Error wile import mapping on Informatica
When I try to import mapping on Informatica, I encounters an error as I explained below, so that I couldn't get the workflow also
Completed importing selected objects, there were some errors, please see output for more information,
Validating Target definition TGT_TBL_FUNCTION
06/03/16 10:45:40****Importing source definition:Credit_trans_txt
Validating source definition Credit_trans_txt
06/03/16 10:45:40****Importing source definition: src_sample
Object referenced by the shortcut does not exist in the original folder
Object will be imported under the current folder
06/03/16 10:45:40****Importing mapping:m_sample
Error: Could not find transformation definition for :Flat file src_sample
**Failed to import:m_sample
Kindly provide me a solution.
The actual reason is this 'shortcut does not exist in the original folder'. So source Flat file src_sample doesnt exist in your target repository's folder.
To elaborate, in your source repo, you may have a folder named 'shared' where this file definition src_sample exists. And you used it as shortcut in your own folder (say 'myfolder') where m_sample resides. While copying to target repository same shortcut should exist.
Depending on situation, solution may be like this -
Folder 'shared' may not be shared in target repository. Modify it as shared and then check if src_sample exists or not. If yes, just copy m_sample again. If no then follow step2.
copy that object definition src_sample first to target repository in 'shared' folder. Then try to copy m_sample
Related
I am currenly working on an erlang project and stuck in reading the file. I want to read a text file which is in the /src folder where all the erlang and a text file are in the same structure. Then too, I am not being able to read the file despite of specifying file paths. Any help would be appreciated.
start() ->
{ok,DataList} = file:consult("Calls.txt"),
io:format("** Calls to be made **"),
io:fwrite("~w~n",[DataList]).
The data file stores contents like : {john, [jill,joe,bob]}.
Try add folder name to the path or try set full patch to the file:
1> {ok,DataList} = file:consult("src/Calls.txt").
Notes: the error {error,enoent} mean that the file does not exist or you don't have a rights to read/write current file, for this case need set 777 rights or similar.
If you need to use src/call.txt, then this simply means that your IDE (or you) has created a src folder in which the calls.txt file has been placed. At the same time, the IDE is using a path that only includes the top level folder (i.e., the root folder for the IDE project). So src/call.txt must be used in that case. This isn’t a problem with Erlang, or even the IDE. It’s just the way your project is set up.
You can do either of two things. Move the calls.txt file up one level in the IDE file manager, so that it can be referenced as calls.txt, not src/call.txt. You can also just change the path to “calls.txt” before you run it from the command line.
enoent means "Error: No Entry/Entity". It means the file couldn't be found. When I try your code, it works correctly and outputs
[{john,[jill,joe,bob]}]
I'm writing some validation code for a bazel build rule and I need to do some path validation. I need to check that a certain file exists in the same directory as the BUILD file. I notice that there's a context attribute build_file_path which points to the BUILD file. I'd like to extract the parent directory from this.
It looks like I can't create a new path object - I don't see a constructor/initializer. It also seems like Starlark doesn't support os.path like python because imports aren't supported.
What's the canonical way to get the parent directory of a string object representing a path in Starlark?
I can't answer your final question, but hopefully the following will help with the initial problem:
You could use the Label of the target for which this instance of the rule is being built and find its package. This will give you a string representing the parent directory of the BUILD file.
i.e. ctx.label.package
load("#bazel_skylib//lib:paths.bzl", "paths")
paths.dirname(path_str)
See https://github.com/bazelbuild/bazel-skylib/blob/main/docs/paths_doc.md
I'm having trouble understanding how to construct proper label forms when dealing with external repositories (directories with their own WORKSPACE).
What is the semantic meaning of characters like /, :, // or #?
For example:
#foo/bar
#foo:bar
//foo
foo
Do they preserve their meaning when used in an external repository? Also, is //external special in any way?
