换句话说,条形码需要布局如下：

[start char][barcode][checksum][stop char]

code128的好处是它比9中的3个简洁得多.

This page帮助我计算算法来计算我的校验和.

该算法的一般概述是：

>条形码的每个字符都会获得分配给它的特定值
取决于角色是什么以及它在哪里
条码.
>上面1)中的所有值都加在一起.
>获取上面2)中总数的模数103值.
>在大多数情况下,校验和字符将是以下ASCII码:(模数值
加上32)如上文3)中所确定的.

有一些细微差别,我最终需要在所有事情的javascript中创建这个算法(没有问题).对于我自己将来的参考和显示一些细微差别这是它的样子：

/*
 * This is the variable part of my barcode,I append this to a 
 * static prefix later,but I need to perform logic to compute the 
 * checksum for this variable. There is logic earlier that enforces 
 * this variable as a 9 character string containing only digits.   
 */ 
var formIncrement = // a 9 char "digit" string variable

/*
 * Dynamically compute the total checksum value (before modulus) 
 * for the variable part of my barcode,I will need to get a modulus 
 * from this total when I am done. If you need a variable number of 
 * characters in your barcodes or if they are not all digits 
 * obvIoUsly something different would have to be done here.  
 */ 
var incrementCS = ((parseInt(formIncrement.charAt(0)) + 16) * 7) +
                  ((parseInt(formIncrement.charAt(1)) + 16) * 8) +
                  ((parseInt(formIncrement.charAt(2)) + 16) * 9) +
                  ((parseInt(formIncrement.charAt(3)) + 16) * 10) +
                  ((parseInt(formIncrement.charAt(4)) + 16) * 11) +
                  ((parseInt(formIncrement.charAt(5)) + 16) * 12) +
                  ((parseInt(formIncrement.charAt(6)) + 16) * 13) +
                  ((parseInt(formIncrement.charAt(7)) + 16) * 14) + 
                  ((parseInt(formIncrement.charAt(8)) + 16) * 15);

/*
 * 452 is the total checksum for my barcodes static prefix (600001),* so it doesn't need to be computed dynamically,I just add it to 
 * the variable checksum total determined above and then get the 
 * modulus of that sum:  
 */ 
var checksum = (452 + incrementCS) % 103


var barcode = "š600001" + formIncrement

/*
 * The 0 and the 95 - 102 cases had to be defined explicitly because 
 * their checksum figures do not line up with the javascript char 
 * codes for some reason (see the code 128 definition table in the 
 * linked page) otherwise we simply need to get the charCode of the 
 * checksum + 32. I also tack on the stop char here. 
 */ 
switch (checksum) {
    case 0 :
    barcode += "€œ";
    break;
    case 95 :
    barcode += "‘œ";
    break;
    case 96 :
    barcode += "’œ";
    break;
    case 97 :
    barcode += "“œ";
    break;
    case 98 :
    barcode += "”œ";
    break;
    case 99 :
    barcode += "•œ";
    break;
    case 100 :
    barcode += "–œ";
    break;
    case 101 :
    barcode += "—œ";
    break;
    case 102 :
    barcode += "˜œ";
    break;
    default :
    barcode += String.fromCharCode(checksum + 32) + "œ";
}

return barcode;

您可能会注意到我的示例中的开始和停止字符(š,œ)似乎与链接页面上显示的字符不匹配.如果我记得我认为这是因为我有一些非标准的code128字体,这些字符转换为正确的字符.

编辑

我在文档中查了回来.看起来我从right here获得了字体.使用上面的算法专门使用该字体,我只是做了一个code128b条形码进行测试,它出来了štestwœ,它扫描得很好.你的校验和算法似乎工作正常,因为我们都有w但看起来如果你得到你的开始和停止代码：ÌtestwÎ.

我指出要查找我正在使用的相同条形码字体,因为我觉得不同品牌的code128字体可能会实现不同的字符来表示开始和停止条形码.