Accessing property values within filename-generator-expression - spring

We have the following snippet in one of our Spring Batch XML files:
<int-file:outbound-channel-adapter
id="out" directory="c:\test"
filename-generator-expression="headers['msg.number']+'.xml'"
channel="in"/>
We would like to add access a value stored in a property file from within filename-generator-expression
The property can be accessed elsewhere in the XML as follows: attribute="${property.name}"
Th question is, how to access it from within filename-generator-expression.
This is obviously wrong:
filename-generator-expression="${property.name} + headers['msg.number']+'.xml'"
What is the correct solution?
Thanks very much

Related

Update parameter value in XML format

I have parameters stored in an XML file. Below is a sample of the file.
<?xml version="1.0" encoding="UTF-8"?>
<root>
<terminal id="A">
<terminalCapacity>3</terminalCapacity>
<terminalMembers id="1">
<memberID>0001</memberID>
<memberCapacity>2</memberCapacity>
</terminalMembers>
</terminal>
<terminal id="B">
<terminalCapacity>4</terminalCapacity>
<terminalMembers id="1">
<memberID>0002</memberID>
<memberCapacity>1</memberCapacity>
</terminalMembers>
<terminalMembers id="2">
<memberID>0003</memberID>
<memberCapacity>3</memberCapacity>
</terminalMembers>
</terminal>
</root>
Each terminalID is associated to a type of simpleModule found in my NED file. The idea is to programmatically update these values throughout the simulation run. The current logic revolves around getting the current parameters in XML format and update the memberCapacity field.
From the Omnet cPar and cXMLElement documentation, I tried using the par("moduleParameter").xmlValue()->getXML() function, but this returns the XML as a string. I also tried using the getAttribute() function, but to no success.
Don't do this. par("moduleParameter").xmlValue() will give you the in memory object tree of the XML document, but that is not meant for modification. Your XML file seems to be just a hierarchical structure and modules and their parameters can mirror that exactly. There is absolutely no reason to reinvent the wheel when you can mirror that with INI file parameters.

Placing file inside folder of S3 bucket

have a spring boot application, where I am tring to place a file inside folder of S3 target bucket. target-bucket/targetsystem-folder/file.csv
The targetsystem-folder name will differ for each file which will be retrived from yml configuration file.
The targetsystem-folder have to created via code if the folder doesnot exit and file should be placed under the folder
As I know, there is no folder concept in S3 bucket and all are stored as objects.
Have read in some documents like to place the file under folder, have to give the key-expression like targetsystem-folder/file.csv and bucket = target-bucket.
But it doesnot work out.Would like to achieve this using spring-integration-aws without using aws-sdk directly
<int-aws:s3-outbound-channel-adapter id="filesS3Mover"
channel="filesS3MoverChannel"
transfer-manager="transferManager"
bucket="${aws.s3.target.bucket}"
key-expression="headers.targetsystem-folder/headers.file_name"
command="UPLOAD">
</int-aws:s3-outbound-channel-adapter>
Can anyone guide on this issue
Your problem that the SpEL in the key-expression is wrong. Just try to start from the regular Java code and imagine how you would like to build such a value. Then you'll figure out that you are missing concatenation operation in your expression:
key-expression="headers.targetsystem-folder + '/' + headers.file_name"
Also, please, in the future provide more info about error. In most cases the stack trace is fully helpful.
In the project that I was working before, I just used the java aws sdk provided. Then in my implementation, I did something like this
private void uploadFileTos3bucket(String fileName, File file) {
s3client.putObject(new PutObjectRequest("target-bucket", "/targetsystem-folder/"+fileName, file)
.withCannedAcl(CannedAccessControlList.PublicRead));
}
I didn't create anymore configuration. It automatically creates /targetsystem-folder inside the bucket(then put the file inside of it), if it's not existing, else, put the file inside.
You can take this answer as reference, for further explanation of the subject.
There are no "sub-directories" in S3. There are buckets and there are
keys within buckets.
You can emulate traditional directories by using prefix searches. For
example, you can store the following keys in a bucket:
foo/bar1
foo/bar2
foo/bar3
blah/baz1
blah/baz2

SpelParseException: After parsing a valid expression, there is still more data in the expression: 'lcurly({)'

