I'm trying to upload a file within the windows pop up after clicking on a button:

After clicking on that button a pop up in shown:

I click on that button by using this:

driver.findElement(By.xpath("//*[@id=\"div-add-file\"]/a")).click();

Then, i used this:

driver.switchTo().activeElement().sendKeys("C:\\Users\\Steve\\Downloads\\01004185FCA003900517097.pdf");

It did not work.

I tried to do something like that avoiding clicking on the button but sending the file path:

driver.findElement(By.xpath("//*[@id=\"div-add-file\"]/a")).sendKeys("C:\\Users\\Maxi\\Downloads\\01004185FCA003900517097.pdf");

but it wont work.

Am i doing anything wrong?

CodePudding user response：

Uploading a file with Selenium is done by sending a path to the uploaded file to a special element on the page. This element can normally be located by this XPath: //input[@type='file']
So your command could be something like the following:

driver.findElement(By.xpath("//input[@type='file']")).sendKeys("C:\\Users\\Maxi\\Downloads\\01004185FCA003900517097.pdf");

Maybe you will need to add some delay to make the page loaded. If so WebDriverWait should be used.

WebDriverWait wait = new WebDriverWait(webDriver, timeoutInSeconds);
wait.until(ExpectedConditions.presenceOfElementLocated(By.xpath("//input[@type='file']"))).sendKeys("C:\\Users\\Maxi\\Downloads\\01004185FCA003900517097.pdf");

Sometimes several more that 1 inputs matching the above locator will be on the same page and sometimes there will be no such element.
In case there are more than 1 elements matching the above locator we will need to find the precise, unique locator.
In case of no such element we will have to find another approach to upload file.