I want to change the temporary path of imagemagick from a path in C: to a path on E:
How can I do this?
Do you mean temporary image's path? What does variable MAGICK_TEMPORARY_PATH looks like?
See: http://www.imagemagick.org/script/resources.php
Related
I am a total noob to LUA.
I need a script that will just copy a file path as text to the clipboard. That's it. I absolutely cannot figure it out. I keep getting the error:
attempt to call a nil value (global 'set_clipboard')
Here is the file path I am trying to copy to the clipboard:
D:_Google Drive_Acting\VO\Room Tone\roomtone.wav
This must be a simple script, right?
For Windows you can do this:
filename = 'my_filename.txt'
io.popen('clip','w'):write(filename):close()
There is no built-in function for that.
In Mac OS, you can do this
function set_clipboard(text)
io.popen('pbcopy','w'):write(text):close()
end
Apparently, in Windows you can use clip instead of pbcopy. I don't know about Linux.
Let's say I have x.y file in /mydir/a/b (on Linux)
When I run waf, it does not find the file.
def configure(context):
pass
def build(build_context):
build_context(source='/mydir/a/b/x.y',
rule='echo ${SRC} > ${TGT}',
target='test.out')
Result: source not found: '/mydir/a/b/x.y' in bld(features=[], idx=1, meths=['process_rule', 'process_source'] ...
Ok, maybe you want a relative path, Waf? And you are not telling me?
def build(context):
path_str = '/mydir/a/b'
xy_node = context.path.find_dir(path_str)
if xy_node is None:
exit ("Error: Failed to find path {}".format(path_str))
# just refer to the current script
orig_path = context.path.find_resource('wscript')
rel_path = xy_node.path_from(orig_path)
print "Relative path: ", rel_path
Result: Error: Failed to find path /mydir/a/b
But that directory exists! What's up with that?
And, by the way, the relative path for some subdirectory (which it can find) is one off. e.g. a/b under current directory results in relative path "../a/b". I'd expect "a/b"
In general there are (at least) two node objects in each context:
- path: is pointing to the location of the wscript
- root: is pointing to the filesystem root
So in you case the solution is to use context.root:
def build(context):
print context.path.abspath()
print context.root.abspath()
print context.root.find_dir('/mydir/a/b')
Hmm, looks like I found an answer on the waf-users group forum, answered by Mr. Nagy himself:
The source files must be present under the top-level directory. You
may either:
create a symlink to the source directory
copy the external source files into the build directory (which may cause problem if there is a structure of folders to copy)
set top to a common folder such as '/' (may require superuse permissions, so it is a bad idea in general)
The recommendation in conclusion is to add a symlink to the outside directory during the configuration step. I wonder how that would work, if I need this on both, Linux and Windows...
Just pass the Node to the copy rule instead of passing the string representing the path:
def build(build_context):
source_node = build_context.root.find_node('/mydir/a/b/x.y')
build_context(source=source_node,
rule='echo ${SRC} > ${TGT}',
target='test.out')
Waf will be able to find the file even if outside of the top level directory.
I got a problem with setting a path to image within the resource file (.rc).
For some reasone it was not possible to concatenate defined string and the text.
e.g.
File1:
#define Path "Brand_1"
File2:
#include File1
Logo BITMAP Path "\Logo.bmp"
Borland resource compiler (5.4) throws error message: 39: Cannot open file: Brand_1
EDIT:
My question would be: Is is possible to combine the path for loading image using resource string variable and a string (file name).
Also, project I'm working on relates to a file (Logo.bmp) being present in two locations. I would like to have a switch (.bat file) to generate a different resouce file depending on requirements.
Thanks.
BRCC32 accepts -i as search path seperated by semicolon, so you could create a bat file like this
compile_res.bat
brcc32 -ic:\mypath1;c:\mypath2 resource_script
and you define your resource_script as normal, for ex:
resource_script.rc
myImg BITMAP Logo.bmp
myDOC RCDATA mydoc.doc
when you run the compile_res.bat, it will run the brcc32.exe with the search path, and having the bat file saves you from retyping the search path every time.
You're not concatenating anything. You're compiling to Logo BITMAP "Brand_1" "\Logo.bmp", and "Brand_1" isn't a valid path to a bitmap file.
#define in the resource compiler acts sort of like find/replace in a text processor - not exactly, but close enough in this case.
You might get by (untested) with removing the quotes and space between them, as long as there are no space characters in either the path or filename; otherwise, you're probably out of luck. (Not sure what you're trying to accomplish, anyway.)
I want to have one directory for all object files and create Common.pri file that set OBJECTS_DIR like that
OBJECTS_DIR = $$PWD/../
But when build project i can't find obj file in given directory.If I write this direct in .pro file I get the expected result.I successfully include Common.pri file. I checked that with
!include( ../../Common.pri)::warning(Fail to include Common.pri)
How to achieve what i want.I can't find anything in google
The PWD variable specifies the full path leading to the directory containing the current file being parsed, that is, in your case the full path leading to the Common.pri file and NOT the .pro file. I would place a warning($$OBJECTS_DIR) function in both the .pri and the .pro file to verify the value of OBJECTS_DIR variable.
The message box crops up could not find part of the path of the file"file path" when i try to open a file that has space in its file path. I have used LocalPath instead of AbsolutePath and it works fine for me, but its only limited to WinApps, i needed a more generic solution. Some thing like Uri unescaped data path. I am not sure about the syntax.
In Java: URI uri = new File("spaces in file name").toURI();
If the file can be fetched depends on your implementation of the software. Try to replace the spaces with %20
What kind of development are we talking about here? JAVA GUI or WebApps? C/C++?
You should try enclosing your URI with quotes
new Uri("\"C:\some path\some file\"");