I am working with selenium automation and I am not able to upload file with relative path please see the below code.
driver.findElement(By.xpath("//span[text()='Theme']")).click();
File filepath=new File("\ntwinelogin.jpg");
WebElement fileobj = driver.findElement(By.name("toplogoupload"));
fileobj.sendKeys("\ntwinelogin.jpg");
String Filepath=filepath.getAbsolutePath();
Filepath.trim();
Use System.getProperty("user.dir") for your current project directory path.
System.getProperty("user.dir")+"\ntwinelogin.jpg";
You can use below function for relative path in Python:
Import
import sys, os
Use code as below :
ROOT_DIR = os.path.dirname(os.path.abspath(__file__))
screenshotpath = os.path.join(os.path.sep, ROOT_DIR,'YOURFOLDERNAME'+ os.sep)
print screenshotpath
make sure you create folder where .py file is present
Java
System.getProperty("user.dir")
Step 1 : First store the file in one variable e.g. String path = "C:\\users\\home\\newhtml.html";
Step 2 : Pass the variable to sendkeys() method
driver.findelement(By.xapth("")).sendkeys(path);
I recently required a platform-agnostic solution to this problem. Since we were using Maven, I implemented the following solution.
Maven directory structure
src
|
- main
| |
| - java
| |
| - com
| |
| - example
| |
| - integration
| |
| - Test1.java
| |
| - Test2.java
| ...
|
- resources
|
- File1.csv
|
- File2.csv
...
Test
// Load a file on the classpath as a resource using the ClassLoader.
URL url = getClass().getClassLoader().getResource("File1.csv");
// Convert the URL into a URI.
URI uri = file.toURI();
// Load the file from the URI.
File file = new File(uri);
// Get the absolute path to the file.
String path = file.getAbsolutePath();
// Use the absolute path with Selenium.
input.sendKeys(path);