I am trying to condtionally create a component using #ConditionalOnExpression("not ${service.synchronous} && not ${service.disabled}").
I based this on Spring Boot SpEL ConditionalOnExpression check multiple properties, which provides a multi-property conditional as follows: #ConditionalOnExpression("${properties.first.property.enable:true} && ${properties.second.property.startServer:false}")
However, I keep getting:
Caused by: org.springframework.expression.spel.SpelParseException: EL1041E: After parsing a valid expression, there is still more data in the expression: 'lcurly({)'
Those properties are always set in my .properties file so I did not provide a default value with the colon notation. What am I doing wrong?
You will need to provide the default values for your properties like in the example you followed, so update the expression to be:
#ConditionalOnExpression("not ${service.synchronous:false} && not ${service.disabled:true}")
In most such cases the properties your app is reading are not what you expect them to be.
Set a breakpoint on all constructors of SpelParseException. In the debugger you will see the expression that is parsed, that will give show you exactly which properties you are really using.
Maybe you have to go search a little in the stack until you find the right location where you can see the expression.
My mistake was that I had not imported the test properties file in a Spring test.
After I added #TestPropertySource("classpath:/application.properties") to the test class, the properties from the properties file were used.

Spring, property file, empty values

I have configured spring security with a ldap server (but continue reading, it's not a problem if you have no knowledge about it, this is really a spring problem). All runs like a charm. Here is the line I use for that:
<ldap-server ldif="" root="" manager-dn="" manager-password="" url="" id="ldapServer" />
If I fill ldif and root attributes, it will run an embeded server:
<ldap-server ldif="classpath://ldap.ldif" root="dc=springframework,dc=org" manager-dn="" manager-password="" url="" id="ldapServer" />
If I fill other fields, it will run a distant server:
<ldap-server ldif="" root="" manager-dn="dc=admin,dc=springframeworg,dc=org" manager-password="password" url="ldap://myldapserver.com/dc=springframeworg,dc=org" id="ldapServer" />
All this stuff run correctly. Now I want to use Spring mechanism to load such parameters from a property file:
So I replace attribute values like this:
<ldap-server ldif="${ldap.ldif.path}" root="${ldap.ldif.root}" manager-dn="${ldap.server.manager.dn}" manager-password="${ldap.server.manager.password}" url="${ldap.server.url}" id="ldapServer" />
and create a property file with:
ldap.server.url=
ldap.server.manager.dn=
ldap.server.manager.password=
ldap.ldif.path=
ldap.ldif.root=
Now, the funny part of the problem. If I fill the following properties in the file:
ldap.server.url=ldap://myldapserver.com/dc=springframeworg,dc=org
ldap.server.manager.dn=dc=admin,dc=springframeworg,dc=org
ldap.server.manager.password=password
ldap.ldif.path=
ldap.ldif.root=
It runs a distant server as expected.
If I fill the property file like this:
ldap.server.url=
ldap.server.manager.dn=
ldap.server.manager.password=
ldap.ldif.path= classpath:ldap.ldif
ldap.ldif.root= dc=springframeworg,dc=org
It does not run, complaining that the ldap url is missing. But the problem is that if I change the spring configuration from:
<ldap-server ldif="${ldap.ldif.path}" root="${ldap.ldif.root}" manager-dn="${ldap.server.manager.dn}" manager-password="${ldap.server.manager.password}" url="${ldap.server.url}" id="ldapServer" />
to (by just removing the reference to the variable ${ldap.server.url})
<ldap-server ldif="${ldap.ldif.path}" root="${ldap.ldif.root}" manager-dn="${ldap.server.manager.dn}" manager-password="${ldap.server.manager.password}" url="" id="ldapServer" />
It runs !
My thoughs are that spring does not replace the attribute value with the property config one if this one is empty. But I find it strange.
Can you give me some clue to understand that ? And what's the best to do to configure my ldap server via a property file ?
EDIT: this is due to a poor design choice (look at accepted answer), an issue has been opened on jira :
https://jira.springsource.org/browse/SEC-1966
Ok, I think this is a spring security bug.
If I debug and look at the class LdapServerBeanDefinition, there is a method called "parse". Here is an extract:
public BeanDefinition parse(Element elt, ParserContext parserContext) {
String url = elt.getAttribute(ATT_URL);
RootBeanDefinition contextSource;
if (!StringUtils.hasText(url)) {
contextSource = createEmbeddedServer(elt, parserContext);
} else {
contextSource = new RootBeanDefinition();
contextSource.setBeanClassName(CONTEXT_SOURCE_CLASS);
contextSource.getConstructorArgumentValues().addIndexedArgumentValue(0, url);
}
contextSource.setSource(parserContext.extractSource(elt));
String managerDn = elt.getAttribute(ATT_PRINCIPAL);
String managerPassword = elt.getAttribute(ATT_PASSWORD);
if (StringUtils.hasText(managerDn)) {
if(!StringUtils.hasText(managerPassword)) {
parserContext.getReaderContext().error("You must specify the " + ATT_PASSWORD +
" if you supply a " + managerDn, elt);
}
contextSource.getPropertyValues().addPropertyValue("userDn", managerDn);
contextSource.getPropertyValues().addPropertyValue("password", managerPassword);
}
...
}
If I debug here, all variables (url, managerDn, managerPassword...) are not replaced by the value specified in the property file. And so, url has the value ${ldap.server.url}, managerDn has the value ${ldap.server.manager.dn} and so on.
The method parse creates a bean, a context source that will be used further. And when this bean will be used, place holders will be replaced.
Here, we got the bug. The parse method check if url is empty or not. The problem is that url is not empty here because it has the value ${ldap.server.url}. So, the parse method creates a context source as a distant server.
When the created source will be used, it will replace the ${ldap.server.url} by empty value (like specified in the property file). And....... Bug !
I don't know really how to solve this for the moment, but I now understand why it bugs ;)
I cannot explain it, but I think you can fix your problem using defaulting syntax, available since Spring 3.0.0.RC1 (see).
In the chageg log you can read: PropertyPlaceholderConfigurer supports "${myKey:myDefaultValue}" defaulting syntax
Anyway, I think that the problem is because "" is valid value, but no value in the property file don't.
I think that url="" works because url attribute is of type xs:token in spring-security XSD and empty string is converted to null (xs:token is removing any leading or trailing spaces, so "" can be recognized as no value). Maybe the value of ${ldap.server.url} is resolved as empty string and that is why you've got an error.
You can try use Spring profiles to define different configurations of ldap server (see Spring Team Blog for details about profiles)
I believe there is an issue here while using place holders. The following will most probably solve the problem:
Create a class which extends PropertyPlaceHolderConfigurer and override its method convertPropertyValue()
in the method you can return the property as empty string if you find anything other than a string which is of type LDAP url i.e. ldap://myldapserver.com/dc=springframeworg,dc=org
Also you need to configure your new specialization of class PropertyPlaceHolderConfigurer in the context file.
Hope this helps.
You can define empty String in the application.properties file as following:
com.core.estimation.stopwords=\ \

