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How do i write unit test for function with void return type

I am new to writing Unit tests. I saw few examples of jUnit , but they all were for functions returning a value.
I have to write unit test for following method. Also we are using objects and methods in this method not belonging to this class like ConfigParser and ParseConfig. How do i go about writing Unit tests for such menthods?
#RequestMapping(value = "/generate-json" , consumes = MediaType.APPLICATION_JSON_VALUE, method = RequestMethod.POST)
#PageType(pageType = "PlaceHolderPageType")
public void execute(#RequestBody String payload) throws JSONException {
JSONObject json = new JSONObject(payload);
JSONObject configJSON = generateJSONSchema(json.toString());
String capabilityName = (String) configJSON.get(JSONConfigurationConstants.CAPABILITY_NAME);
String fileLocation = FormInputFieldConstants.JSON_PATH + capabilityName + JSONConfigurationConstants.FILE_EXTENSION;
//Write JSON file
try (OutputStreamWriter file = new OutputStreamWriter(new FileOutputStream(fileLocation), StandardCharsets.UTF_8)) {
file.write(configJSON.toString());
file.flush();
file.close();
} catch (IOException e) {
e.printStackTrace();
}
URI uriSchemaLocation = new File(fileLocation).toURI();
ConfigParser configParser = new ConfigParser(uriSchemaLocation);
configParser.parseConfig(TestHelper.getMockedJSONObject(fileLocation));
}

This is the method that you intend to test, correct? I am surprised to see that it contains test code (configParser.parseConfig(TestHelper.getMockedJSONObject(fileLocation));)...
What exactly do you want to test?
You could for example verify the content of the file that gets written by loading that file. However, that accounts only for a part of the method logic.
I don't see any #Override annotation and therefore assume that you are not overriding a method. You could change the method signature to return the parsed config and check it with assertions.
There might be other ways to retrieve the configuration; this depends on the logic of configParser.parseConfig(...).
You could also consider extracting the last tree lines to another method and test that method.
To sum up, you can either change the method to return a result, or extract chunks to own methods and test these, or retrieve the result from another source (in case something like configParser.getParsedConfig() exists).

Insert a string in the middle of text file without replacing [duplicate]

This question already has answers here:
inserting data in the middle of a text file through java
(2 answers)
Closed 9 years ago.
Suppose i have a text file named Sample.text.
i need advice on how to achieve this:
Sample.txt before running a program:
ABCD
while running the program, user will input string to be added starting at the middle
for example: user input is XXX
Sample.txt after running a program:
ABXXXCD

Basically you've got to rewrite the file, at least from the middle. This isn't a matter of Java - it's a matter of what file systems support.
Typically the way to do this is to open both the input file and an output file, then:
Copy the first part from the input file to the output file
Write the middle section to the output file
Copy the remainder of the input file to the output file
Optionally perform file renaming if you want the new file to have the same eventual name as the original file

The basic idea is to read the file contents into memory, say at program start, manipulate the string as desired, then write the entire thing back to the file.
So you would open and read in Sample.txt. In memory you have a string = "ABCD"
in your program execution, accept user input of XXX. Insert that into your string with your favorite string manipulation method. Now string = "ABXXXCD"
Finally you would overwrite Sample.txt with your updated string and close it.
If you were worried about corruption or something, you might save it to a secondary file, then verify its contents, delete the original, and rename the new to be the same as the original.

Actually i have did something like what you want, here try this code, its not a complete but it should give you a clear idea:
public void addString(String fileContent, String insertData) throws IOException {
String firstPart = getFirstPart(fileContent);
Pattern p = Pattern.compile(firstPart);
Matcher matcher = p.matcher(fileContent);
int end = 0;
boolean matched = matcher.find();
if (matched) {
end = matcher.end();
}
if(matched) {
String secondPart = fileContent.substring(end);
StringBuilder newFileContent = new StringBuilder();
newFileContent.append(firstPart);
newFileContent.append(insertData);
newFileContent.append(secondPart);
writeNewFileContent(newFileContent.toString());
}
}

Normally a new file would be created, but the following probably suffices (for non-gigabyte files). Mind the explicit encoding UTF-8; which you can ommit for the encoding of the operating system.
public static void insertInMidstOfFile(File file, String textToInsert)
throws IOException {
if (!file.exists()) {
throw new FileNotFoundException("File not found: " + file.getPath());
// Because file open mode "rw" would create it.
}
if (textToInsert.isEmpty()) {
return;
}
long fileLength = file.length();
long startPosition = fileLength / 2;
long remainingLength = fileLength - startPosition;
if (remainingLength > Integer.MAX_VALUE) {
throw new IllegalStateException("File too large");
}
byte[] bytesToInsert = textToInsert.getBytes(StandardCharsets.UTF_8);
try (RandomAccessFile fh = new RandomAccessFile(file, "rw")) {
fh.seek(startPosition);
byte[] remainder = new byte[(int)remainingLength];
fh.readFully(remainder);
fh.seek(startPosition);
fh.write(bytesToInsert);
fh.write(remainder);
}
}
Java 7 or higher.

