I create my contexts by dropping [myname].xml files in the contexts/ directory but in my CFML code I want to dynamically find the value of [myname], ie the name of the context/webapp (or failing that the filename of the xml file or the original value of the resourceBase property before path translation occurs).
I can get data about the context (like the array of virtualhosts) using the object returned from getPageContext().getConfig().getServletContext().getContextHandler().getCurrentWebAppContext() but if the context name is in there I haven't worked out how to get at it.
Use getDisplayName on that object you have?
It defaults to null (would be useful if it was the filename), but you can specify it in the context XML file with <Set name="DisplayName">bob</Set>
(If you have lots of XML files to deal with, do a script to loop through each file and plonk that with the filename inside the Configure tag.)
Related
UPDATE: After doing some more poking around, it looks as though the problem has to do with where CF is looking for the DTD file referenced in the XML.
We have the DTDs, but it looks as though CF isn't finding them, so it isn't sure how to parse the XML according to the DTD. I determined this by having it parse XML without any DTD, and it worked as expected and as I wanted - returning a parsed xmlDoc, not a string.
Is there some way of setting the default directory for where CF should look for the DTD specified in the XML?
We're running CF2021, and xmlParse(file), which should return a parsed XML object is instead returning the file contents as a string, inside a wddx encoded object. We have just migrated from a CF2018 server running on a remote hosting service to CF2021 running on an AWS box.
In order to return the XML object we need, I need to run xmlParse on the file, then wddx2cfml on the object, then xmlParse again on the string.
Is there a reason why xmlParse, which should return a parsed XML object, is instead behaving this way?
We pass the system file location to the method. Call it docPath, and it'd look something like g:\appName\xmlFiles\20230125.xml
Then we have, in cfscript:
doc = xmlParse(docPath);
When I dump that to a file, I get what I described above. When I change it to the following, I get what want:
docFile = xmlParse(docPath); cfwddx(action="wddx2cfml", input="#docFile#", output="xmlString"); xmlDoc = xmlParse(xmlString);
But I don't understand why this is necessary, and I'm concerned about having to change it everywhere in the code that we use xmlParse. For the record, this also occurs in tagged CF as well as cfscript, so it's not that.
Putting the dtd files in CF's WEB-INF folder solved the problem. CF was able to match the DTD with the DOCTYPE and properly parse the XML.
How could I possible open a symbolic link and get the content of the file instead of the file it is pointing to?
By doing:
with open('/home/symlink.txt', 'rb') as f:
data=f.read()
If the symbolic link points to /foo/faa.txt, the variable data will contain the content of faa.txt. This is a big security and file problem from my server because I'm generating zip archives.
If for example, a folder contains multiple symbolic links with different names to avoid duplicating files, the zip archive will contain multiple files instead of multiple symbolic links!
I hope to be clear enough!
An extra explanation:
The point of this is to allow downloading symlinks in a django server. The way of returning files is the following one:
response = HttpResponse()
response.write(data))
return response
This means that data must contain the content that the user will download. I can not just give it a path. So what I need to do is to give it a symbolic link. The problem is that reading a symbolic link makes python read the content where it is pointing to instead of its real content. In a few words, the user downloads the real file instead of the symbolic link!
A possible solution to this would be to get the path where the symlink points to, and then generate the link in the buffer. Is this possible?
It looks like there are 2 questions here: How can you read a symlink from the filesystem, and how can you store this in a .zip file such that it will be recreated when you unzip it.
Reading a symlink
The contents of a symlink are defined here:
http://man7.org/linux/man-pages/man7/symlink.7.html
A symbolic link is a special type of file whose contents are a string that is the pathname of another file, the file to which the link refers
You can read that path by using os.readlink (https://docs.python.org/2/library/os.html#os.readlink) - this is analogous to C's readlink function.
It's also important to note that these symlinks aren't distinguished by their content or file attributes, but by the fact that the file entry on disk points to a string rather than a file object:
In other words, a symbolic link is a pointer to another name, and not to an underlying object.
This means that there isn't really a "file" you could store in the ZIP. So how do the existing zip & unzip utilities do it?
Storing a symlink in a zip file
The spec for the ZIP format is here: https://pkware.cachefly.net/webdocs/casestudies/APPNOTE.TXT
Note that section 4.5.7 (defining UNIX Extra Field) says:
The variable length data field will contain file type specific data. Currently the only values allowed are the original "linked to" file names for hard or symbolic links, and the major and minor device node numbers for character and block device nodes. [...] Link files will have the name of the original file stored.
This means that to store a symlink, all you need to do is add the UNIX extra field block to the data you are writing (these appear to live immediately after the filename is written, and you need to set the extra field length accordingly), and populate its "Variable length data field" with the path you get from readlink. The content you store for the node will be empty.
If you're using a library to generate the zip data (recommended!), it will probably have an abstraction available for that. If not, I'd suggest you put in a feature request!
Of course, most existing zip and unzip utilities follow the same definition, which is why you are able to zip and unzip symbolic links as if they were regular files.
