I'm trying to pass a file from a form to a hashing function, the only problem is the hashing function takes a file path as an argument, not a file object. So my idea is to choose a file from a normal file input field and then some how extracting the file path from the chosen file.
I tried searching for the file path through the documentation, but i failed to find anything
Related
I have created a batch script that requires the location of 2 files to be passed into it. I would like it to work regardless of whether the user gives the full file path or relative file path of the files to be read in.
I've tried the following
set Var1=%~dp0%1
set Var2=%~dp0%2
which only works if relative file path is given. Without %~dp0obviously only works if full path is given.
Is there a way for either the full or relative path to be given and for my batch file to work.
Thanks
I am trying to archive a file. included in the image is the settings of my system file task. I'm trying to send a file whose name is stored in a variable within the loop, and send that file into the requisite folder.
The file is supposed to land in F:\DATA\ARCHIVE\WELLBORE\ using the filename it was given. However, it keeps landing in F:\DATA\ARCHIVE with the NAME WELLBORE, which is not what I am looking for.
What do I have missing?
Thanks
Problem
The issue is when you select Move File so you have to select the source full file path and the destination full file path, so the destination variable should contains the destination full path (with file name) and if the variable contains a folder path, the folder name will be considered as a filename and it will throw an exception.
Solution
Create a new variable #[User::DestinationFile] and assign the following expression to it:
#[User::ArchiveFolder] + RIGHT( #[User::WellBoreFile] , FINDSTRING(REVERSE( #[User::WellBoreFile] ) , "\\", 1) - 1)
This expression will add the filename to the destination path. And use this new variable as a Destination
References
SSIS EXPRESSION TO GET FILE NAME FROM FULL PATH
SSIS Expression to get filename from FilePath
Microsoft Docs article*
I'm trying to build a file based integration where files are dumped in one of the subdirectories of a main directory for processing. I need to get the name of the sub-directory to know which client the file is for. So if I have:
/uploads/foo/bar.txt
I need to process that file and know that it's for client "foo". I'm not sure how to get that part and set it as a header for the processor that processes the bar.txt file. I've got it picking up files and processing, now I need to add in this piece.
Anyone have ideas for me?
You can get most of this information in the header of the exchange. In your situation as you are consuming the file the following items are avlable:
CamelFileName: Name of the consumed file as a relative file path with
offset from the starting directory configured on the endpoint.
CamelFileNameOnly: Only the file name (the name with no leading
paths).
CamelFileAbsolute: A boolean option specifying whether the consumed
file denotes an absolute path or not. Should normally be false for
relative paths. Absolute paths should normally not be used but we
added to the move option to allow moving files to absolute paths. But
can be used elsewhere as well.
CamelFileAbsolutePath: The absolute path to the file. For relative
files this path holds the relative path instead.
CamelFilePath: The file path. For relative files this is the starting directory + the relative filename. For absolute files this is the absolute path.
CamelFileRelativePath: The relative path.
CamelFileParent: The parent path.
CamelFileLength: A long value containing the file size.
CamelFileLastModified: A Date value containing the last modified
timestamp of the file.
You can query these headers for the information you are looking for using the following example as guidelines:
<log message ="${header.CamelFileAbsolutePath}"/>
See the file component documents at the Camel website for more details.
File path=new File(".") ? What does this code line do ?
File path=new File(args[0]) Does it gives absolute or relative path to the file with name given in args[0]?
your first line will create a Java File object which will point to the current directory i.e. the directory you started your Java program from. You can call isDirectory() on it and you should get true. Read the Java 7 javadocs here.
your second line will create a Java File object which will point to args[0]. It does not verify that the file exists or is readable or writable. This will be checked when you start trying to work with the file.
You can determine the absolute path of a File object using method getAbsolutePath().
I'm trying to load files into MAGMA and am running into some trouble. Ostensibly, the command load "filename";should be sufficient. I've attempted, but keep getting the same result:
>> load "filename";
^
User error: Could not open file "filename" (No such file or directory)
The file is saved in my documents folder, so I'm not sure what the issue is. Do I have to specify the path? Save the file in a particular place?
I've tried reformatting, using both txt and rtf files, so I don't think that's the issue.
For loading file in MAGMA you can place your file in installed place folder. For example: C:\Program Files (x86)\Magma
Also if your file have an special format you should mention it.
Suppose You want loading a txt file with name a. with load"a"; you face with error. You must type load"a.txt";.
Try using GetCurrentDirectory() command to find your current directory location. And then you can use SetPath() to change where MAGMA has to be to search for your file. This will fix it.