/ is a separator for package and target names.
relative/package/to/my:target
//absolute/package/to:my/file/target.java
A package is defined as a directory containing a BUILD or BUILD.bazel file.
: is the lexeme for selecting a rule or file target in a package.
//my/package:my_java_binary
Selects the target my_java_binary defined in <workspace root>/my/package/BUILD
//my/package:file.go
Selects the file <workspace root>/my/package/file.go if <workspace root>/my/package/BUILD exists, and if there's a rule in that BUILD file that references it.
//:my/nested/file.txt
Selects the file <workspace root>/my/nested/file.txt if <workspace root>/BUILD exists, but not in the my and my/nested subdirectories.
// is the location of the current or closest parent directory containing a WORKSPACE file.
Otherwise known as workspace root.
# is used for referencing a repository by its name when used to the left of //
#io_bazel_rules_scala//scala:scala.bzl: look into your WORKSPACE file for a repository named io_bazel_rules_scala. Usually defined using http_archive or git_repository.
#//my/package:target: # alone refers to the current workspace.
As of Bazel 0.16.0, # can be used in package names.
Do they preserve their meaning when used in an external repository?
Yes, think of the #<repository> syntax as a namespace mechanism.
Also, is //external special in any way?
Yes, it's used for the bind function, which is not recommended anymore. bind lets you give a target an alias in //external.
please take a look at the bin-win target in my repository here:
https://github.com/thinlizzy/bazelexample/blob/master/demo/BUILD#L28
it seems to be properly packing the executable inside a file named bin-win.tar.gz, but I still have some questions:
1- in my machine, the file is being generated at this directory:
C:\Users\John\AppData\Local\Temp_bazel_John\aS4O8v3V\execroot__main__\bazel-out\x64_windows-fastbuild\bin\demo
which makes finding the tar.gz file a cumbersome task.
The question is how can I make my bin-win target to move the file from there to a "better location"? (perhaps defined by an environment variable or a cmd line parameter/flag)
2- how can I include more files with my executable? My actual use case is I want to supply data files and some DLLs together with the executable. Should I use a filegroup() rule and refer its name in the "srcs" attribute as well?
2a- for the DLLs, is there a way to make a filegroup() rule to interpret environment variables? (e.g: the directories of the DLLs)
Thanks!
Look for the bazel-bin and bazel-genfiles directories in your workspace. These are actually junctions (directory symlinks) that Bazel updates after every build. If you bazel build //:demo, you can access its output as bazel-bin\demo.
(a) You can also set TMP and TEMP in your environment to point to e.g. c:\tmp. Bazel will pick those up instead of C:\Users\John\AppData\Local\Temp, so the full path for the output directory (that bazel-bin points to) will be c:\tmp\aS4O8v3V\execroot\__main__\bazel-out\x64_windows-fastbuild\bin.
(b) Or you can pass the --output_user_root startup flag, e.g. bazel--output_user_root=c:\tmp build //:demo. That will have the same effect as (a).
There's currently no way to get rid of the _bazel_John\aS4O8v3V\execroot part of the path.
Yes, I think you need to put those files in pkg_tar.srcs. Whether you use a filegroup() rule is irrelevant; filegroup just lets you group files together, so you can refer to the group by name, which is useful when you need to refer to the same files in multiple rules.
2.a. I don't think so.
The snippet below is taken from a TFS build definition file.
<mtbwa:CopyDirectory Destination="[BuildDetail.DropLocation]" DisplayName="Copy Files to Drop Location" Source="[BinariesDirectory]..\Packages" />
Regarding the source attribute, I want to set the value to a particular folder destination, using an existing variable and then going up one level.
If the variable [BinariesDirectory] was equal to "C:\TFS\Binaries", I am trying to set the source to "C:\TFS\Packages", but it just does not understand ..\ and will actually look for a folder at "C:\TFS\Binaries..\Packages"
How can I go up a folder level?
Thanks in advance
Means that it elavates to parent folder, then submerge to PACKAGES
use double \