The File Name in the Send Port should be set with the mask like ABC.txt.pgp. Since I have used the PGP Encryption Component it is generating the File name like ABC.pgp.txt.pgp. But what I need is just the ABC.txt.pgp. How can be this be done. Any help is greatly appreciated.
Thanks
What you are seeing is the expected behavior. If you are referring to this:
https://code.msdn.microsoft.com/windowsdesktop/BizTalk-Sample-PGP-ebcbc8b2
or one of it's derivatives, it will internally modify FILE.ReceivedFileName to append .pgp if that property is set.
So, if you use just %SourceFileName%, you will likely get the desired result. Otherwise, you will have to explicitly set FILE.ReceivedFileName to ABC.txt somewhere before the PGP component.
You can also modify the source code to remove this behavior.
(Same Answer)
Thanks Johns-305. I included the Message assignment shape before the send shape and used the
SendMessage(FILE.ReceivedFileName) = "ABC.txt";
In the Send Port I used Filename as "%SourceFileName%". Now I get the filename as ABC.txt.pgp in the Send Port
Related
I wrote a small service in go (although I don't think this would be a language specific issue), that caches some results by saving it to a file, and writing a URL query parameter into the filename with "prefix" + param + ".json" using ioutil.WriteFile. The service runs on Ubuntu.
Is it possible to do something malicious, by passing an unexpected string via the query?
Relevant attacks that come to mind are called path injection. For example what if the query parameter is something like ../../etc/passwd (okthis would probably not work as the user running this service would have no permissions, but you get the point). For example it could be possible to overwrite your service code itself.
You should sanitize the parameter before adding it to the filename. The best would be a strict whitelist of letters and numbers that are allowed, anything else should ve removed from the parameter. That way injection would not be possible.
You can also check whether the path you are writing to is actually under an explicitly allowed directory.
I will make a test in python, here is the struct of the project
app1/main.py
while True:
a = input() # passing query
with open("{}.json".format(a), "w") as f:
f.write("Hello world")
now i am a hacker, and i want to change "yourfile.json"
so i passed this
and than, the content of yourfile.json become: Hello world
I need to extract the Call-id info from the calls started by a Call-file (in Asterisk), and use this value as a parameter of another function in order to return the full-cdr from the SIP-Proxy. Either i need to extract the call-id from call file based calls, or i gotta insert a call-id value through the file and use it. However, could not figure out how to do that. Looking forward to your replies, thanks in advance.
p.s. I'm not asking for Caller-id which refers to, as far as i understand, the from-number.
Call-id will be created by chan_sip AFTER you do call via call file.
So no, there are no way put it in file.
Maybe possible read by SIPHeader function AFTER call placed, but more likly you should do patch.
You can add special header like X-something if other part support search by it.
Based on file name prefix, I want to PGP encrypt and sign with different keys. I can use multiple encryption routes and direct them using a Message Router. Do anyone know a way to avoid multiple routs and get the related PGP keys at route running time?
final PGPDataFormat encryptAndSign01 = new PGPDataFormat();
encryptAndSign01.setKeyFileName(conf.pgpPublicKeyFile);
encryptAndSign01.setKeyUserid(conf.pgpEncryptUser01);
encryptAndSign01.setSignatureKeyFileName(conf.pgpPrivateKeyFile);
encryptAndSign01.setSignatureKeyUserid(conf.pgpSignUser01);
encryptAndSign01.setSignaturePassword(conf.pgpSignUser01Passphrase);
from("encrypt01")
.marshal(encryptAndSign01)
.to("file:tmp/output?fileName=${file:name}.pgp");
...
from("file:tmp/output?include=output.*.csv")
.choice()
.when(...)
.to(direct:encrypt01)
.when(...)
.to(direct:encrypt02);
You can specify the parameters of encryption by using the message header just like this
But I think it could be more easy if you use different routes to do the job.
How can I Parameterize an http call parameter in soapui to read parameters from a txt file for each iteration.
If needed can the parameters be encoded(url or gzip) before the call was sent?
Any help (pointers/links/code) is greatly appreciated? Thank You
Use groovy script test step to read data from txt file and store the data in TestCase property .
Something like this would work:
String fileContents = new File('/path/to/file').text;
testRunner.testCase.setPropertyValue(property_name, fileContents);
More information about groovy script steps here.
You can access this property as ${#TestCase#property_name} in your requests. Then you can use template parameters for your request url - I've already answered about it here.
If i'm not wrong you are asking about parametrization of URL which you send as HTTP Request for your Rest call. Let me explain you with an example :
Suppose you are looking for a resource and invoking the WebService using the GET method by making use of the ResourceID already present in the DB...Parametrize it as below :
http://${#Project#HOST}:${#Project#PORT}/rest/${#Project#WebApplicationName}/Resource/${#TestCase#ResourceID}
where HOST, PORT, WebApplicationName are the Project Level properties and ResourceID is a Test Case Level property(as it may change with the test cases i.e., dynamic in nature).
This is my approach of parametrization instead of taking it from a local file. Hope this helps!
Good afternoon,
I'm trying to set the output file name in a send port and the available file name macros won't quite work.
I need something of the form "file.YYYYMMDD_HHMMSS". There's a datetime combined macro (with the wrong format), a time only macro, but no date only macro.
I don't have an orchestration for this process.
Is there any way I can set the file name from within a map or a pipeline?
Thanks!
You can achieve filename in this format YYYYMMDD_HHMMSS using following.
Use expression shape in orchestration and create four variables.
varDateTime=System.DateTime.Now; //varDateTime is datetime type
strDate=varDateTime.ToString("yyyyMMdd"); //strDate is string type
strTime=varDateTime.ToString("HHmmss"); //strTime is string type
outboundfilename= ""+strDate + " _ " +strTime +" "; //outboundfilename is the string variable.
Use the following code in Message Assignment Shape
msgOutbound(FILE.ReceivedFileName)=outboundfilename+".xml";
//File extension can be any based on your requirement. I have xml file type.
You don't need custom pipeline to do this.
You can do it with a custom pipeline or an orchestration but not with a map. Info on building a custom pipeline can be found here:
BizTalk MSDN Blogs
This can be set by setting the FILE.ReceivedFileName (instead of BTS.ReceiveFileName).
You can create a custom pipeline component to be placed on the send side (say encode stage), and then set the BTS.ReceiveFileName property to the custom file name value that needs to be set.
After this is done, you can use the %SourceFileName% macro in the Send Port. You can also refer to this MSDN forum link for more details