I'm trying to run a command in Python by concatenating two strings, each string has an absolute path in it. It's throwing an error. Below is the code and error for reference. Kindly correct me where am I going wrong.
Code:
def test_execute_resources_temp_dir(self):
sourcePath = Path("Gemini_logo.png").absolute()
print(sourcePath)
with tempfile.TemporaryDirectory() as tmpdirname:
destination = Path(tmpdirname).resolve("Gemini_logo_converted.jpg")
print('created temporary directory', destination)
command = "magick convert" + sourcePath +" "+ destination
formatConversion = FormatConversion()
formatConversion.executeCommand(command)
Error:
TypeError: can only concatenate str (not "WindowsPath") to str test_FormatConversion.py:65: TypeError
You can replace sourcePath with str(sourcePath).
Similarly, you can replace destination with str(destination)
command = "magick convert" + str(sourcePath) +" "+ str(destination)
This worked out for me
def test_execute_resources_temp_dir(self):
os.chdir('..')
sourcePath = Path("src\\resources\\Gemini_logo.png").absolute()
print(sourcePath)
with tempfile.TemporaryDirectory() as tmpdirname:
# os.mkdir(os.path.basename(tmpdirname))
destination = Path(tmpdirname+"/Gemini_logo_converted1.jpg")
print('created temporary directory', destination)
command = **"magick convert"+" "+ str(sourcePath) + " " + str(destination)**
Related
I am stuck from a couple of days on an issue in my micro Address Book project. I have a function that writes all records from a SQLite3 Db on file in order to open in via OS module, but as soon as I try to open the file, Python gives me the following error:
Error while opening tempfile. Error:startfile: filepath should be string, bytes or os.PathLike, not _io.TextIOWrapper
This is the code that I have to write records on file and to open it:
source_file_name = open("C:\\workdir\\temp.txt","w")
#Fetching results from database and storing in result variable
self.cur.execute("SELECT id, first_name, last_name, address1, address2, zipcode, city, country, nation, phone1, phone2, email FROM contacts")
result = self.cur.fetchall()
#Writing results into tempfile
source_file_name.write("Stampa Elenco Contatti\n")
for element in result:
source_file_name.write(str(element[0]) + "|" + str(element[1]) + "|" + str(element[2]) + "|" + str(element[3]) + "|" + str(element[4]) + "|" + str(element[5]) + "|" + \
str(element[6]) + "|" + str(element[7]) + "|" + str(element[8]) + "|" + str(element[9]) + "|" + str(element[10]) + "|" + str(element[11]) + "\n")
#TODO: Before exiting printing function you MUST:
# 1. filename.close()
# 2. exit to main() function
source_file_name.close()
try:
os.startfile(source_file_name,"open")
except Exception as generic_error:
print("Error while opening tempfile. Error:" + str(generic_error))
finally:
main()
Frankly I don't understand what this error means, in my previous code snippets I've always handled text files without issues, but I realize this time it's different because I am picking my stream from a database. Any ideas how to fix it?
Thanks in advance, and sorry for my english...
Your problem ultimately stems from poor variable naming. Here
source_file_name = open("C:\\workdir\\temp.txt","w")
source_file_name does not contain the source file name. It contains the source file itself (i.e., a file handle). You can't give that to os.startfile(), which expects a file path (as the error also says).
What you meant to do is
source_file_name = "C:\\workdir\\temp.txt"
source_file = open(source_file_name,"w")
But in fact, it's much better to use a with block in Python, as this will handle closing the file for you.
It's also better to use a CSV writer instead of creating the CSV manually, and it's highly advisable to set the file encoding explicitly.
import csv
# ...
source_file_name = "C:\\workdir\\temp.txt"
with open(source_file_name, "w", encoding="utf8", newline="") as source_file:
writer = csv.writer(source_file, delimiter='|')
source_file.write("Stampa Elenco Contatti\n")
for record in self.cur.fetchall():
writer.writerow(record)
# alternative to the above for loop on one line
# writer.writerows(self.cur.fetchall())
I have this part code which i don't understand the for loop part:
also what does file path holds and how it is different than all file path?
all_files_path = glob.glob("ADNI1_Screening_1.5T/ADNI/*/*/*/*/*.nii")
for file_path in all_file_path:
print(os.system("./runROBEX.sh " + file_path + " stripped/" +file_path.split("/")[-1]))
I am trying to split a path into parent and the name.
