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...get file name for the bug report?

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Option 1

Use OnCustomFileName event handler:

 

uses
  EEvents;
 
procedure ControlNames(const ACustom: Pointer; AExceptionInfo: TEurekaExceptionInfo; const AFileType: TEurekaLogFileType; var AFileName: Stringvar ACallNextHandler: Boolean);
var
  BugReportFileName: String;

begin
  // ftSavedBugReport indicate bug report
  // you can also use other values
  if AFileType = ftSavedBugReport then
  begin
    BugReportFileName := AFileName;
    // you can also change file name (AFileName) here
  end;
end;
 
initialization
  RegisterEventCustomFileName(nil, ControlNames);
end.

 

 

Option 2

Assuming you have EurekaLog's information object (e.g. event handler):

 

{ ... } AExceptionInfo: TEurekaExceptionInfo; { ... }
var
  BugReportFileName: String;
begin
  BugReportFileName := AExceptionInfo.ExpandEnvVars(AExceptionInfo.Options.OutputLogFile(''));
end;

 

 

Option 3

Only as last resort measure - use global settings:

 

uses
  EModules; // for CurrentEurekaLogOptions
 
BugReportFileName := CurrentEurekaLogOptions.ExpandEnvVars(CurrentEurekaLogOptions.OutputLogFile(''));

 

Note: be aware, that global code can not expand some variables (such as %BugID% - because those does not exist outside of exception processing code).

 

 

See also:




Send feedback... Build date: 2018-11-26
Last edited: 2018-11-12
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