条形码代码 39 JavaScript 编码

我正在尝试通过修改一些D3 javascript来编码代码39条形码。

问题是我不明白代码 39 条形码是如何编码的。看起来他们使用 12 位二进制文件对字符进行编码。

如何将像"598649"这样的字符串解码为12位二进制等效项，最好使用D3.js？

characters = {
'100100100101': '$',
'100100101001': '/',
'100101001001': '+',
'100101011011': '-',
'100101101011': 'X',
'100101101101': '*',
'100110101011': 'V',
'100110101101': ' ',
'100110110101': 'Z',
'101001001001': '%',
'101001011011': '7',
'101001101011': '4',
'101001101101': '0',
'101010011011': 'G',
'101010110011': 'Q',
'101011001011': 'D',
'101011001101': 'J',
'101011010011': 'N',
'101011011001': 'T',
'101100101011': '2',
'101100101101': '9',
'101100110101': '6',
'101101001011': 'B',
'101101001101': 'I',
'101101010011': 'L',
'101101011001': 'S',
'101101100101': 'F',
'101101101001': 'P',
'110010101011': 'U',
'110010101101': '0',
'110010110101': 'Y',
'110011010101': 'W',
'110100101011': '1',
'110100101101': '8',
'110100110101': '5',
'110101001011': 'A',
'110101001101': 'H',
'110101010011': 'K',
'110101011001': 'R',
'110101100101': 'E',
'110101101001': 'O',
'110110010101': '3',
'110110100101': 'C',
'110110101001': 'M',
}

这是我到目前为止一直在使用的一些代码。以前的代码生成随机二进制文件，然后将键与值匹配。我想做相反的事情，所以我不确定我是否应该创建一个键/值颠倒的第二个数组。

    //var num= prompt("Enter num")
    //console.log(parseInt(num, 10).toString(2))
    //var decnum=892938
    //alert(decnum.toString(2))
    // var b = parseInt( a, 2 )


    // var str = '34566'
    // var found = str.match(/[01]{12}/g).map(['110100101101','101100101101', '101100101011', '101100101101', '110110010101', '110100101101'])
    //str.match(/[01]{12}/g).map(characters)
    //console.log(characters[v])
    //console.log(found)

    var str = ['110100101101','101100101101', '101100101011', '101100101101', '110110010101', '110100101101']
    //var barcode_array = str.match(/[01]{12}/g).map(Object.keys(characters))
    //console.log(barcode_array)
    //split into array
    var barcode_input = ['110100101101','101100101101', '101100101011', '101100101101', '110110010101', '110100101101']
    //map to binary characters
    //console.log(barcode_input.range(Object.keys(characters)))

这是小提琴https://jsfiddle.net/6szmwkgm/6/

代码修改自克里斯托弗·曼宁 http://bl.ocks.org/christophermanning/4324236作为参考，这是另一个代码 39 脚本 http://notionovus.com/blog/code-39-barcode-biscuits/

通过将 characters 对象用作简单的查找表来完成您正在构建的编码。获取每个字符并查找其线条模式，构建这些字符的数组，然后对其进行处理。

您

正在处理的代码比您似乎需要的要多得多。我用它作为起点，然后将其削减到最低限度。

请注意，我没有方便的扫描仪来测试它，所以如果它给您带来任何麻烦，请告诉我。

下面是一个注释的代码片段：

<!DOCTYPE html>
<html>
<head>
  <meta http-equiv="Content-Type" content="text/html;charset=utf-8">
  <script type="text/javascript" src="//cdnjs.cloudflare.com/ajax/libs/d3/3.5.5/d3.min.js"></script>
  <style type="text/css">
    body {
      display: block;
      margin: 0;
      background-color: #FFF;
    }
    
    text {
      font-family: monospace;
      font-size: 1.5em;
      text-anchor: middle;
    }
    
    path {
      fill-opacity: 0;
    }
  </style>
</head>
<body>
  <script type="text/javascript">
		var svg = d3.select("body")
		  .append("svg")
		  .attr("width", 300)
		  .attr("height", 300);
        // input barcode
		var barcode = "598649";
        // some configuration items
		var config = {
		  "height": 100,
		  "x": 10,
		  "margin": 20
		};
    
        // map each character to it's pattern
        // note I reversed it from the sample
        // to make lookups on each letter easier
		var characters = {
		  "0": "110010101101",
		  " ": "100110101101",
		  "1": "110100101011",
		  "2": "101100101011",
		  "3": "110110010101",
		  "4": "101001101011",
		  "5": "110100110101",
		  "6": "101100110101",
		  "7": "101001011011",
		  "8": "110100101101",
		  "9": "101100101101",
		  "$": "100100100101",
		  "%": "101001001001",
		  "*": "100101101101",
		  "+": "100101001001",
		  "-": "100101011011",
		  "/": "100100101001",
		  "A": "110101001011",
		  "B": "101101001011",
		  "C": "110110100101",
		  "D": "101011001011",
		  "E": "110101100101",
		  "F": "101101100101",
		  "G": "101010011011",
		  "H": "110101001101",
		  "I": "101101001101",
		  "J": "101011001101",
		  "K": "110101010011",
		  "L": "101101010011",
		  "M": "110110101001",
		  "N": "101011010011",
		  "O": "110101101001",
		  "P": "101101101001",
		  "Q": "101010110011",
		  "R": "110101011001",
		  "S": "101101011001",
		  "T": "101011011001",
		  "U": "110010101011",
		  "V": "100110101011",
		  "W": "110011010101",
		  "X": "100101101011",
		  "Y": "110010110101",
		  "Z": "100110110101"
		};
        // build you data array
		var data = [];
		// Code 39 barcodes start with a *
		data.push('100101101101');
		barcode.split("").forEach(function(d, i) {
          // for each character, append the patter to our array
		  data.push(characters[d]); //<-- look up for each character
		});
		// Code 39 barcodes end with a *
		data.push('100101101101');
		
        // set up line function
		var line = d3.svg.line()
		  .x(function(d) {
			return d.cx = d.x
		  })
		  .y(function(d) {
			return d.cy = d.y
		  })
		  
        // group all the barcode paths
		var g = svg.append('g')
		  .attr('class', 'barcode');
        // set up the data attributes necessary
		var path = g.selectAll('path')
		  .data(data.map(function(c) {
			  return c + '0' // put a space between each character
			}).join('').split('')
			.map(function(d, i) {
			  return [{
				c: d,
				x: config['margin']+ (i * 1.2),
				y: config['margin']
			  }, {
				c: d,
				x: config['margin'] + (i * 1.2),
				y: config['margin'] + config['height']
			  }]
			})
		  );
        
		path.enter()
		  .append('path');
        // draw the barcode
		path
		  .style("stroke", function(d) {
			return d[0].c == '1' ? '#000' : '#FFF'
		  })
		  .style("stroke-width", 1.2)
		  .attr("d", line);
		path.exit().remove();
		
        // add the text underneath
		var text = svg.selectAll('text')
		  .data([barcode])
		
		text.exit().remove();
		
		text.enter()
		  .append('text');
		
		text
		  .text(barcode)
		  .style('fill', 'black')
		  .attr('y', config.margin + config.height + 15)
      .attr('x', g.node().getBBox().width/2 + config.margin);
  </script>
</body>
</html>

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