What is the difference between:
include("./somepath/class.php");
and
include("somepath/class.php");
There shouldn't be any difference, directory wise, since the former assumes relative directory structure. The only potential pitfall could be if "somepath" were actually a command to be run - some users expect to type a command for a local script file and assume it should run, when you actually have to run "./somepath" to invoke it. This, of course, only pertains to commands not on your $PATH.
I don't see any difference. "." means current directory.
. refers to the current working directory.
Sometimes you want to specify explicitly that what you want is in the current working directory, and that it is not from something in your path variable for example.
For example in PHP "somepath/somefile" is appended after paths specified in include_dir (for example: /home/var/; /home/bin/ ....) directive.
The second variant is more specific it says: search in current directory!
Related
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.
I am trying to get a value of the full current directory path from within .bzl rule. I have tried following:
ctx.host_configuration.default_shell_env.PATH returns "/Users/[user_name]/.rbenv/shims:/usr/local/bin:/usr/bin:/bin:...
ctx.bin_dir.path returns bazel-out/local-fastbuild/bin
pwd = ctx.expand_make_variables("cmd", "$$PWD", {}) returns string $PWD - I don't think this rule is helpful for me, but may be just using it wrong.
What I need is the directory under which the cmd that runs Bazel .bzl rule is running. For example: /Users/[user_name]/git/workspace/path/to/bazel/rule.bzl or at least first part of the path prior to the WORKSPACE directory.
I can't use pwd because I need this value before I call ctx.actions.run_shell()
Are there no attributes in Bazel configurations that hold this value?
The goal is to have hermetic builds, so you shouldn't depend on the absolute path.
Feel free to use pwd inside the command of ctx.actions.run_shell() (for reproducible builds, be careful, avoid putting the absolute path in the generated files).
Edit.
Technically, there are some workarounds. For example, you can pass the path via the --define flag:
bazel build :all --define=path=$(pwd)
Then the value will be available using ctx.var["path"].
Based on your comment below, you want the path to declare an output. Let me repeat: You shouldn't use an absolute path to declare the output file. Declare an output in your package. Then ask the tool you call to use that output.
For example, when you call gcc, you can use -o to specify the output. When a tool writes to stdout, use the shell to redirect it. If the tool is really not flexible, you may want to wrap it with your own script (e.g. call the tool and copy the output file).
Using an absolute path here is not the right solution. For example, it should be possible to execute the action on a remote machine (where your absolute path won't make sense.
Zip may be a reasonable solution. It's useful when you cannot know in advance the number or the names of the output files.
How do I get the absolute path that the sandbox worker will be executing in from skylark?
I've got a number of rules where I need to add an argument to the action command in skylark. The argument is always the equivalent of "-fdebug-prefix-map=$(/bin/readlink -f .)=.". I need the path so I can teach my tools to strip off the sandbox path and leave a relative path. What's the best way to get access to that path?
As a workaround you can use the blaze-bin symlink.
However, in general, you cannot (particularly if you use --define).
You should wrap your script in a blaze rule and use blaze run.
Please advise. This SQLPlus call:
SQL > #dba_files_all
...is not working.
SP2-0310: unable to open file "dba_files_all.sql"
How can I resolve the error?
You need to provide the path of the file as string.
Put the path in double quotes and it will work.
For example:
#"C:\Users\Arpan Saini\Zions R2\Reports Statements and Notices\Patch\08312017_Patch_16.2.3.17\DB Scripts\snsp.sql";
I encountered this error when attempting to execute a file in the same folder as the calling function. In my example, this process:
Was executed in SQL Developer;
Has been a long-standing part of my system (moving a setup file with some settings and variable names through various folders; those folder names include the feature IDs and a short description);
Has worked fine in the past;
Did not require any pathing in my case because the files were in the same folder;
Failed on the most recent attempt with the error above (SP2-0310).
The issue in my situation was that the folder name in which it failed included a character (#) that was valid for a Windows file name, but confusing to SQL Developer.
1.Use absolute path:
/u01/app/oaracle/test.sql
2.Check the path to see if script exists:
ls -l /u01/app/oaracle/test.sql
Note that
SQL> #some_file.sql
means that sql app you are using will look for that using "absolute path" so if you want to use "relative path" use following format [add ?]
SQL> #?some_file.sql
else, use "full path" with first command.
All the answers so far imply that absolute paths are required. That aren't. Relative paths in sql is pretty universal in sql tools. Sometimes, you have to configure a lost default configuration such as in the case of SQLDeveloper as explained in this answer:
https://stackoverflow.com/a/24003529/442968
I just run into same error when I was trying to unlock oe schema.
While reading the error, I realized that when I run the following line:
>SQL #?/demo/schema/order_entry/oe_main.sql
The error returned a completely different path
SP2-0310: unable to open file "C:/app/USER/product/18.0.0/dbhomeXE/demo/schema/order_entry/oe_main.sql"
Thus I copied my sql file to the path specified by the error and everything worked. I recommend that you do the same. Check the path in the error and adjust accordingly.
Use absolute path or run sqlplus command from a shell/dos that points to the path of the script. Also, to use a masterscript, refer to subscripts with ##.
verify that your file has an extension .sql not .sql.txt
I make a call just like this:
value = ./simulated_annealing
Which is a C Object file, but Rails tells me it cannot find that file. I
put it in the same dir that the rest of the models files (since it's
called by one of those models), but I guess it should be in any other
place.
I've tried that outside Ruby and it works great.
What do I have to do?
The thing is, when you say:
./simulated_annealing
you're explicitly saying: run file named simulated_annealing which is found in the current directory. That's what the ./ means. If the file's located elsewhere you need to provide the path to it, or add that path to the environment variable $PATH. So, you should replace that line with:
/path/to/simulated_annealing
where /path/to represents the actual path.
The best option is to use an absolute path for running the program. For ex.,
you can create a directory "bin" under your rails application top level
directory. Place your program under "bin" directory. Then you can
execute the program something like:
cmd = "#{RAILS_ROOT}/bin/cbin arg1 arg2"
value = `#{cmd}`