Update database table without uploading a file while using a MultiPart Form - JavaEE, Servlet

I have a servlet which is responsible for enabling a user to update a reports table and upload a report at the same time. I have written code that enables a user upload a document and also be able to update the table with other details e.g date submitted etc.
However not all the times will a user have to upload a document. in this case it should be possible for a user to still edit a report's details and come back later to upload the file. i.e the user can submit the form without selecting a file and it still updates the table.
This part is what is not working. If a user selects a file and makes some changes. The code works. If a user doesn't select a file and tries to submit the form, it redirects to my servlet but it is blank. no stacktrace. No error is thrown.
Below is part of the code I have in my servlet:
if(param.equals("updateschedule"))
{
String[] allowedextensions = {"pdf","xlsx","xls","doc","docx","jpeg","jpg","msg"};
final String path = request.getParameter("uploadlocation_hidden");
final Part filepart=request.getPart("uploadreport_file");
int repid = Integer.parseInt(request.getParameter("repid_hidden"));
int reptype = Integer.parseInt(request.getParameter("reporttype_select"));
String webdocpath = request.getParameter("doclocation_hidden");
String subperiod = request.getParameter("submitperiod_select");
String duedate = request.getParameter("reportduedate_textfield");
String repname = request.getParameter("reportname_textfield");
String repdesc = request.getParameter("reportdesc_textarea");
String repinstr = request.getParameter("reportinst_textarea");
int repsubmitted = Integer.parseInt(request.getParameter("repsubmitted_select"));
String datesubmitted = request.getParameter("reportsubmitdate_textfield");
final String filename = getFileName(filepart);
OutputStream out = null;
InputStream filecontent=null;
String extension = filename.substring(filename.lastIndexOf(".") + 1, filename.length());
if(Arrays.asList(allowedextensions).contains(extension))
{
try
{
out=new FileOutputStream(new File(path+File.separator+filename));
filecontent = filepart.getInputStream();
int read=0;
final byte[] bytes = new byte[1024];
while((read=filecontent.read(bytes))!=-1)
{
out.write(bytes,0,read);
}
String fulldocpath = webdocpath+"/"+filename;
boolean succ = icreditdao.updatereportschedule(repid, reptype, subperiod, repname, repsubmitted,datesubmitted, duedate,fulldocpath, repdesc, repinstr);
if(succ==true)
{
response.sendRedirect("/webapp/Pages/Secured/ReportingSchedule.jsp?msg=Report Schedule updated successfully");
}
}
catch(Exception ex)
{
throw new ServletException(ex);
}
}
I'm still teaching myself javaee. Any help will be appreciated. Also open to other alternatives. I have thought of using jquery to detect if a file has been selected then use a different set of code. e.g
if(param.equals("updatewithnofileselected"))
{//update code here}
but I think there must be a better solution. Using jdk6, servlet3.0.

try this one.
MultipartParser parser = new MultipartParser(request, 500000000, false, false, "UTF-8");
Part part;
while ((part = parser.readNextPart()) != null) {
if(part.isParam()){
if(part.isFile()){
if(part.getName().equals("updatewithnofileselected")){
//update code here.
} else if(part.getName().equals("updateschedule")) {
//updateschedule
}
}
}
}
I used this one when I am using Multipart-form and it's working fine.

how to get input file name in hadoop cascading

In map-reduce I would extract the input file name as following
public void map(WritableComparable<Text> key, Text value, OutputCollector<Text,Text> output, Reporter reporter)
throws IOException {
FileSplit fileSplit = (FileSplit)reporter.getInputSplit();
String filename = fileSplit.getPath().getName();
System.out.println("File name "+filename);
System.out.println("Directory and File name"+fileSplit.getPath().toString());
process(key,value);
}
How can I do the similar with cascading
Pipe assembly = new Pipe(SomeFlowFactory.class.getSimpleName());
Function<Object> parseFunc = new SomeParseFunction();
assembly = new Each(assembly, new Fields(LINE), parseFunc);
...
public class SomeParseFunction extends BaseOperation<Object> implements Function<Object> {
...
#Override
public void operate(FlowProcess flowProcess, FunctionCall<Object> functionCall) {
how can I get the input file name here ???
}
Thanks,

I don't use Cascading but I think it should be sufficient to access the context instance, using functionCall.getContext(), to obtain the filename you can use:
String filename= ((FileSplit)context.getInputSplit()).getPath().getName();
However, it seems that cascading use the old API, if the above doesn't work you must try with:
Object name = flowProcess.getProperty( "map.input.file" );

Thank Engineiro for sharing the answer. However, when invoking hfp.getReporter().getInputSplit() method, I got MultiInputSplit type which can't be casted into FileSplit type directly in cascading 2.5.3. After diving into the related cascading APIs, I found a way and retrieved input file names successfully. Therefore, I would like to share this to supplement Engineiro's answer. Please see the following code.
HadoopFlowProcess hfp = (HadoopFlowProcess) flowProcess;
MultiInputSplit mis = (MultiInputSplit) hfp.getReporter().getInputSplit();
FileSplit fs = (FileSplit) mis.getWrappedInputSplit();
String fileName = fs.getPath().getName();

You would do this by getting the reporter within the buffer class, from the provided flowprocess argument in the buffer operate call.
HadoopFlowProcess hfp = (HadoopFlowProcess) flowprocess;
FileSplit fileSplit = (FileSplit)hfp.getReporter().getInputSplit();
.
.//the rest of your code
.

