I need to upload a file using Python and Selenium. When I click the upload HTML element a "File Upload" window is opened and the click() method does not return since it waits to fully load the page. Therefore I cannot continue using pywinauto code to control the window.
The first method clicks the HTML element (an img) to upload a new file:
def add_file(self):
return self.selenium.find_element(By.ID, "add_file").click()
and the second method is using pywinauto to type the path to the file and then click open
def upload(self):
from pywinauto import application
app = application.Application()
app.connect_(title_re = "File Upload")
app.file_upload.TypeKeys("C:\\Path\\To\\FIle")
app.file_upload.Open.Click()
How can I force add_file method to return and to be able to run the upload method?
Solve it. There was an iframe dealing with the upload but was hidden and didn't see it in the first place. The iframe contains an input of type file also hidden. To solve it make the iframe visible using javascript:
selenium.execute_script("document.getElementById('iframe_id').style.display = 'block';")
then switch to the iframe and make the input visible also:
selenium.switch_to_frame(0)
selenium.execute_script("document.getElementById('input_field_id').type = 'visible';")
and simply send the path to the input:
selenium.find_element(By.ID, 'input_field_id').send_keys("path\\\\to\\\\file")
For windows use 4 '\\\\' as path separator.
Related
I m making an Excel automation via pywinauto library. But there is a hard challange for me due to using Excel Oracle add-ins called Smartview.
I need to click 'Private Connections' button, however i can't find any little info in app.Excel.print_control_identifiers() Private Connections
So i tried to use inspector.exe for find ui element regarding private connections button, however i couldn't find any little solvetion inside of inspector.exe's result inspector's result
Then i used another program called UISpy, however i can only find private connection's pane inside of the program. UISpy's result
i tried to find an answer but i couldn't find out anything. So, can you help me to click here?
By the way here is my code :
import pywinauto
from pywinauto import application
from pywinauto.keyboard import send_keys
from pywinauto.controls.common_controls import TreeViewWrapper
program_path = r"C:\Program Files\Microsoft Office\root\Office16\EXCEL.EXE"
file_path = r"C:\Users\AytugMeteBeder\Desktop\deneme.xlsx"
app = application.Application(backend="uia").start(r'{} "{}"'.format(program_path, file_path))
# sapp = application.Application(backend="uia").connect(title = 'deneme.xlsx - Excel')
time.sleep(7)
myExcel = app.denemeExcel.child_window(title="Smart View", control_type="TabItem").wrapper_object()
myExcel.click_input()
Panel = app.denemeExcel.child_window(title="Panel", control_type="Button").wrapper_object()
Panel.click_input()
time.sleep(1)
app.denemeExcel.print_control_identifiers()
Hi team,
I have created program where I need to upload file. By clicking on "Upload" , I reach to this window. but here I stuck now. I need to enter path automatically. please help if you have any examples.
driver1 = webdriver.Chrome(executable_path= r'C:\\Users\\Ast1\\Desktop\\chromedriver.exe')
Sharepoint ="https://ast.sharepoint.com/sites/ExtTeam_GPS/Shared%20Documents/Forms/AllItems.aspx?viewid=3638f543%2D486c%2D4bca%2D85a7%2D8d86da0a8eb5&id=%2Fsites%2FExtTeam%5FNO%2FShared%20Documents%2FPowerBI%20Datasets"
driverLink= Sharepoint
driver1.get(driverLink)
Em = WebDriverWait(driver1,120).until(EC.presence_of_element_located((By.ID,'i0116')))
Em.send_keys("astemp1#ast.com")
elm =WebDriverWait(driver1,120).until(EC.presence_of_element_located((By.ID,'idSIButton9')))
elm.click()
driver1.implicitly_wait(100)
#elm1 = WebDriverWait(driver1,100).until(EC.presence_of_element_located((By.PARTIAL_LINK_TEXT,'Upload')))
elm2 = driver1.find_element_by_xpath('//*[#id="appRoot"]/div[1]/div[2]/div[3]/div/div[2]/div[2]/div[2]/div[1]/div/div/div/div/div/div/div[1]/div[2]/button')
elm2.click()
driver1.implicitly_wait(20)
elm3 = driver1.find_element_by_name('Files')
elm3.click()
driver1.implicitly_wait(20)
#Follwing code is not working.
Elm4 = driver1.find_element(By.ID('File name'))
Elm4.send_keys(r'C:\A1\AA1\M1\Input\TblGpsNew.xlsx')
#driver1.implicitly_wait(100)
If the page where you have the upload button (entire page), if the below xpath
//input[#type='file']
exists at least one time 1/1 matches then, you can directly send the keys.
Explicitly no need to click on upload button and then interact with the windows pop up.
driver1.find_element(By.XPATH("//input[#type='file']")).send_keys("C:\A1\AA1\M1\Input\TblGpsNew.xlsx")
I'm using python 3 with selenium, I have to download an image
HTML:
<img id="labelImage" name="labelImage" border="0" width="672" height="456" alt="labelImage" src="/shipping/labelAction.handle?method=doGetLabelFromCache&isDecompressRequired=false&utype=null&cacheKey=774242409034SHIPPING_L">
Python code:
found = browser.find_element_by_css_selector('img[alt="labelImage"]')
src = found.get_attribute('src')
urllib.request.urlretrieve(src, 'image.png')
that image file is empty, if I try to switch extension to html, shows me message below:
"We're sorry, we can't process your request right now. It appears you don't have permission to view this webpage"
The error you recieve when attempt to download comes from the fact the urllib call is a brand new session for their server - it does not have the cookies and authentication your browser does. E.g. it is the same as if you open incognito mode in the browser, and paste in the address bar the src attribute - for the server you are a new client, that hasn't fill in the form, logged in, etc.
