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Read lines from a file into a Bash array [duplicate]
I am trying to read a file containing lines into a Bash array.
6 Answers
6
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Loaded nib but the 'view' outlet was not set
I added a new nib file to my project, and tried to load it.
32 Answers
32
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Get real path from URI, Android KitKat new storage access framework [duplicate]
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column, sel, new String[]{ id }, null);
String filePath = "";
int columnIndex = cursor.getColumnIndex(column[0]);
if (cursor.moveToFirst()) {
filePath = cursor.getString(columnIndex);
}
cursor.close();
Reference: I'm not able to find the post that this solutio...
String replacement in batch file
We can replace strings in a batch file using the following command
4 Answers
4
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Read and parse a Json File in C#
...affing" about with code samples and etc., trying to read a very large JSON file into an array in c# so I can later split it up into a 2d array for processing.
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Set environment variables from file of key/value pairs
TL;DR: How do I export a set of key/value pairs from a text file into the shell environment?
33 Answers
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Can I have an IF block in DOS batch file?
In a DOS batch file we can only have 1 line if statement body? I think I found somewhere that I could use () for an if block just like the {} used in C-like programming languages, but it is not executing the statements when I try this. No error message either. This my code:
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List all files in one directory PHP [duplicate]
What would be the best way to list all the files in one directory with PHP? Is there a $_SERVER function to do this? I would like to list all the files in the usernames/ directory and loop over that result with a link, so that I can just click the hyperlink of the filename to get there. Thanks!
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How do I convert this list of dictionaries to a csv file?
...keys = toCSV[0].keys()
with open('people.csv', 'w', newline='') as output_file:
dict_writer = csv.DictWriter(output_file, keys)
dict_writer.writeheader()
dict_writer.writerows(toCSV)
EDIT: My prior solution doesn't handle the order. As noted by Wilduck, DictWriter is more appropriate ...
How can I get file extensions with JavaScript?
...s is the accepted answer; wallacer's answer is indeed much better:
return filename.split('.').pop();
My old answer:
return /[^.]+$/.exec(filename);
Should do it.
Edit: In response to PhiLho's comment, use something like:
return (/[.]/.exec(filename)) ? /[^.]+$/.exec(filename) : undefined;
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