I want to get the project path and then use it, to find a file.
The problem is that it not returned in the correct format.
For example this is the response:
X:\Project_11_01_2021\test
and to use it I need it to be
X:\\Project_11_01_2021\\test
I use it in selenium to select the path to upload a file
.sendKeys(
path);
Is their a way to get the value and use it in other function?
String path2 = System.getProperty("user.dir");
I used it in windows computer
So if your problem is just, that the path is formatted wrong that is no problem. You just use pythons string replace function and do something like this:
path.replace("\", "\\")
and you should be ready to go. If not, I don't really understand your problem...
Related
We have a folder where we dump lot of files. Our program needs to read one of the specific files with the latest version. The file name would be something like "2016-03-04-12-46-48_ABC_123456_1.xml".
Insted of reading all the files and then iterating to find the exact file i have used following code with a regular expression
File folder = new File("C:\\some_folder")
folder.listFiles((FilenameFilter) new AwkFilenameFilter("(\\d){4}-(\\d){2}-(\\d){2}-(\\d){2}-(\\d){2}-(\\d){2}_ABC_" + <ID_String> +"_(\\d){1,2}"))
But for some reason the reqular expression is not working. Can someone please help with this?
It seems you are missing the file extension in the regex:
(\\d){4}-(\\d){2}-(\\d){2}-(\\d){2}-(\\d){2}-(\\d){2}_ABC_" + <ID_String> +"_(\\d){1,2}\\.xml
Try out this one.
"\d{4}-\d{2}-\d{2}-\d{2}-\d{2}-\d{2}_ABC_"+<ID string>+"_\d{1}.xml"
It works perfectly for me.
I have the path to my XML file on my computer, but how can I use selenium (web automation tool) to inject the XML file ?
Usually how it is done (manually) is navigate to the URL and COPY AND PASTE the entire XML text into the provided text box..
Any ideas how to inject the file using automation ? There is no way to "drag" the XML file to the text box and I believe the way I'm thinking that it will work is very complicated.
I think this is actually what you want -
File xml = new File("xmlpath");
String url = xml.getAbsolutePath();
url = url.replace('\\', '/');
url = url.replace(" ", "%20");
String actual = "file:/" + url;
selenium.open(actual);
Then you should be able to get the xml using String theXML = selenium.getText("//rootxmlnode"); Then do what you will with it.
Check out the topic of Data Driven Testing to get you started. Something like this should get you going.
Selenium tool allows you to create an automatically generated code in Java.
So, you need to place any text in the provided text box and generate this Java-test code.
Next step is modifying of the generated test. You have to manually write a simplest code, which will read your XML file, get it contents and paste into the text box. The last thing is replacement (in the generated Java code of test!) of the mentioned above text-block to the contents of read XML.
A simplest way for reading file into a string is using Apache commons-io library.
For example: FileUtils.readFileToString(File file, String encoding) gives you a string object with contents of the file.
I have a problem here, I have a String that contains a value of C:\Users\Ewen\AppData\Roaming\MyProgram\Test.txt, and I want to remove the C:\Users\Ewen\AppData\Roaming\MyProgram\ so that only Test is left. So the question is, how can i remove any part of the string.
Thanks for your time! :)
If you're working strictly with file paths, try this
String path = "C:\\Users\\Ewen\\AppData\\Roaming\\MyProgram\\Test.txt";
File f = new File(path);
System.out.println(f.getName()); // Prints "Test.txt"
Thanks but I also want to remove the .txt
OK then, try this
String fName = f.getName();
System.out.println(fName.substring(0, fName.lastIndexOf('.')));
Please see this for more information.
The String class has all the necessary power to deal with this. Methods you may be interested in:
String.split(), String.substring(), String.lastIndexOf()
Those 3, and more, are described here: http://docs.oracle.com/javase/1.4.2/docs/api/java/lang/String.html
Give it some thought, and you'll have it working in no time :).
I recommend using FilenameUtils.getBaseName(String filename). The FilenameUtils class is a part of Apache Commons IO.
According to the documentation, the method "will handle a file in either Unix or Windows format". "The text after the last forward or backslash and before the last dot is returned" as a String object.
String filename = "C:\\Users\\Ewen\\AppData\\Roaming\\MyProgram\\Test.txt";
String baseName = FilenameUtils.getBaseName(filename);
System.out.println(baseName);
The above code prints Test.
Hi, I have a big problem. I'm making a java program and I have to call an exe file in a folder that have whitespace. This program also has 2 arguments that always have whitspace in the path.
Example:
C:\Users\Program File\convert image\convert.exe C:\users\image exe\image.jpeg C:\Users\out put\out.bmp
I have to do this in Windows but i want generalize it for every OS.
My code is:
Runtime run = Runtime.getRuntime();<br/>
String path_current = System.getProperty("user.dir");<br/>
String [] uno = new String[]{"cmd","/c",path_current+"\\\convert\\\convert.exe",path_current+"\\\f.jpeg", path_current+"\\\fr.bmp"};<br/>
Process proc2 = run.exec(uno);<br/>
proc2.waitFor();<br/>
This does not work. I tried removing the String array and inserting a simple String with "\"" before and after the path but that didn't work. How do I resolve this?
you may want to use :
http://commons.apache.org/io/api-1.4/org/apache/commons/io/FilenameUtils.html#separatorsToSystem(java.lang.String)
see also this answer :
Is there a Java utility which will convert a String path to use the correct File separator char?
Remove "cmd" and "/c", and use a single forward slash instead of your triple backslaches.
If I have a directory called temp with the following files:
a_file1.jpg
a_file2.jpg
b_file1.jpg
b_file2.jpg
It's possible to get all files like this:
VFS.getManager().resolveFile("temp").getChildren();
But, what I actually want to do is get a_file1.jpg and a_file2.jpg. Maybe like:
VFS.getManager().resolveFile("temp/a*").getChildren();
But this throws an exception:
org.apache.commons.vfs.FileSystemException: Could not list the contents of "temp/a*" because it is not a folder.
So, does anyone know how to resolve a set of files based on a regex with VFS?
You could use the findFiles method, with a FileFilterSelector.
You'll need to create your own FileFilter that accepts the files that match your desired regex.