我需要从一些文本生成PDF417条形码。我有一个API(不是我创建的)，它根据数据、行数和列数(以及与问题无关的其他参数)生成PDF417条形码。
我的PDF417条形码使用文本编码。这意味着一个码字最多可以容纳2个字符。现在，列的数量必须固定，因为我在一个非常有限的空间打印这个条形码。
以下是我从这份文件中推断出来的内容(参考第38页-条形码的大小):
- 让每行码字数，CWPerRow = 7.
- 给定文本所需的码字数，ReqCW = strlen(text)/2.
- 所需行数= ReqCW/CWPerRow
那么，当列数已知时，如何计算某些给定文本所需的行数呢?
您可以查看一些PDF417实现的源代码，例如ZXing。
文本编码不是每个码字只有两个字符。如果您使用大写字母和空格以外的任何字符，编码器将添加额外的字符来切换字符集等。你必须对文本进行编码，看看它会变成多少个码字。
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 "tr #$%&*+,-./0123456789:=^".indexOf(ch) > -1;
}
private static boolean isPunctuation(char ch) {
return "tnr!"$'()*,-./:;<>?@[\]_`{|}~".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;
}
}
"Hello, world!" requires 9 code-words, and 2 error correction code-words. This becomes 2 rows.
我把Markus Jarderot的回答做了一个Python移植。计算保持不变。
import string
SUBMODE_ALPHA = string.ascii_uppercase + ' '
SUBMODE_LOWER = string.ascii_lowercase + ' '
SUBMODE_MIXED = "tr #$%&*+,-./0123456789:=^"
SUBMODE_PUNCTUATION = "tnr!"$'()*,-./:;<>?@[\]_`{|}~"
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()