I need to generate a PDF417 barcode from some text. I have an API (that I didn't create) that generates a PDF417 barcode given the data, number of rows and number of columns (among other parameters irrelevant to the question).
My PDF417 barcode uses text encoding. This means 1 codeword can hold up to 2 characters. Now, the number of columns HAS to be fixed because I'm printing this barcode in a very constrained space.
The following is what I have inferred from this document (Refer page 38 - Sizing a barcode):
- Let number of codewords per row, CWPerRow = 7.
- Number of codewords required for some given text, ReqCW = strlen(text) / 2.
- Number of rows required = ReqCW / CWPerRow
When I test the above algorithm, nothing displays. When I use the same API when the data is very small and the number of rows = 25, the barcode prints just fine (verified by various barcode scanners).
So, how do I calculate the number of rows required for some given text when the number of columns is known?
You could look at the source-code of some PDF417 implementation, such as ZXing.
The text encoding isn't just two characters per code-word. If you use any other character than uppercase letters and space, the encoder will add extra characters to switch character-sets etc. You really have to encode the text to see how many code-words it will become.
public class Test
{
public static void main(String[] args)
{
String msg = "Hello, world!";
int columns = 7;
int sourceCodeWords = calculateSourceCodeWords(msg);
int errorCorrectionCodeWords = getErrorCorrectionCodewordCount(0);
int rows = calculateNumberOfRows(sourceCodeWords, errorCorrectionCodeWords, columns);
System.out.printf("\"%s\" requires %d code-words, and %d error correction code-words. This becomes %d rows.%n",
msg, sourceCodeWords, errorCorrectionCodeWords, rows);
}
public static int calculateNumberOfRows(int sourceCodeWords, int errorCorrectionCodeWords, int columns) {
int rows = ((sourceCodeWords + 1 + errorCorrectionCodeWords) / columns) + 1;
if (columns * rows >= (sourceCodeWords + 1 + errorCorrectionCodeWords + columns)) {
rows--;
}
return rows;
}
public static int getErrorCorrectionCodewordCount(int errorCorrectionLevel) {
if (errorCorrectionLevel < 0 || errorCorrectionLevel > 8) {
throw new IllegalArgumentException("Error correction level must be between 0 and 8!");
}
return 1 << (errorCorrectionLevel + 1);
}
private static boolean isAlphaUpper(char ch) {
return ch == ' ' || (ch >= 'A' && ch <= 'Z');
}
private static boolean isAlphaLower(char ch) {
return ch == ' ' || (ch >= 'a' && ch <= 'z');
}
private static boolean isMixed(char ch) {
return "\t\r #$%&*+,-./0123456789:=^".indexOf(ch) > -1;
}
private static boolean isPunctuation(char ch) {
return "\t\n\r!\"$'()*,-./:;<>?@[\\]_`{|}~".indexOf(ch) > -1;
}
private static final int SUBMODE_ALPHA = 0;
private static final int SUBMODE_LOWER = 1;
private static final int SUBMODE_MIXED = 2;
private static final int SUBMODE_PUNCTUATION = 3;
public static int calculateSourceCodeWords(String msg)
{
int len = 0;
int submode = SUBMODE_ALPHA;
int msgLength = msg.length();
for (int idx = 0; idx < msgLength;)
{
char ch = msg.charAt(idx);
switch (submode)
{
case SUBMODE_ALPHA:
if (isAlphaUpper(ch))
{
len++;
}
else
{
if (isAlphaLower(ch))
{
submode = SUBMODE_LOWER;
len++;
continue;
}
else if (isMixed(ch))
{
submode = SUBMODE_MIXED;
len++;
continue;
}
else
{
len += 2;
break;
}
}
break;
case SUBMODE_LOWER:
if (isAlphaLower(ch))
{
len++;
}
else
{
if (isAlphaUpper(ch))
{
len += 2;
break;
}
else if (isMixed(ch))
{
submode = SUBMODE_MIXED;
len++;
continue;
}
else
{
len += 2;
break;
}
}
break;
case SUBMODE_MIXED:
if (isMixed(ch))
{
len++;
}
else
{
if (isAlphaUpper(ch))
{
submode = SUBMODE_ALPHA;
len++;
continue;
}
else if (isAlphaLower(ch))
{
submode = SUBMODE_LOWER;
len++;
continue;
}
else
{
if (idx + 1 < msgLength)
{
char next = msg.charAt(idx + 1);
if (isPunctuation(next))
{
submode = SUBMODE_PUNCTUATION;
len++;
continue;
}
}
len += 2;
}
}
break;
default:
if (isPunctuation(ch))
{
len++;
}
else
{
submode = SUBMODE_ALPHA;
len++;
continue;
}
break;
}
idx++; // Don't increment if 'continue' was used.
}
return (len + 1) / 2;
}
}
Output:
"Hello, world!" requires 9 code-words, and 2 error correction code-words. This becomes 2 rows.
I've made a Python port of Markus Jarderot's answer. The calculation remains the same.
import string
SUBMODE_ALPHA = string.ascii_uppercase + ' '
SUBMODE_LOWER = string.ascii_lowercase + ' '
SUBMODE_MIXED = "\t\r #$%&*+,-./0123456789:=^"
SUBMODE_PUNCTUATION = "\t\n\r!\"$'()*,-./:;<>?@[\\]_`{|}~"
def calculateNumberOfRows(sourceCodeWords, errorCorrectionCodeWords, columns):
rows = ((sourceCodeWords + 1 + errorCorrectionCodeWords) / columns) + 1
if columns * rows >= sourceCodeWords + 1 + errorCorrectionCodeWords + columns:
rows -= 1
return rows
def getErrorCorrectionCodewordCount(errorCorrectionLevel):
if 0 > errorCorrectionLevel > 8:
raise ValueError("Error correction level must be between 0 and 8!")
return 1 << (errorCorrectionLevel + 1)
def calculateSourceCodeWords(msg):
length = 0;
submode = SUBMODE_ALPHA
msgLength = len(msg)
idx = 0
while(idx < msgLength):
ch = msg[idx]
length += 1
if not ch in submode:
old_submode = submode
if submode == SUBMODE_ALPHA:
for mode in (SUBMODE_LOWER, SUBMODE_MIXED):
if ch in mode:
submode = mode
elif submode == SUBMODE_LOWER:
if ch in SUBMODE_MIXED:
submode = SUBMODE_MIXED
elif submode == SUBMODE_MIXED:
for mode in (SUBMODE_ALPHA, SUBMODE_LOWER):
if ch in mode:
submode = mode
if idx + 1 < len(msg) and msg[idx + 1] in SUBMODE_PUNCTUATION:
submode = SUBMODE_PUNCTUATION
elif submode == SUBMODE_PUNCTUATION:
submode = SUBMODE_ALPHA
if old_submode != submode:
# submode changed
continue
length += 1
idx += 1 # Don't increment if 'continue' was used.
return (length + 1) / 2
def main():
msg = "Hello, world!"
columns = 7
sourceCodeWords = calculateSourceCodeWords(msg)
errorCorrectionCodeWords = getErrorCorrectionCodewordCount(0)
rows = calculateNumberOfRows(sourceCodeWords, errorCorrectionCodeWords, columns)
print("\"%s\" requires %d code-words, and %d error correction code-words. This becomes %d rows.\n"
%( msg, sourceCodeWords, errorCorrectionCodeWords, rows))
if __name__ == '__main__':
main()
来源：https://stackoverflow.com/questions/16898890/in-a-pdf417-barcode-where-number-of-columns-are-fixed-how-would-i-calculate-the