我需要从一些文本生成PDF417条码。我有一个API(我没有创建),该API在给定数据,行数和列数(以及与问题无关的其他参数)的情况下生成PDF417条形码。
我的PDF417条码使用文本编码。这意味着1个代码字最多可容纳2个字符。现在,由于我要在非常有限的空间中打印此条形码,因此必须固定列数。
以下是我从本文档中得出的结论(请参阅第38页-调整条形码大小):
当我测试上述算法时,什么也没有显示。当数据很少且行数= 25时使用相同的API时,条形码可以很好地打印(已通过各种条形码扫描仪验证)。
那么,当列数已知时,如何计算某些给定文本所需的行数?
您可以查看一些PDF417实现的源代码,例如ZXing。
文本编码不仅是每个代码字两个字符。如果您使用大写字母和空格以外的任何其他字符,则编码器将添加额外的字符以切换字符集等。您实际上必须对文本进行编码,以查看它将变成多少个代码字。
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 "\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; } }
输出:
"Hello, world!" requires 9 code-words, and 2 error correction code-words. This becomes 2 rows.