You may want to try something else - in the selenium/the browser session, taking a screenshot of just the <img> element. That op is with variable success, Chrome for instance added support for it only recently, and in some situations it fails:
found = browser.find_element_by_css_selector('img[alt="labelImage"]')
try:
found.screenshot('element.png')
except Exception as ex: # FIXME: anti-pattern - I don't recall the exact exception - when you run the code, change it to the proper one
print('The correct exception is {}'.format(ex))
browser.get_screenshot_as_file('page.png')
If taking the element's screenshot fails, you'll get one of the whole page - which you can then trim to the element.
From Selenium, I managed to automate my tasks for a website.
However I ran into a problem:I need to upload a file to my web-page before submitting the form.
It is NOT an option to write the uploaded file into it's input element, because it is more complicated than this.
So basically I need to launch the FileUpload dialog by clicking a button, sendKeys there, and then close it by clicking on Ok.
I am wondering if this is even possible using just Selenium?
I am using it from python (so I don't have access to Robot class)
I tried so far:
element.click()
time.sleep(5)
alert = driver.switch_to.alert
alert.send_keys("path.to.myfile.txt")
alert.accept()
(nothing happens - I mean, the file open dialog works fine, but it does not send the keys )
I also tried:
alert = driver.switch_to.alert
buildu = ActionChains(driver).send_keys('path.to.my.file.txt')
buildu.perform()
(also not working)
Maybe I am looking at it wrong...
Maybe the alerts is not a good approach?
Do you have any idea?
I would prefere not having to use AUTOIT (for my own reasons)
So my goal is to click on a link element (DONE) then the link opens the File Upload open file dialog (DONE), then I need to be able to enter a text in the only textBox of the new window, and click on the Ok button
EDIT
This is the Open File dialog that shows up.
All I want to do is send the filename directly to the window (the textBox is focused when the dialog shows up, so no need to make more actions). After I send the keys (text) I need to be able to click on the Open button
You can create a Java program that will paste the filename and press the enter key. I had this same problem exactly. This is how I implemented:
package myaots_basic;
import java.io.*;
import java.awt.AWTException;
import java.awt.Robot;
import java.awt.Toolkit;
import java.awt.datatransfer.StringSelection;
import java.awt.event.KeyEvent;
public class trial {
public static void main(String[] args) throws AWTException {
System.out.println("HELLO WORLD");
StringSelection attach1 = new StringSelection ("C:\\My Office Documents\\Selinium projects\\Data\\attachment1.doc");
Toolkit.getDefaultToolkit().getSystemClipboard().setContents(attach1, null);
Robot rb1 = new Robot();
rb1.delay(3000);
rb1.keyPress(KeyEvent.VK_CONTROL);
rb1.keyPress(KeyEvent.VK_V);
rb1.keyRelease(KeyEvent.VK_V);
rb1.keyRelease(KeyEvent.VK_CONTROL);
rb1.delay(500);
rb1.keyPress(KeyEvent.VK_ENTER); // press Enter
rb1.keyRelease(KeyEvent.VK_ENTER);
}
}
Name it to trial.jar. Make this class a an executable.
Then in your python code just add a simple step:
import subprocess
subprocess.call("java -jar trial.jar", shell=True)
I don't know if this is restricted to windows 10 or works slightly different in other Windows versions but you can use the following code/idea.
Open the windows shell with win32com.client.Dispatch("WScript.Shell")
Send Tab keys to navigate through the open file dialog
Paste your absolute file path into the top of the dialog and also into the file selection text field and send enter key when
tab-navigating to the "open" button
Make a sleep of at least 1 second between each action.. otherwise Windows blocks this action.
import win32com.client as comclt
...
def handle_upload_file_dialog(self, file_path):
sleep = 1
windowsShell = comclt.Dispatch("WScript.Shell")
time.sleep(sleep)
windowsShell.SendKeys("{TAB}{TAB}{TAB}{TAB}{TAB}")
time.sleep(sleep)
windowsShell.SendKeys("{ENTER}")
time.sleep(sleep)
windowsShell.SendKeys(file_path)
time.sleep(sleep)
windowsShell.SendKeys("{TAB}{TAB}{TAB}{TAB}{TAB}")
time.sleep(sleep)
windowsShell.SendKeys(file_path)
time.sleep(sleep)
windowsShell.SendKeys("{TAB}{TAB}")
time.sleep(sleep)
windowsShell.SendKeys("{ENTER}")
...
I have a python script that tries to upload a file from my PC to a web application.
I press via WebDriver the specific upload button in the browser and then a Win7 explorer window opens for me to navigate and select the desired file to upload.
How could I manipulate this window with pywinauto?
optional: could this be done in linux as well (with an appropriate library I suppose) ?
This is my sample code:
wd.find_element_by_css_selector("img.editLecturesButtons.fromVideo").click()
#switch to the lightbox
wd.switch_to_frame(int("1"))
#hit upload
wd.find_element_by_xpath("//*[#id='fileUpload']").click()
#TODO
import os,pywinauto.application
file = os.path.normpath("C:\Users\me\Desktop\image.jpg")
....
I agree with Mark, you should try the Webdriver methods. Regard to pywinauto, code may looks like:
import pywinauto
pwa_app = pywinauto.application.Application()
w_handle = pywinauto.findwindows.find_windows(title=u'Open', class_name='#32770')[0]
window = pwa_app.window_(handle=w_handle)
ctrl = window['Name']
ctrl.SetText(file)
ctrl = window['OK']
ctrl.Click()
This sollution only for Windows, since pywinauto uses win32 api.