Robot Framework how to set file path? - robotframework

I am using Choose File keyword to upload file. In documentation is written that I can use ${CURDIR} to set the path to my file, but it means that this file has to be in the same directory ${CURDIR}/filename.txt. But how to set the path to the file that exists in another directory?

Use ${EXECDIR} so you go back to the root and go from there.

Related

How to get the files in the upper level directory?

There are two files c:/a/b/mycode.r and c:/a/myfunction.r on my drive. When I run mycode.r, how to read the myfunction.r by using the relative path with source()? I tried source("../myfunction.r"), but it says No such file or directory. How to solve this problem?
Try to set chdir=TRUE.
This will change the working directory to the directory of the file being sourced.
`source("../myfunction.r", chdir=TRUE)`

jar -creation of log file in relative path using log4j

Log file is created if an absolute path is provided in the log4j.properties in a runnable jar created using eclipse. However, if relative file path is given , file is not created
log4j.appender.appLogger=org.apache.log4j.RollingFileAppender
log4j.appender.appLogger.File=.//logs//MyApp.log
Please correct this in your log4j.properties.
log4j.appender.appLogger.File=/logs/MyApp.log
For example:
In Ubuntu(Linux).
/home/bizruntime/Desktop/logs/MyApp.log
In Windows.
C:\\Desktop\logs\MyApp.log
If i have my log file in some folder of Desktop i need to specific the exact path.

IAR project variable like $PROJ_DIR$

Is there any possibility in IAR to add additional project variable like $PROJ_DIR$ to specify my project environment?
I like to keep my project portable and adaptable.
Of course there are, according to the manuals:
Variable / Description
$CONFIG_NAME$ The name of the current build configuration, for example Debug or Release.
$CUR_DIR$ Current directory
$CUR_LINE$ Current line
$DATE$ Today’s date
$EW_DIR$ Top directory of IAR Embedded Workbench, for example c:\program files\iar systems\embedded workbench 6.n
$EXE_DIR$ Directory for executable output
$FILE_BNAME$ Filename without extension
$FILE_BPATH$ Full path without extension
$FILE_DIR$ Directory of active file, no filename
$FILE_FNAME$ Filename of active file without path
$FILE_PATH$ Full path of active file (in Editor, Project, or Message window)
$LIST_DIR$ Directory for list output
$OBJ_DIR$ Directory for object output
$PROJ_DIR$ Project directory
$PROJ_FNAME$ Project filename without path
$PROJ_PATH$ Full path of project file
$TARGET_DIR$ Directory of primary output file
$TARGET_BNAME$ Filename without path of primary output file and without extension
$TARGET_BPATH$ Full path of primary output file without extension
$TARGET_FNAME$ Filename without path of primary output file
$TARGET_PATH$ Full path of primary output file
$TOOLKIT_DIR$ Directory of the active product, for example c:\program files\iar systems\embedded workbench 6.n\arm
$USER_NAME$ Your host login name
$ENVVAR$ The environment variable ENVVAR. Any name within $_ and _$ will
be expanded to that system environment variable.
If you go to Tools > Configure Custom Argument Variables you can add variables that you can address with $VARIABLE_NAME$. Not sure if that's exactly what you were looking for.
As a caution, if you are using IarBuild.exe to build from the command line, the workspace or global values set from "Configure Custom Arguments Variables" are not included in the project files (.ewp) and thus is not expanded by IarBuild.exe at build time. This is not an issue if you only use the IDE to build.
I found a reason for my problem (but it givs another one):
I define a windows "path-variable" like LIB_PATH and put it in the IAR project file with $_LIB_PATH_$.This works as long until i save the IAR-project. Then IAR sets all paths realive to $PROJ_DIR$ :-(

Creating a file using Pico under a different directory

If I am trying to create a file name "Data01" using the Pico editor with the contents of "Data File Number 01" on one line.
However my current directory is
csc209/asgn06
I need to create these files in
csc209/asgn06/test01 without changing directories in one command. Is there a way to do this?
Just prefix the filename with a relative path. If the directory exists:
pico test01/Data01
Otherwise, first create it with:
mkdir test01
then proceed as above.

How do I create a directory from hook_install of a module in Drupal 7?

Actually, I know how to create a directory = mkdir.
However, I need to pass it a full absolute path in the server, and I don't know how to get the relevant path in hook_install.
I tried using base_path(), but it returned '/'.
The path I want is sites/default/files (inside "files" I create the directory).
Thanks.
Please use the proper APIs.
$directory = file_default_scheme() . '://yourdir';
file_prepare_directory($directory, FILE_CREATE_DIRECTORY | FILE_MODIFY_PERMISSIONS);
Did you try mkdir('sites/default/files/yourdir')? If so, where does that create the directory? If not, why not?
Relative directory names are resolved based on the working directory. The working directory in Drupal is usually the directory in which your install.php and index.php is located.

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