file

opening a file based on user input c++

让人想犯罪 __ 提交于 2021-02-11 08:01:31
问题 I am trying to make a program that would open a file based on the users input. Here`s my code: #include <iostream> #include <fstream> #include <string> using namespace std; int main() { string filename; ifstream fileC; cout<<"which file do you want to open?"; cin>>filename; fileC.open(filename); fileC<<"lalala"; fileC.close; return 0; } But when I compile it, it gives me this error: [Error] no match for 'operator<<' (operand types are 'std::ifstream {aka std::basic_ifstream<char>}' and 'const

write mouse position to .txt file [duplicate]

♀尐吖头ヾ 提交于 2021-02-11 07:33:33
问题 This question already has answers here : .write not working in Python (5 answers) Closed 5 months ago . My little project is to write coordinates to file and my code look like this: import pyautogui import time c = open("coord.txt", "w") while True: x, y = pyautogui.position() positionStr = str(x).rjust(4) + str(y).rjust(4) print(positionStr) c.write(positionStr) time.sleep(1) It not work becouse it show in terminal the coordinates but file coord.txt is still empty 回答1: You are just missing

How should a file: URI corresponding to a Windows path name look like?

被刻印的时光 ゝ 提交于 2021-02-11 07:01:47
问题 I am trying to load a MySQL jar dynamically in code but I am unsure about the format of the Windows path name. Is what I am using below correct, for loading a .jar from a thumbdrive? URL u = new URL("jar:file:G:/mysql-connector-java-5.1.15.jar!/"); URLClassLoader ucl = new URLClassLoader(new URL[] { u }); Now, this is not the same as the traditional path that you see in Java tutorials: URL url = new URL("file:/g:/mysql-connector-java-5.1.15.jar"); For the answer, I am looking for

Append word to input file name for output file name

徘徊边缘 提交于 2021-02-11 05:15:57
问题 I am trying to read in some data, process the data, and write the results to a CSV saved with original file name + the word "folded". I'm using sys.argv to pass the input filename, and thought that I could just create a new variable such as filename = sys.argv[1]+'_folded.csv but I ended up with file.csv_folded.csv . How can I do this so my output file is saved as file_folded.csv? Simplified code example: import sys import networkx as nx G = nx.read_edgelist(sys.argv[1], delimitier=',')

Append word to input file name for output file name

白昼怎懂夜的黑 提交于 2021-02-11 05:12:39
问题 I am trying to read in some data, process the data, and write the results to a CSV saved with original file name + the word "folded". I'm using sys.argv to pass the input filename, and thought that I could just create a new variable such as filename = sys.argv[1]+'_folded.csv but I ended up with file.csv_folded.csv . How can I do this so my output file is saved as file_folded.csv? Simplified code example: import sys import networkx as nx G = nx.read_edgelist(sys.argv[1], delimitier=',')

Append word to input file name for output file name

冷暖自知 提交于 2021-02-11 05:12:14
问题 I am trying to read in some data, process the data, and write the results to a CSV saved with original file name + the word "folded". I'm using sys.argv to pass the input filename, and thought that I could just create a new variable such as filename = sys.argv[1]+'_folded.csv but I ended up with file.csv_folded.csv . How can I do this so my output file is saved as file_folded.csv? Simplified code example: import sys import networkx as nx G = nx.read_edgelist(sys.argv[1], delimitier=',')

Append word to input file name for output file name

混江龙づ霸主 提交于 2021-02-11 05:10:18
问题 I am trying to read in some data, process the data, and write the results to a CSV saved with original file name + the word "folded". I'm using sys.argv to pass the input filename, and thought that I could just create a new variable such as filename = sys.argv[1]+'_folded.csv but I ended up with file.csv_folded.csv . How can I do this so my output file is saved as file_folded.csv? Simplified code example: import sys import networkx as nx G = nx.read_edgelist(sys.argv[1], delimitier=',')

How can I overwrite an assets image in Flutter having a source image?

好久不见. 提交于 2021-02-11 00:27:18
问题 I'm fairly new to Dart and Flutter, and I'm having trouble to overwrite an existing assets image from a source image. My attempt: try { File localFile = File('assets/images/myImage.png'); localFile.writeAsBytesSync(originFile.readAsBytesSync()); catch (e) { log(e.toString()); } I get: [log] FileSystemException: Cannot open file, path = 'assets/images/myImage.png' (OS Error: No such file or directory, errno = 2) I did define the assets folder in pubspec.yaml : assets: - assets/images/ Ok, so I

How can I overwrite an assets image in Flutter having a source image?

大城市里の小女人 提交于 2021-02-11 00:26:14
问题 I'm fairly new to Dart and Flutter, and I'm having trouble to overwrite an existing assets image from a source image. My attempt: try { File localFile = File('assets/images/myImage.png'); localFile.writeAsBytesSync(originFile.readAsBytesSync()); catch (e) { log(e.toString()); } I get: [log] FileSystemException: Cannot open file, path = 'assets/images/myImage.png' (OS Error: No such file or directory, errno = 2) I did define the assets folder in pubspec.yaml : assets: - assets/images/ Ok, so I

How can I overwrite an assets image in Flutter having a source image?

。_饼干妹妹 提交于 2021-02-11 00:25:13
问题 I'm fairly new to Dart and Flutter, and I'm having trouble to overwrite an existing assets image from a source image. My attempt: try { File localFile = File('assets/images/myImage.png'); localFile.writeAsBytesSync(originFile.readAsBytesSync()); catch (e) { log(e.toString()); } I get: [log] FileSystemException: Cannot open file, path = 'assets/images/myImage.png' (OS Error: No such file or directory, errno = 2) I did define the assets folder in pubspec.yaml : assets: - assets/images/ Ok, so I