BIRT: Specifying XML Datasource file as parameter does not work

Using BIRT designer 3.7.1, it's easy enough to define a report for an XML file data source; however, the input file name is written into the .rptdesign file as constant value, initially. Nice for the start, but useless in real life. What I want is start the BIRT ReportEngine via the genReport.bat script, specifying the name of the XML data source file as parameter. That should be trivial, but it is surprisingly difficult...
What I found out is this: Instead of defining the XML data source file as a constant in the report definition you can use params["datasource"].value, which will be replaced by the parameter value at runtime. Also, in BIRT Designer you can define the Report Parameter (datasource) and give it a default value, say "file://d:/sample.xml".
Yet, it doesn't work. This is the result of my Preview attempt in Designer:
Cannot open the connection for the driver: org.eclipse.datatools.enablement.oda.xml.
org.eclipse.datatools.connectivity.oda.OdaException: The xml source file cannot be found or the URL is malformed.
ReportEngine, started with 'genReport.bat -p "datasource=file://d:/sample.xml" xx.rptdesign' says nearly the same.
Of course, I have made sure that the XML file exists, and tried different spellings of the file URL. So, what's wrong?
What I found out is this: Instead of defining the XML data source file as a constant in the report definition you can use params["datasource"].value, which will be replaced by the parameter value at runtime.
No, it won't - at least, if you specify the value of &XML Data Source File as params["datasource"].value (instead of a valid XML file path) at design time then you will get an error when attempting to run the report. This is because it is trying to use the literal string params["datasource"].value for the file path, rather than the value of params["datasource"].value.
Instead, you need to use an event handler script - specifically, a beforeOpen script.
To do this:
Left-click on your data source in the Data Explorer.
In the main Report Design pane, click on the Script tab (instead of the Layout tab). A blank beforeOpen script should be visible.
Copy and paste the following code into the script:
this.setExtensionProperty("FILELIST", params["datasource"].value);
If you now run the report, you should find that the value of the parameter datasource is used for the XML file location.
You can find out more about parameter-driven XML data sources on BIRT Exchange.
Since this is an old thread but still usefull, i ll add some info :
In the edit datasource, add some url to have sample data to create your dataset
Create your dataset
Then remove url as shown
add some script

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