Comments for benjamin.ramey http://blog.benramey.com Things I and my friends have learned while solving problems. Mon, 21 Sep 2009 16:51:29 +0000 hourly 1 http://wordpress.org/?v=3.0 Comment on Replacing newline characters in VB.NET strings by Brad Gearon http://blog.benramey.com/2009/09/01/replacing-newline-characters-in-vb-net/comment-page-1/#comment-18 Brad Gearon Mon, 21 Sep 2009 16:51:29 +0000 http://blog.benramey.com/?p=117#comment-18 The "newline character" you mentioned doesn't work because the spaces you see are basically a carriage return and a linefeed character: "\r\n". This is specific for the operating system and implementation of the .NET Framework. Technically the carriage return actually denotes that the cursor should move to the beginning of the current line and the linefeed moves to the next one. Replacing just the "\n" will obviously leave the carriage return and therefore still format as a new line. In this case, you must remove both the "\r\n" which you've obviously found, but this can be simplified. This can be done in a few ways. First of all you could use the vbCrLf constant like so: Replace(stringVariable, vbCrLf, "") This is exactly Chr(13) + Chr(10) and will also create a variable in memory to use for the empty string. Thus, another way might be more suitable. Second you could use Environment.NewLine and String.Empty like so: Replace(stringVariable, Environment.NewLine, String.Empty) This will grab the environment specific newline sequence which in this case is "\r\n" and use the existing empty string constant to replace the matches. I guess the environment specific one to look for would be up to the operating system the value was entered in though, essentially. The “newline character” you mentioned doesn’t work because the spaces you see are basically a carriage return and a linefeed character: “\r\n”. This is specific for the operating system and implementation of the .NET Framework. Technically the carriage return actually denotes that the cursor should move to the beginning of the current line and the linefeed moves to the next one. Replacing just the “\n” will obviously leave the carriage return and therefore still format as a new line. In this case, you must remove both the “\r\n” which you’ve obviously found, but this can be simplified. This can be done in a few ways.

First of all you could use the vbCrLf constant like so:
Replace(stringVariable, vbCrLf, “”)
This is exactly Chr(13) + Chr(10) and will also create a variable in memory to use for the empty string. Thus, another way might be more suitable.

Second you could use Environment.NewLine and String.Empty like so:
Replace(stringVariable, Environment.NewLine, String.Empty)
This will grab the environment specific newline sequence which in this case is “\r\n” and use the existing empty string constant to replace the matches.

I guess the environment specific one to look for would be up to the operating system the value was entered in though, essentially.

]]> Comment on Having Fun With sed by jonlabass http://blog.benramey.com/2009/07/14/having-fun-with-sed/comment-page-1/#comment-12 jonlabass Thu, 13 Aug 2009 21:59:39 +0000 http://blog.benramey.com/?p=113#comment-12 More fun includes commenting a line out if you don't know what the whole line is: sed 's/.*string/#&/' More fun includes commenting a line out if you don’t know what the whole line is:

sed ‘s/.*string/#&/’

]]> Comment on Delete multiple files/folders in multiple depths by jonlabass http://blog.benramey.com/2009/05/27/delete-multiple-filesfolders-in-multiple-depths/comment-page-1/#comment-2 jonlabass Wed, 08 Jul 2009 19:50:35 +0000 http://blog.benramey.com/?p=93#comment-2 You can also accomplish the same thing with find: find ./ -name ".svn" -exec rm -r {} \; You can also accomplish the same thing with find:

find ./ -name “.svn” -exec rm -r {} \;

]]>