I tried this. Will it work?
FileUpload1.SaveAs(Server.MapPath("resume\\"+FileUpload1.FileName + " "));
StatusLabel.Text = "Upload status: File uploaded;
Try this:
Server.MapPath("data.txt")
Related
Katalon is popular in automation testing. I have already used it in our project and it works amazingly.
Now, What I want to achieve is to create a test case where it opens a terminal (using mac) and type in some commands to run it like for example:
cd /documents/pem/key.pem
connect to -my server via SSH#method
sudo su
yum install php7
yum install mysql
You are not alone, and with custom keywords you can achieve what you want. Here is an example showing a test of a command line app. You could do the same thing to call any command line script you wish. Think of a runCmd keyword, or a runCmdWithOutput to grab the output and run various asserts on it.
#Keyword
def pdfMetadata(String input) {
KeywordUtil.logInfo("input: ${input}")
def csaHome = System.getenv("CSA_HOME")
def cmd = "cmd /c ${csaHome}/bin/csa -pdfmetadata -in \"${projectPath}${input}\"";
runCmd(cmd)
}
def runCmd(String cmd) {
KeywordUtil.logInfo("cmd: ${cmd}")
def proc = cmd.execute();
def outputStream = new StringBuffer();
def errStream = new StringBuffer()
proc.waitForProcessOutput(outputStream, errStream);
println(outputStream.toString());
println(errStream.toString())
if(proc.exitValue() != 0){
KeywordUtil.markFailed("Out:" + outputStream.toString() + ", Err: " + errStream.toString())
}
}
You can then use this in a test case:
CustomKeywords.'CSA.pdfMetadata'('/src/pdf/empty.pdf')
Here is another custom keyword! It is takes the file name and path, and if you don't give it a path, it search for the file in the project root directory. It export the batch file's output in a batch_reports folder in your project folder, you need to create that in advance.
#Keyword
def runPostmanBatch(String batchName , String batchPath){
// source: https://www.mkyong.com/java/how-to-execute-shell-command-from-java/
String firstParameter = "cmd /c " + batchName;
String secondParameter = batchPath;
if (batchPath == ""){
secondParameter = RunConfiguration.getProjectDir();
}
try {
KeywordUtil.logInfo("Executing " + firstParameter + " at " + secondParameter)
Process process = Runtime.getRuntime().exec(
firstParameter , null, new File(secondParameter));
StringBuilder output = new StringBuilder();
BufferedReader reader = new BufferedReader(
new InputStreamReader(process.getInputStream()));
String line;
while ((line = reader.readLine()) != null) {
output.append(line + "\n");
}
int exitVal = process.waitFor();
Date atnow = new Date()
String now = atnow.format('yy-MM-dd HH-mm-ss')
String report_path = RunConfiguration.getProjectDir() + "/postman_reports/" + RunConfiguration.getExecutionSourceName() + "_" + now + ".txt"
BufferedWriter writer = new BufferedWriter(new FileWriter(report_path));
writer.write(output.toString());
writer.close();
KeywordUtil.logInfo("postman report at: " + report_path)
if (exitVal == 0) {
println("Success!");
println(output);
KeywordUtil.markPassed("Ran successfully")
} else {
KeywordUtil.markFailed("Something went wrong")
println(exitVal);
}
} catch (IOException e) {
e.printStackTrace();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
I've done some research. I did not found any resources or any people that is looking for the same thing I am. I think this is officially, No. The answer to this is, it is not possible.
It is possible to run Katalon Studio from the command line.
There's a short tutorial here.
And it will be possible to override Profile Variables via command line execution mode from v5.10 (currently in beta).
An example given on Katalon forum is:
Simply pass the parameters in command line using: -g_XXX = XXX
Below is an example of override an URL variable:
-g_URL=http://demoaut.katalon.com
I am using the following method to browse for a file:
OpenFileDialog.ShowDialog()
PictureNameTextEdit.Text = OpenFileDialog.FileName
Is there a way get ONLY the file name?
The FileName method returns the entire path and file name.
i.e. I want Foo.txt instead of C:\SomeDirectory\Foo.txt
Use Path.GetFileName(fullPath) to get just the filename part, like this:
OpenFileDialog.ShowDialog()
PictureNameTextEdit.Text = System.IO.Path.GetFileName(OpenFileDialog.FileName)
OpenFileDialog.ShowDialog()
PictureNameTextEdit.Text = System.IO.Path.GetFileName(OpenFileDialog.FileName)
C++ code for obtain filename and complete path in OpenFileDialog:
textBox1->Text = OpenFileDialog1->FileName; //complete path
textBox1->Text = System::IO::Path::GetFileName(OpenFileDialog1->FileName); //filename
Suppose that I did select word2010 file named as "MyFileName.docx"
This is for ONLY the selected file extension "including the dot mark, f.e (.docx)"
MsgBox(System.IO.Path.GetExtension(Opendlg.FileName))
And this for the selected File name without extension: (MyFileName)
MsgBox(System.IO.Path.GetFileNameWithoutExtension(Opendlg.FileName))
and you can try the other options for the "PATH Class" like: GetFullPath,GetDirectoryName ...and so on.
if you want just the selected name without Extension you can try this code
Imports System.IO
PictureNameTextEdit.Text = Path.GetFileNameWithoutExtension(OpenFileDialog1.FileName)
thanx
//Following code return file name only
string[] FileFullPath;
string FileName;
objOpenFileDialog.Title = "Select Center Logo";
objOpenFileDialog.ShowDialog();
FileFullPath = objOpenFileDialog.FileNames[0].ToString().Split('\\');
FileName = FileFullPath[FileFullPath.Length - 1]; //return only File Name
//Use following code if u want save other folder ,
// following code save file to CenterLogo folder which inside bin folder//
System.IO.File.Copy(OFD.FileName, Application.StartupPath +
"/CenterLogo/" + FileName, true);
Use SafeFileName instead of FileName and it will return a name (and extension) without path.