When trying
String path = "/root/file"
File file = new File(path)
println("Name: " + file.name)
println("Parent: " + file.parent)
We get
Name: file
Parent: /root
With the Windows path C:\\root\\file.exe we get
Name: C:\root\file.exe
Parent: null
Is this the intended behaviour? And if it is how do I get the same result for a Windows path? (If possible please without using regular expressions)
use .replace to change the "\" to "/
String path = "C:\\root\\file.exe"
path = path.replace("\\","/")
File file = new File(path)
println("Name: " + file.name)
println("Parent: " + file.parent)
I have the following line in my code
def genList = (args[]?.size() >=4)?args[3]: "
when I run my whole code I get the following error
expecting anything but ''\n''; got it anyway at line: 9, column: 113
here I am adding the whole code so you can see what I am doing
def copyAndReplaceText(source, dest, targetText, replaceText){
dest.write(source.text.replaceAll(targetText, replaceText))
}
def dire = new File(args[0])
def genList = (args[]?.size() >=4)?args[3]: " // check here if argument 4 is provided, and generate output if so
def outputList = ""
dire.eachFile {
if (it.isFile()) {
println it.canonicalPath
// TODO 1: copy source file to *.bak file
copy = { File src,File dest->
def input = src.newDataInputStream()
def output = dest.newDataOutputStream()
output << input
input.close()
output.close()
}
//File srcFile = new File(args[0])
//File destFile = new File(args[1])
//File srcFile = new File('/geretd/resume.txt')
//File destFile = new File('/geretd/resumebak.txt')
File srcFile = it
File destFile = newFile(srcFile + '~')
copy(srcFile,destFile)
// search and replace to temporary file named xxxx~, old text with new text. TODO 2: modify copyAndReplaceText to take 4 parameters.
if( copyAndReplaceText(it, it+"~", args[1], args[2]) ) {
// TODO 3: remove old file (it)
it.delete()
// TODO 4: rename temporary file (it+"~") to (it)
// If file was modified and parameter 4 was provided, add modified file name (it) to list
if (genList != null) {
// add modified name to list
outputList += it + "\n\r"
}
}
}
}
// TODO 5: if outputList is not empty (""), generate to required file (args[3])
if (outputList != ""){
def outPut = new File(genList)
outPut.write(outputList)
}
Thank you
Just close your double quotes
def genList = (args?.size() >=4)?args[3]: ""
The specific OP question was already answered, but for those who came across similar error messages in Groovy, like:
expecting anything but '\n'; got it anyway
expecting '"', found '\n'
It could be caused due to multi-line GString ${content} in the script, which should be quoted with triple quotes (single or double):
''' ${content} ''' or """ ${content} """
Why do you have a single " at the end of this line: def genList = (args[]?.size() >=4)?args[3]: "?
You need to make it: def genList = (args[]?.size() >=4)?args[3]: ""
You need to add a ; token at the end of def outputList = ""
Also get rid of the " at the end of def genList = (args[]?.size() >=4)?args[3]: "
Hi I have recently made this script to rename files I scan for work with a prefix and a date. It works pretty well however it would be great if it could make a directory in the current directory with the same name as the first file then move all the scanned files there. E.g. First file is renamed to 'Scanned As At 22-03-2012 0' then a directory called 'Scanned As At 22-03-2012 0' (Path being M:\Claire\Scanned As At 22-03-2012 0) is made and that file is placed in there.
I'm having a hard time figuring out the best way to do this. Thanks in advance!
import os
import datetime
#target = input( 'Enter full directory path: ')
#prefix = input( 'Enter prefix: ')
target = 'M://Claire//'
prefix = 'Scanned As At '
os.chdir(target)
allfiles = os.listdir(target)
count = 0
for filename in allfiles:
t = os.path.getmtime(filename)
v = datetime.datetime.fromtimestamp(t)
x = v.strftime( ' %d-%m-%Y')
os.rename(filename, prefix + x + " "+str(count) +".pdf")
count +=1
Not quite clear about your requirement. If not rename the file, only put it under the directory, then you can use the following codes (only the for-loop of your example):
for filename in allfiles:
if not os.isfile(filename): continue
t = os.path.getmtime(filename)
v = datetime.datetime.fromtimestamp(t)
x = v.strftime( ' %d-%m-%Y')
dirname = prefix + x + " " + str(count)
target = os.path.join(dirname, filename)
os.renames(filename, target)
count +=1
You can check help(os.renames).