In haxe macro for every expression we can get it's position in form of http://api.haxe.org/haxe/macro/Position.html :
{
file:String, // filename - relative to source path
min:Int, // position of first character in file
max:Int // position of last character in file
}
I want to get line number and position in line for min and max variables.
I definitely can do this by opening the file
FileSystem.absolutePath(Context.resolvePath(posInfo.file));
and calculating line number, but haxe already does this, it's much better to get this info from compiler. Is it possible?
In the current versions of Haxe you can use PositionTools.toLocation
class Macro {
public static macro function log(args:Array<Expr>):Expr {
var loc = PositionTools.toLocation(Context.currentPos());
var locStr = loc.file + ":" + loc.range.start.line;
args.unshift(macro $v{locStr});
return macro SomeExtern.logFunc($a{args});
}
}
to have Macro.log("hi!") translate into SomeExtern.logFunc("Main:5", "hi!")
I know a few projects do that manually (like checkstyle)
Load the file content, find the carriage returns (\n, \r or \n\r) mark the character position for each new line, lookup your pos.min against those positions
I guess it might be more problematic if you have multibyte characters in the file ...
In haxe 4 PositionTools.toLocation function was added.
Related
I'm using AutoHotkey for this as the code is the most understandable to me. So I have a document with numbers and text, for example like this
120344 text text text
234000 text text
and the desired output is
12:03:44 text text text
23:40:00 text text
I'm sure StrReplace can be used to insert the colons in, but I'm not sure how to specify the position of the colons or ask AHK to 'find' specific strings of 6 digit numbers. Before, I would have highlighted the text I want to apply StrReplace to and then press a hotkey, but I was wondering if there is a more efficient way to do this that doesn't need my interaction. Even just pointing to the relevant functions I would need to look into to do this would be helpful! Thanks so much, I'm still very new to programming.
hfontanez's answer was very helpful in figuring out that for this problem, I had to use a loop and substring function. I'm sure there are much less messy ways to write this code, but this is the final version of what worked for my purposes:
Loop, read, C:\[location of input file]
{
{ If A_LoopReadLine = ;
Continue ; this part is to ignore the blank lines in the file
}
{
one := A_LoopReadLine
x := SubStr(one, 1, 2)
y := SubStr(one, 3, 2)
z := SubStr(one, 5)
two := x . ":" . y . ":" . z
FileAppend, %two%`r`n, C:\[location of output file]
}
}
return
Assuming that the "timestamp" component is always 6 characters long and always at the beginning of the string, this solution should work just fine.
String test = "012345 test test test";
test = test.substring(0, 2) + ":" + test.substring(2, 4) + ":" + test.substring(4, test.length());
This outputs 01:23:45 test test test
Why? Because you are temporarily creating a String object that it's two characters long and then you insert the colon before taking the next pair. Lastly, you append the rest of the String and assign it to whichever String variable you want. Remember, the substring method doesn't modify the String object you are calling the method on. This method returns a "new" String object. Therefore, the variable test is unmodified until the assignment operation kicks in at the end.
Alternatively, you can use a StringBuilder and append each component like this:
StringBuilder sbuff = new StringBuilder();
sbuff.append(test.substring(0,2));
sbuff.append(":");
sbuff.append(test.substring(2,4));
sbuff.append(":");
sbuff.append(test.substring(4,test.length()));
test = sbuff.toString();
You could also use a "fancy" loop to do this, but I think for something this simple, looping is just overkill. Oh, I almost forgot, this should work with both of your test strings because after the last colon insert, the code takes the substring from index position 4 all the way to the end of the string indiscriminately.
I am trying to write a script that will auto sort files based on the 7th and 8th digit in their name. I get the following error: "Argument must be a string scalar or character vector". Error is coming from line 16:
Argument must be a string scalar or character vector.
Error in sort_files (line 16)
movefile (filelist(i), DirOut)
Here's the code:
DirIn = 'C:\Folder\Experiment' %set incoming directory
DirOut = 'C:\Folder\Experiment\1'
eval(['filelist=dir(''' DirIn '/*.wav'')']) %get file list
for i = 1:length(filelist);
Filename = filelist(i).name
name = strsplit(Filename, '_');
newStr = extractBetween(name,7,8);
if strcmp(newStr,'01')
movefile (filelist(i), DirOut)
end
end
Also, I am trying to make the file folder conditional so that if the 10-11 digits are 02 the file goes to DirOut/02 etc.
First, try avoid using the eval function, it is pretty much dreaded as slow and hard to understand. Specially if you need to create variables. Instead do this:
filelist = dir(fullfile(DirIn,'*.wav'));
Second, the passage:
name = strsplit(Filename, '_');
Makes name a list, so you can access name{1} or possibly name{2}. Each of these are strings. But name isn't a string, it is a list. extractBetween requires a string as an input. That is why you are getting this problem. But note that you could have simply done:
newStr = name(7:8);
If name was a string, which in Matlab is a char array.
EDIT:
Since it has been now claimed that the error occurs on movefile (filelist(i), DirOut), the likely cause is because filelist(i) is a struct. Wheres a filena name (char array) should have been given at input. The solution should be replacing this line with:
movefile(fullfile(filelist(i).folder, filelist(i).name), DirOut)
Now, if you want to number the output folders too, you can do this:
movefile(fullfile(filelist(i).folder, filelist(i).name), [DirOut,filesep,name(7:8)])
This will move a file to /DirOut/01. If you wanted /DirOut/1, you could do this:
movefile(fullfile(filelist(i).folder, filelist(i).name), [DirOut,filesep,int2str(str2num(name(7:8)))])
I'm trying to parse a file that (apparently) ends its lines with carriage returns, but they aren't being matched as such in Swift, despite having the same UTF8 value. I can see possible fixes for the problem, but I'm curious as to what these characters actually are.
