我试图用这种方式生成随机的ean-8条形码。我已经生成了10000000到99999999之间的随机数,为ean-8代码生成随机的8位数字。它给我一个错误。
Exception in thread "main" java.lang.IllegalArgumentException: Checksum is bad (1). Expected: 7
at org.krysalis.barcode4j.impl.upcean.EAN8LogicImpl.handleChecksum(EAN8LogicImpl.java:85)
at org.krysalis.barcode4j.impl.upcean.EAN8LogicImpl.generateBarcodeLogic(EAN8LogicImpl.java:102)
at org.krysalis.barcode4j.impl.upcean.UPCEANBean.generateBarcode(UPCEANBean.java:93)
at org.krysalis.barcode4j.impl.ConfigurableBarcodeGenerator.generateBarcode(ConfigurableBarcodeGenerator.java:174)
at barcode2.BARCODE2.main(BARCODE2.java:42)
Java Result: 1
这里是代码。
import java.awt.image.BufferedImage;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.OutputStream;
import java.util.Random;
import org.apache.avalon.framework.configuration.Configuration;
import org.apache.avalon.framework.configuration.ConfigurationException;
import org.apache.avalon.framework.configuration.DefaultConfiguration;
import org.krysalis.barcode4j.BarcodeException;
import org.krysalis.barcode4j.BarcodeGenerator;
import org.krysalis.barcode4j.BarcodeUtil;
import org.krysalis.barcode4j.output.bitmap.BitmapCanvasProvider;
public class BARCODE2 {
public static void main(String[] args) throws ConfigurationException, BarcodeException, IOException {
BarcodeUtil util = BarcodeUtil.getInstance();
BarcodeGenerator gen = util.createBarcodeGenerator(buildCfg("ean-8"));
OutputStream fout = new FileOutputStream("ean-8.jpg");
int resolution = 200;
BitmapCanvasProvider canvas = new BitmapCanvasProvider(
fout, "image/jpeg", resolution, BufferedImage.TYPE_BYTE_BINARY, false, 0);
int min = 10000000;
int max = 99999999;
Random r = new Random();
int randomnumber = r.nextInt(max - min + 1) + min;
String barcodecods = String.valueOf(randomnumber);
gen.generateBarcode(canvas, barcodecods);
canvas.finish();
}
private static Configuration buildCfg(String type) {
DefaultConfiguration cfg = new DefaultConfiguration("barcode");
//Bar code type
DefaultConfiguration child = new DefaultConfiguration(type);
cfg.addChild(child);
//Human readable text position
DefaultConfiguration attr = new DefaultConfiguration("human-readable");
DefaultConfiguration subAttr = new DefaultConfiguration("placement");
subAttr.setValue("bottom");
attr.addChild(subAttr);
child.addChild(attr);
return cfg;
}
}
但是当我将用于随机代码的字符串值替换为特定的8位数时,程序运行正常。我该怎么办?我哪里做错了?有没有其他方法可以为ean-8条形码生成生成随机8位数字?
答案 0 :(得分:3)
条形码不仅仅是简单的数字。整个数字包含一个校验位,它由算术过程从其他数字生成。因此,并非所有数字都是有效的条形码。
不同的条形码使用不同的校验位算法。您需要找出您正在使用的库所期望的算法,然后生成满足此要求的条形码。
因此,例如,如果条形码是8位数字,您将生成随机的7位数字并附加正确计算的第8位数字以生成有效的条形码。
注意:校验位是奇偶校验位的十进制等效值。它允许软件在大多数情况下检测代码是否被错误地读取。它并不完美,因为有些错误会产生相同的校验位,但它会大大降低误读的可能性。
答案 1 :(得分:0)
生成一个7位数的随机数,并使用以下方法添加校验位:
public static int checkdigit(String idWithoutCheckdigit) {
// allowable characters within identifier
String validChars = "0123456789ABCDEFGHIJKLMNOPQRSTUVYWXZ_";
// remove leading or trailing whitespace, convert to uppercase
idWithoutCheckdigit = idWithoutCheckdigit.trim().toUpperCase();
// this will be a running total
int sum = 0;
// loop through digits from right to left
for (int i = 0; i < idWithoutCheckdigit.length(); i++) {
// set ch to "current" character to be processed
char ch = idWithoutCheckdigit.charAt(idWithoutCheckdigit.length() - i - 1);
// throw exception for invalid characters
if (validChars.indexOf(ch) == -1)
throw new RuntimeException("\"" + ch + "\" is an invalid character");
// our "digit" is calculated using ASCII value - 48
int digit = ch - 48;
// weight will be the current digit's contribution to
// the running total
int weight;
if (i % 2 == 0) {
// for alternating digits starting with the rightmost, we
// use our formula this is the same as multiplying x 2 and
// adding digits together for values 0 to 9. Using the
// following formula allows us to gracefully calculate a
// weight for non-numeric "digits" as well (from their
// ASCII value - 48).
weight = (2 * digit) - (digit / 5) * 9;
} else {
// even-positioned digits just contribute their ascii
// value minus 48
weight = digit;
}
// keep a running total of weights
sum += weight;
}
// avoid sum less than 10 (if characters below "0" allowed,
// this could happen)
sum = Math.abs(sum) + 10;
// check digit is amount needed to reach next number
// divisible by ten
return (10 - (sum % 10)) % 10;
}