back to a Date object. I have tried many things and the best I could find was to use DateFormater for parsing and formating but it has a setback. When you format a date you lose seconds information. I tried to find if there is an option to format it and display the seconds (even better would be to the millisecond level since that's the resolution the Date object allows you to have) but I came up short. Any ideas?

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  • Parse RSS pubDate to Date object in java

    - by Derk
    How can I parse a pubDate from a RSS feed to a Date object in java. The format in the RSS feed: Sat, 24 Apr 2010 14:01:00 GMT What I have at the moment: DateFormat dateFormat = DateFormat.getInstance(); Date pubDate = dateFormat.parse(item.getPubDate().getText()); But this code throws an ParseException with the message Unparseable date

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  • How to edit and display Date in ASP.Net MVC 2

    - by Picflight
    I am storing a date in my database and want the user to be able to edit this date in the Edit View. In the past with ASP.Net web forms I have used 3 dropdownlists for the Month, Day and Year to get the date from the user and to bind it on display. I want to do the same in ASP.Net MVC and not sure how to do it? I am not using any jQuery or Javascript, the design calls for simple postback.

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  • grails date from params in controller

    - by nils petersohn
    y is it so hard to extract the date from the view via the params in a grails controller? i don't want to extract the date by hand like this: instance.dateX = parseDate(params["dateX_value"])//parseDate is from my helper class i just want to use instance.properties = params you know :) in the model the type is java.util.Date and in the params is all the information (dateX_month, dateX_day, ...) i searched on the net and found nothing on this :( i hoped that grails 1.3.0 could help but still the same thing. i can't and will not belief that extracting the date by hand is nessesary!

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  • Core Data data type for just the date - not including time

    - by Jason
    I am new at Core Data, and it seems like it is a great way to manage the data store. However I am also very memory-conscious due to the fact that the iPhone doesn't have that much of it. I was a little surprised to see that the data types are so limited - eg. there is a Date type which includes also the time, but no Date type for just the date! All the time information takes up precious bytes of memory, if I just wanted an attribute with the date (e.g. 2/15/2010 rather than 2/15/2010 02:34:48), how could I do this? Is it possible?

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  • number of months between two dates - using boost's date

    - by MartinP
    I've used boost::gregorian::date a bit now. I can see that there are the related months & years & weeks duration types. I can see how to use known durations to advance a given date. Qu: But how can I get the difference between two dates in months (or years or weeks) ? I was hoping to find a function like: template<typename DURATION> DURATION date_diff<DURATION>(const date& d1,const date& d2); There would need to be some handling of rounding too.

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  • convert variable with mixed date formats to one format in r

    - by jalapic
    A sample of my dataframe: date 1 25 February 1987 2 20 August 1974 3 9 October 1984 4 18 August 1992 5 19 September 1995 6 16-Oct-63 7 30-Sep-65 8 22 Jan 2008 9 13-11-1961 10 18 August 1987 11 15-Sep-70 12 5 October 1994 13 5 December 1984 14 03/23/87 15 30 August 1988 16 26-10-1993 17 22 August 1989 18 13-Sep-97 I have a large dataframe with a date variable that has multiple formats for dates. Most of the formats in the variable are shown above- there are a couple of very rare others too. The reason why there are multiple formats is that the data were pulled together from various websites that each used different formats. I have tried using straightforward conversions e.g. strftime(mydf$date,"%d/%m/%Y") but these sorts of conversion will not work if there are multiple formats. I don't want to resort to multiple gsub type editing. I was wondering if I am missing a more simple solution? Code for example: structure(list(date = structure(c(12L, 8L, 18L, 6L, 7L, 4L, 14L, 10L, 1L, 5L, 3L, 17L, 16L, 11L, 15L, 13L, 9L, 2L), .Label = c("13-11-1961", "13-Sep-97", "15-Sep-70", "16-Oct-63", "18 August 1987", "18 August 1992", "19 September 1995", "20 August 1974", "22 August 1989", "22 Jan 2008", "03/23/87", "25 February 1987", "26-10-1993", "30-Sep-65", "30 August 1988", "5 December 1984", "5 October 1994", "9 October 1984"), class = "factor")), .Names = "date", row.names = c(NA, -18L), class = "data.frame")

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