open a form from current directory - visual-foxpro

How can I open a form stored in the current directory
do form ADDBS( JUSTPATH(SYS(16,0))) +"\form5.scx" WITH thisform.grid1.Column1.Text1.Value TO aa
I tried this but says that the file doesn't exist

do form form5 with ...
VFP uses relative pathing and also search paths. If it is in current directory then would be the first one to be picked by VFP.

ADDBS( JUSTPATH(SYS(16,0)))
this adds already a slash ("\") to the path and you add another slash with "\form5.scx"
can do
do form ADDBS( JUSTPATH(SYS(16,0)))+"form5.scx"
or use relative path or locfile() function

to test what i said copy this code and execute it: you can see there is 2 "\"
addbs() works only on what its applied not in the second independent term of the path.
local m.myvar
text to m.myvar noshow
messagebox(ADDBS( JUSTPATH(SYS(16,0))) +"\form5.scx")
_cliptext=ADDBS( JUSTPATH(SYS(16,0))) +"\form5.scx"
endtext
strtofile(m.myvar,"test.prg")
do test

Related

Move files based on filename using command prompt

I am trying to move files from a location to another one based on filename ONLY by using command line. The below works pretty fine but I do not want to specify both, the file name and the extension:
MOVE C:\firstlocation\file.sdltm "C:\secondlocation\file.sdltm"
The reason behind this is that I have the following files in the firstlocation:
file.sdltm
file.tmx
and my aim is to move both files named "file" to the secondlocation regardless the extension.
Sorry if this question has been asked before, - couldn't find any that matches my criteria.

clang:how can fdebug-prefix-map use new path relative to user home path `~`?

I try rewrite the source file path to ~/src/lib by using fdebug-prefix-map.
I can confirm DW_AT_decl_file is rewritten to something like ~/src/lib/path.
But the result is lldb can't find the source file. If I change to a absolute path, it works fine.
How can I solve this?
You can use the target.source-map setting to remap location of source files. From (lldb) apropos source-map:
Source path remappings are used to track the change of location between a source file when built, and where it exists on the current system. It consists of an array of duples, the first element of each duple is some part (starting at the root) of the path to the file when it was built, and the second is where the remainder of the original build hierarchy is rooted on the local system. Each element of the array is checked in order and the first one that results in a match wins.
The usage looks something like:
(lldb) settings append target.source-map /foo /bar
Note that you use append here instead of set, because otherwise you'd overwrite the mapping every time you add an entry. You can check the mapping with:
(lldb) settings show target.source-map

Create a directory structure from a path in gradle/groovy

I am implementing a diff package generation task in my project's gradle build from the git command line output. Currently I have a method which will give me a list of changed files from git diff --name-only. What I would like to do is create a directory structure in a new directory which matches the paths of each file. For example: inputting the string repo/dir/file.java would create in an output directory if not already created and inside it the directories head/repo/dir with the current file.java and prev/repo/dir with the previous file.java.
My current plan is to split the string repo/dir/file.java on the forward slash, and create directories until the last element of the split result, then write the file there. but nothing I have been able to come up with in gradle is nice or clean. I am wondering if there is a nicer way to create directories from a string like that.
My current plan is to split the string repo/dir/file.java on the forward slash, and create directories until the last element of the split result
Rather than splitting your string manually, you could try using File.mkdirs():
File newDirectoryStructureParent = new File('some/path/to/parent/dir')
def s = 'repo/dir/file.java'
def newContainer = new File(s, newDirectoryStructureParent).getParent()
newContainer.mkdirs()
everyone
In this part of my code you can just work around Path not File!
At the first you can define Path and second need check that path exist or not, if not mkdirs can make it ;)
Its help when you unknown about that path exist or not /
File fullPath = new File('/tmp/Test1')
if (!fullPath.exists())
fullPath.mkdirs()

how to change web server folder location in xampp

I have installed xampp and by default my website files are stored in htdocs. How do I change the webdirectory path?
Edit the file /opt/lampp/etc/httpd.conf and change the lines containing current web directory path. It should start with DocumentRoot. Update all the places having the current path to the path you would like it to be.
It can be done by editing the "httpd.conf" file present in this path "\xampp\apache\conf\".
Change the line that starts with "Directory ...".
Note that all the lines that start with a hash (#) are just comments and are there just to describe the commands present there in the file.
search for the file "httpd.conf" and change the path in that file.
change localhost's name ? thus instead of having localhostdomain/webdirectory
you may have newdomain/webdirectory ?

Placing the semicolon in the Windows PATH environment variable

Where should the trailing semicolon in the Windows PATH environment variable be placed when adding a new folder?
Is it
[oldPATH];C:\My Folder
[oldPATH]C:\My Folder;
[oldPATH];C:\My Folder;
?
I saw different practices.
This is not really a syntax thing, actually. The correct answer here is: Place the semicolon so the result is a valid PATH.
This usually means one of the following:
set PATH=%PATH%;C:\Foo\Bar
set PATH=C:\Foo\Bar;%PATH%
because usually PATH doesn't end in a semicolon, so you have to add one appropriately to not mangle an existing path in it.
Just look at how PATH looks like and consider what you need to do if you add another path. This means you have to add a separator (the semicolon) and the path itself.
The first one. At least thats what Windows does on mine, so if Windows does it that way then that will probably be best :)
The first one:
[oldPATH]; C:\My Folder.
If you want to be sure, you can use the formula:
"%PATH%;C:\My Folder".
If it is only to execute something in, for example, a BAT script, use:
SET PATH "%PATH%;C:\My Folder".
(this one will be working just as a temporal variable)
To add a permanent User Enviroment Variable through command line:
SETX PATH "%PATH%;C:\My Folder".
Your oldPATH may end with semicolon, so when using fourth style [newPath];[OldPath] you don't add double semicolons.
path=%cd%;%path%
Note that windows doesn't care whether you write commands upper- or lowercase.

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