Need to get Start Menu Paths in Java

Using java, I would like some code that could get me the paths for:
1) Start Menu for Current User
2) Start Menu for All User
I need the answer for both WinXP and Win7. So hopefully there is a general answer that can get me both.

You have no other choice but to write a DLL and call native Windows API:
SHGetFolderPath(NULL, CSIDL_PROGRAMS, NULL, SHGFP_TYPE_CURRENT, &szPathBuffer)
SHGetFolderPath(NULL, CSIDL_COMMON_PROGRAMS, NULL, SHGFP_TYPE_CURRENT, &szPathBuffer)
If you really need the root of Start menu, use CSIDL_STARTMENU and CSIDL_COMMON_STARTMENU.
The full list of known folders: CSIDL.
If you target Windows Vista and above, use SHGetKnownFolderPath function instead of SHGetFolderPath.
You can use JNA library to call native Windows API without writing native code yourself but pure Java code.

Okay, I figured out a solution, but maybe someone else has a more eligant one.
I plan on doing something like "Runtime.getRuntime().exec(command);" and the command will be a "reg query" to query the following registry keys:
Current User can referenced by: HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\Explorer\Shell Folders\Start Menu
All users can be referenced by: HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\Explorer\Shell Folders\Common Start Menu
These are the same for both Win7 and WinXP. If anyone else knows of a better solution, I'll be happy to look at it too.

In my program I used a simple System.getProperty("user.home") + "/Start Menu/Programs" This gave me the user's Start Menu folder.
It worked on windows 7 and windows 10. I tried this because in order to get a user's desktop, all I had to do was call System.getProperty("user.home") + "/Desktop". SO I figured that it might work for the Start Menu as well, and seemed to have worked fine. I can delete and write files to the Start Menu just like I can with the desktop. Whether this is the right way to do something like this or not, I have no idea. But I'm just sharing what worked for me.

Another option is managing Start Menu items from vbs API.
I made a Java Wrapper for that.
// Install Start Menu
WindowsUtils.installStartMenuItem(WindowsUtils.SPECIALFOLDER_Programs,"my_start_menu", "explorer.exe", "http://www.google.es","Acceso directo a google");
// Uninstall Start Menu
WindowsUtils.uninstallStartMenuItem(WindowsUtils.SPECIALFOLDER_Programs, "my_start_menu");

i recently found this
public class VBSUtils {
public static String SF_ALLUSERSDESKTOP = "AllUsersDesktop";
public static String SF_ALLUSERSSTARTMENU = "AllUsersStartMenu";
public static String SF_ALLUSERSPROGRAMS = "AllUsersPrograms";
public static String SF_ALLUSERSSTARTUP = "AllUsersStartup";
public static String SF_DESKTOP = "Desktop";
public static String SF_FAVORITES = "Favorites";
public static String SF_MYDOCUMENT = "MyDocuments";
public static String SF_PROGRAMS = "Programs";
public static String SF_RECENT = "Recent";
public static String SF_SENDTO = "SendTo";
public static String SF_STARTMENU = "StartMenu";
private VBSUtils() { }
public static String getSpecialFolder(String folder) {
String result = "";
try {
File file = File.createTempFile("realhowto",".vbs");
file.deleteOnExit();
FileWriter fw = new java.io.FileWriter(file);
String vbs = "Set WshShell = WScript.CreateObject(\"WScript.Shell\")\n"
+ "wscript.echo WshShell.SpecialFolders(\"" + folder + "\")\n"
+ "Set WSHShell = Nothing\n";
fw.write(vbs);
fw.close();
Process p = Runtime.getRuntime().exec("cscript //NoLogo " + file.getPath());
BufferedReader input =
new BufferedReader
(new InputStreamReader(p.getInputStream()));
result = input.readLine();
input.close();
}
catch(Exception e){
e.printStackTrace();
}
return result;
}
public static void main(String[] args){
System.out.println(VBSUtils.getSpecialFolder(VBSUtils.SF_ALLUSERSSTARTMENU));
System.out.println(VBSUtils.getSpecialFolder(VBSUtils.SF_ALLUSERSDESKTOP));
System.out.println(VBSUtils.getSpecialFolder(VBSUtils.SF_DESKTOP));
System.out.println(VBSUtils.getSpecialFolder(VBSUtils.SF_PROGRAMS));
//System.out.println(VBSUtils.getSpecialFolder(VBSUtils.SF_STARTUP));
}
}