Here's some sample code, with the output below. (CR is set using Character("\r"), although I've tried it using "\r" as well.
try f.forEach() { c in
print(c, terminator:" ") // DBG
if (c == "\r") {
print("Carriage return found!")
}
print(String(c).utf8.first!, terminator:" ")//DBG
print(String(describing:pstate)) // DBG
...
case .field:
switch c {
case CR,LF :
self.endline()
pstate = .eol
When it reaches the end of line (which shows up as such in my text editors), I get this:
. 46 field
0 48 field
13 field
I 73 field
It doesn't seem to be matching using == or in the switch statement. Is there another approach I should be using for this character?
(I'll note that the parsing works fine with files that terminate in newlines.)
I determined what the problem was. By looking at c.unicodeScalars I discovered that the end of line character was in fact "\r\n", not just "\r". As seen in my code I was only taking the first when printing it out as UTF-8. I don't know if that's something from String.forEach or in the file itself.
I know that there are tests to determine if something is a newline. Swift 5 has them directly (c.isNewline), and there is also the CharacterSet approach as noted by Bill Nattaner.
I'm happier with something that will work in my switch statement (and thus I'll define each one explicitly), but that might change if I expect to deal with a wider variety of files.
I'm a little hazy as to what the f.forEach represents, but if your variable c is of type Character then you could replace your if statement with:
if "\(c)".rangeOfCharacter( from: CharacterSet.newlines ) != nil
{
print("Carriage return found!")
}
That way you won't have to invent a list of all-possible new line characters.
I'm trying to use iText PDFSweep RegexBasedCleanupStrategy to redact some words from pdf, however I only want to redact the word but not appear in other word, eg.
I want to redact "al" as single word, but I don't want to redact the "al" in "mineral".
So I add the word boundary("\b") in the Regex as parameter to RegexBasedCleanupStrategy,
new RegexBasedCleanupStrategy("\\bal\\b")
however the pdfAutoSweep.cleanUp not work if the word is at the end of line.
In short
The cause of this issue is that the routine that flattens the extracted text chunks into a single String for applying the regular expression does not insert any indicator for a line break. Thus, in that String the last letter from one line is immediately followed by the first letter of the next which hides the word boundary. One can fix the behavior by adding an appropriate character to the String in case of a line break.
The problematic code
The routine that flattens the extracted text chunks into a single String is CharacterRenderInfo.mapString(List<CharacterRenderInfo>) in the package com.itextpdf.kernel.pdf.canvas.parser.listener. In case of a merely horizontal gap this routine inserts a space character but in case of a vertical offset, i.e. a line break, it adds nothing extra to the StringBuilder in which the String representation is generated:
if (chunk.sameLine(lastChunk)) {
// we only insert a blank space if the trailing character of the previous string wasn't a space, and the leading character of the current string isn't a space
if (chunk.getLocation().isAtWordBoundary(lastChunk.getLocation()) && !chunk.getText().startsWith(" ") && !chunk.getText().endsWith(" ")) {
sb.append(' ');
}
indexMap.put(sb.length(), i);
sb.append(chunk.getText());
} else {
indexMap.put(sb.length(), i);
sb.append(chunk.getText());
}
A possible fix
One can extend the code above to insert a newline character in case of a line break:
if (chunk.sameLine(lastChunk)) {
// we only insert a blank space if the trailing character of the previous string wasn't a space, and the leading character of the current string isn't a space
if (chunk.getLocation().isAtWordBoundary(lastChunk.getLocation()) && !chunk.getText().startsWith(" ") && !chunk.getText().endsWith(" ")) {
sb.append(' ');
}
indexMap.put(sb.length(), i);
sb.append(chunk.getText());
} else {
sb.append('\n');
indexMap.put(sb.length(), i);
sb.append(chunk.getText());
}
This CharacterRenderInfo.mapString method is only called from the RegexBasedLocationExtractionStrategy method getResultantLocations() (package com.itextpdf.kernel.pdf.canvas.parser.listener), and only for the task mentioned, i.e. applying the regular expression in question. Thus, enabling it to properly allow recognition of word boundaries should not break anything but indeed should be considered a fix.
One merely might consider adding a different character for a line break, e.g. a plain space ' ' if one does not want to treat vertical gaps any different than horizontal ones. For a general fix one might, therefore, consider making this character a settable property of the strategy.
Versions
I tested with iText 7.1.4-SNAPSHOT and PDFSweep 2.0.3-SNAPSHOT.
Instead of:
stringstream szBuffer;
szBuffer>>string;
myFunc(string);
How do I do like:
muFunc(szBuffer.NextString());
I dont want to create a temp var just for passing it to a function.
If you want to read the whole string in:
// .str() returns a string with the contents of szBuffer
muFunc(szBuffer.str());
// Once you've taken the string out, clear it
szBuffer.str("");
If you want to extract the next line (up to the next \n character), use istream::getline:
// There are better ways to do this, but for the purposes of this
// demonstration we'll assume the lines aren't longer than 255 bytes
char buf[ 256 ];
szBuffer.getline(buf, sizeof(buf));
muFunc(buf);
getline() can also take in a delimiter as a second parameter (\n by default), so you can read it word by word.