I have my vugen script like this: ITEMDATA,
"Name=File", "Value=C:\\testdata\\readme1.txt",
and I want to parametrrize the readme.txt part of the script with a unique filename for each vuser. How can I do that?
The filename information is simply a string. You should be able to parameterize it like any other string, such as
ITEMDATA, "Name=File", "Value={myfilename}",
or
char myfilename[120]
...
sprintf( myfilename,
"Value=C:\\testdata\\%s",
lr_eval_string("{readfromafileparameterfilename}") );
...
ITEMDATA, "Name=File", myfilename,
Related
I'm trying to use groovy to update Jenkins job config.xml by the following code
def updateParameter(String key, String value){
println "changing defult value as $value for key $key"
def xml = new XmlSlurper().parseText(jobConfig)
xml.properties.'hudson.model.ParametersDefinitionProperty'.'parameterDefinitions'.'hudson.model.StringParameterDefinition'.each {
println 'found parameter: ' + it.name
if(it.name.text() == key){
println('default value changed')
it.defaultValue=value
}
}
jobConfig = XmlUtil.serialize(xml)
}
When running jobConfig = XmlUtil.serialize(xml), it changes the format, which is pretty, but I lost link break in pipeline plugin, so pipeline script doesn't work anymore. Is there a way to convert GPathResult to String without format changing?
Best Regards,
Eric
It is all my fault, the line breaks were removed when I read the xml. It seems XmlUtil.serialize(xml) doen't format the text of a xml tag, which is good :)
Best Regards,
Eric
For example for a string like :
"https://www.example.com/myname:abcd"
I want only the part before ":"
i.e i want an output as below:
"https://www.example.com/myname"
What you want is to extract the namespace of the resource.
resource.getNamespace() will do the trick ;)
In SPARQL:
REPLACE("https://www.example.com/myname:abcd", ":[^:]*$", "")
In code, use a plain java string operation.
I have a file containing attribute dxl. I have created a template that creates a module exactly the way I want it, with new attributes and views and such. One of the attributes needs to be a dxl attribute,but I cannot find a good way to create a new dxl attribute from a dxl script using code contained in a separate file. I thought I might try something like this:
String s = #include "filepath"
But that obviously doesn't work. Is there a way to get the contents of a separate file into a string?
Thanks
You can do this using a Stream.
Stream inFile = read "filepath"
String s, sContent = ""
while(true) {
inFile >> s
sContent = sContent "\n" s
if(end of inFile) break
}
close inFile
This will fill the string sContent with your DXL file contents. Then you can use it to create the attribute.
Updated Code based on feedback.
I would like to set the filename for an item I create in an Excel Document Library. Howeverm when I try to interfere with the standard save with my own filename, it wants to save to my LOCAL MACHINE. I would happily like to supply the PATH if that is necessary, but I really DONT WANT TO HARD CODE IT.
Are there any properties I can use to parse this info? Meta data? Template?
Private Sub Workbook_BeforeSave(ByVal SaveAsUI As Boolean, Cancel As Boolean)
If SaveAsUI Then
Sheets("Purchase Order").Range("Description,VendorInfo,QUANTITY1,PRODUCT1,ITEM1,PRICE1,submitter,ShipVia,Terms,MACRO_ALERT").Interior.ColorIndex = xlColorIndexNone
Sheets("Purchase Order").Range("Location").Interior.Color = RGB(184, 204, 228)
Sheets("Purchase Order").Range("MACRO_ALERT").Value = ""
Dim MyFileName As String
Dim MyFilePath As String
' MyFilePath = "http://server/dept/purchasetracking/"
MyFilePath = Application.Path
MyFileName = "PO_" & Format(Now(), "yyyymmdd-hhnnss")
MsgBox (MyFilePath & MyFileName)
ActiveWorkbook.SaveAs Filename:=MyFilePath & MyFileName ', FileFormat:=52
' ActiveWorkbook.SaveAs Filename:=MyFileName
End If
End Sub
I like to use function GetSetting() and statement SaveSetting to save such informations to the registry (last used path, etc.).
When the file is opened from a SHP folder, ActiveWorkbook.Path starts with "http://". In this case you could save this path from a Sub Workbook_Open() procedure into the registry to retain the SHP path information. This can be later on used to offer a good path/filename to the user.
Hope that helps .... Good Luck MikeD
References:
http://www.grails.org/plugin/amazon-s3
http://svn.codehaus.org/grails-plugins/grails-amazon-s3/trunk/grails-app/services/org/grails/s3/S3AssetService.groovy
http://svn.codehaus.org/grails-plugins/grails-amazon-s3/trunk/grails-app/domain/org/grails/s3/S3Asset.groovy
By "happy" names, I mean the real name of the file I'm uploading... for instance, if I'm putting a file called "foo.png" I'd expect the url to the file to be /foo.png. Currently, I'm just getting what appears to be a GUID (with no file extension) for the file name.
Any ideas?
You can set the key field on the S3Asset object to achieve what you need.
I'll update the doco page with more information on this.
With length, inputstream and fileName given from the uploaded file, you should achieve what you want with the following code :
S3Service s3Service = new RestS3Service(new AWSCredentials(accessKey, secretKey))
S3Object up = new S3Object(s3Service.getBucket("myBucketName"), fileName)
up.setAcl AccessControlList.REST_CANNED_PUBLIC_READ
up.setContentLength length
up.setContentType "image/jpeg"
up.setDataInputStream inputstream
up = s3Service.putObject(bucket, up)
I hope it helps.
Actual solution (as provided by #leebutts):
import java.io.*;
import org.grails.s3.*;
def s3AssetService;
def file = new File("foo.png"); //assuming this file exists
def asset = new S3Asset(file);
asset.mimeType = extension;
asset.key = "foo.png"
s3AssetService.put(asset);