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web.py templator传递的JSON在javascript中未正确解析_Javascript_Json_Web.py - Fatal编程技术网

web.py templator传递的JSON在javascript中未正确解析

web.py templator传递的JSON在javascript中未正确解析,javascript,json,web.py,Javascript,Json,Web.py,我正在使用web.py和templetor,并将一个充满JSON的变量传递给嵌入在templetor文件中的javascript。这些JSON将由D3消耗 Python Code for row in rs: #z['time'] = row.gps z[

我正在使用web.py和templetor,并将一个充满JSON的变量传递给嵌入在templetor文件中的javascript。这些JSON将由D3消耗

 Python Code
           for row in rs:
                    #z['time'] = row.gps                                                                      
                    z['year'] = count + 2000
                    z['value'] = row.br
                    allrows.append(z)
                    count = count+1
                    if count > 5: break;
            p = json.dumps(allrows)
            return render.index(p)
现在应该在templetor文件中使用:

$def with (data)
<!DOCTYPE html>
<body>
    <meta charset="utf-8">
    <script type="text/javascript" src="http://d3js.org/d3.v3.min.js"></script>

<div id="bar-demo"></div>

<script type="text/javascript" id="bar-demo-js">
var data = JSON.parse($data)
$def带(数据)
var data=JSON.parse($data)
这似乎遇到了错误——(根据firefox) SyntaxError:无效的属性id [在此错误上中断]


var data=JSON.parse([{“value”:151.479999999999,“year”:…

您不必调用JSON.parse,因为它已经是对象,而不是JavaScript中的字符串。只要var data=$data就足够了