I have a multi file template in resharper and I can use $NAME$ macro to get the name of the original file to use to name the other files in the template. But I also want to use the $NAME$ of the original file in the content of the other file template.
Is this possible? I can't see a macro which seems suitable for the internal variables as onlt the Current File Name seems available.
Anyone know if this is possible or how I might workaround this?
As a workaround, you may create a parameter $FILENAME$ (macro "Current file name without extension") in the first file e.g. in the comments, like:
class Foo
{
//$FILENAME$
}
Then you may call this parameter in other files of the multifile template - this parameter will contain the name of the first file since the first file will be generated before other ones.
Unfortunately, there isn't a macro that will give you this. I've added a feature request that you can vote on and track (and more specific detail as to what your requirements are would be useful) - http://youtrack.jetbrains.com/issue/RSRP-415055
It is possible to write your own macros as part of a plugin, but there isn't a sure-fire way of getting the name of the first document in the created file set. The IHotspotSessionContext instance that is passed to the macro via IHotspotSession.Context property includes an enumerable of IDocument, from which you can get IDocument.Moniker, which will be the full path for file based documents. However, there's no guarantee of the order of the enumerable - it's backed by a hashset. You might be able to rely on implementation details (small set, no removes) to be able to use the first document as the original, but there is really no guarantee of this.
I have a mapping like
SA-->SQ--->EXPR--->TGT
The source will be of the same structure and the tartget also.
There are a bunch of files(with the same structure) which will go through this mapping .
So i want to use a parameter file through which i will give the file names for every run manually.
How to use the param file in session for Source filename attribute
Please suggest..
you could use indirect source type, wherein your source file is basically a list of files, and in turn the session reads each of the files one by one.
the parameter file could reference a source file name (the list) as
$InputFile_myName=/a/b/c.list
In line with what Raghav says, indicate the name of a file that will hold a list of input files in the 'Source filename' property box for the SQ in question in the Mapping tab, making the file 'Source filetype' be 'Indirect', specified in the Session Properties. If you already know ahead of time the names of the input files, you can specify them in that file and deploy that file with the workflow to the location you indicate in the 'Source file directory' property box. However if you won't know the names of the input files until run-time but know the files' naming standard (e.g: "Input_files_name_ABC_" where "" represents variable text, such as a numeric value incremented per input file generated by some other process), then one way to deal with that is to use a Pre-Session Command specifiable in the 'Components' tab of the Session. Create one that will build a new file at the location and with the name specified for the Indirect input file referenced above by using the Unix shell (or if running on Windows, the cmd shell) to list the files conforming to the naming standard for them and redirect the listing output to that file.
Tricky thing is that there must be one or more files listed in that Indirect type of input file. If that file is empty, the workflow will fail (abend). An Indirect file type must have in it listed at least one file (even if that file is empty) and that file must exist. The workflow fails if the indirect file reader gets no files to read from or if a file listed in it is not present on the server to be read from. One way to get around this is to make sure an empty file is present at all times that conforms to the naming standard. This can be assured by creating a "touchfile" before executing the listing command to build the Indirect file type listing file. In Unix, you'd use the 'touch {path}/{filename}' command ({filename} could be, for example, "Input_files_name_ABC_TOUCHFILE"), or on Windows you'd redirect an empty string to a file likewise named via cmd shell process. Either way, that will help you avoid an abend. Cleaning up that file is easy to do: a Post-Session command can be used to delete the empty touchfile. Likewise, you can do the same for the Indirect type of file if desired.
My page has a form on it that interacts with a CFC via an ajax post. When the user changes the text and clicks save it should update the DB, which it does, and rename a photo with the value they typed in. Everything works except that I keep getting the error:
The value of the attribute source, which is currently /Applications/ColdFusion8/wwwroot/theClient/staging/admin/images/Old_Image.jpg, is invalid.
The file exists and I've changed it to read/write for everyone but ti still gives me the error.
Any ideas?
I should point out that a var in my CFC runs a replace for spaces and changes them out for underscores so, there's no spaces in the filename...
If I remember right this is probably a problem with the destination file name rather then the source filename. I think it's a bug with the error message in CF.
The problem might be that you are using a relative path to the source file. The problem is, relative links in the source attribute are relative to the ColdFusion temporary directory. Try giving an absolute path or using the EXPANDPATH function to create an absolute path to your resource like this:
<cffile action="rename" source="#expandPath('/Applications/ColdFusion8/wwwroot/theClient/staging/admin/images/Old_Image.jpg')#" destination="#expandPath('/Applications/ColdFusion8/wwwroot/theClient/staging/admin/images/NEW_Image.jpg')#">
From the CF 9 documentation on CFFILE action="rename"
If not an absolute path (starting with a drive letter and a colon, or a forward or backward slash), it is relative to the ColdFusion temporary directory, which is returned by the GetTempDirectory function.
I had the same issue and in the end it was something else: the ColdFusion user had no "modify" permission in the folder. It had read/write but not modify. (sigh!)