Use this code to put the filename in PictureNameTextEdit:
OpenFileDialog.ShowDialog()
PictureNameTextEdit.Text = OpenFileDialog.SafeFileName
I am trying to download an Excel file from Podio space. I get access to space and get the files within space and the using fileAPI .
The problem is with the file that gets downloaded. I am not able to open it with Excel.
Here is the code:
List <File> aFileList = aFileAPI.getOnSpace(spaceId, null, null);
for (int i=0; i<aFileList.size(); i++){
File thisFile = aFileList.get(i);
MimeType thiFileMimeType = thisFile.getMimetype();
System.out.println("thisFile = " +thisFile.toString());
System.out.println("thisFile.getDescription() =" + thisFile.getDescription());
System.out.println("thisFile.getName() =" + thisFile.getName());
System.out.println("thisFile.getId() =" + thisFile.getId());
System.out.println("thisFile.getSize() =" + thisFile.getSize());
System.out.println("thisFile.getCreatedOn() =" + thisFile.getCreatedOn());
System.out.println("thisFile.getMimetype() =" + thisFile.getMimetype());
try {
aFileAPI.downloadFile(thisFile.getId(), new java.io.File("C:/Users/alisa/Desktop/"+ thisFile.getName()), null);
}
catch (IOException e) {
e.printStackTrace();
}
I think the API is creating the file without regard to MIME and I don't see anywhere to set it. Please help!!
file name: abc.aspx
my code:
string Path = #"E:\documents\Data20160129110355.xls";
HttpContext.Current.Response.ContentType = "application/vnd.ms-excel";
HttpContext.Current.Response.AppendHeader("Content-Disposition", "inline; filename=Data20160129110355.xls");
HttpContext.Current.Response.TransmitFile(Path);
HttpContext.Current.Response.Flush();
HttpContext.Current.Response.Close();
On click of "Open" it will open the file successfully but the header of the file are not showing "Data20160129110355.xls". It's showing abc.aspx.
How to solve this problem please reply.
Hope this will help:
Content-Disposition:What are the differences between "inline" and "attachment"?
replace 'inline' with 'attachment'.
UP: try this:
HttpContext.Current.Response.AppendHeader("Content-Disposition", "attachment; filename=Data20160129110355.xls");
I am using the following method to browse for a file:
OpenFileDialog.ShowDialog()
PictureNameTextEdit.Text = OpenFileDialog.FileName
Is there a way get ONLY the file name?
The FileName method returns the entire path and file name.
i.e. I want Foo.txt instead of C:\SomeDirectory\Foo.txt
Use Path.GetFileName(fullPath) to get just the filename part, like this:
OpenFileDialog.ShowDialog()
PictureNameTextEdit.Text = System.IO.Path.GetFileName(OpenFileDialog.FileName)
OpenFileDialog.ShowDialog()
PictureNameTextEdit.Text = System.IO.Path.GetFileName(OpenFileDialog.FileName)
C++ code for obtain filename and complete path in OpenFileDialog:
textBox1->Text = OpenFileDialog1->FileName; //complete path
textBox1->Text = System::IO::Path::GetFileName(OpenFileDialog1->FileName); //filename
Suppose that I did select word2010 file named as "MyFileName.docx"
This is for ONLY the selected file extension "including the dot mark, f.e (.docx)"
MsgBox(System.IO.Path.GetExtension(Opendlg.FileName))
And this for the selected File name without extension: (MyFileName)
MsgBox(System.IO.Path.GetFileNameWithoutExtension(Opendlg.FileName))
and you can try the other options for the "PATH Class" like: GetFullPath,GetDirectoryName ...and so on.
if you want just the selected name without Extension you can try this code
Imports System.IO
PictureNameTextEdit.Text = Path.GetFileNameWithoutExtension(OpenFileDialog1.FileName)
thanx
//Following code return file name only
string[] FileFullPath;
string FileName;
objOpenFileDialog.Title = "Select Center Logo";
objOpenFileDialog.ShowDialog();
FileFullPath = objOpenFileDialog.FileNames[0].ToString().Split('\\');
FileName = FileFullPath[FileFullPath.Length - 1]; //return only File Name
//Use following code if u want save other folder ,
// following code save file to CenterLogo folder which inside bin folder//
System.IO.File.Copy(OFD.FileName, Application.StartupPath +
"/CenterLogo/" + FileName, true);
Use SafeFileName instead of FileName and it will return a name (and extension) without path.
Use this code to put the filename in PictureNameTextEdit:
OpenFileDialog.ShowDialog()
PictureNameTextEdit.Text = OpenFileDialog.SafeFileName