2010-04-27 54 views
0

我试图检索一些JSON来进入一个海军报曲线图。我知道json是正确的,因为我硬编码它检查,但我很确定我没有通过,因为它没有显示出来。 这里的JavaScript:问题传递到JSON jquery的曲线图(海军报)

var total = $.ajax({ 
    type: "POST", 
    async: false, 
    url: "../api/?key=xxx&api=report&crud=return_months&format=json" 
}).responseText; 
//var total = $.evalJSON(total); 
var plot = $.plot($("#placeholder"),total); 

这里的JSON:

[ { data: [[1,12], [2,43], [3,10], [4,17], ], label: "E-File"}, { data: [[1,25], [2,35], [3,3], [4,5], ], label: "Bank Products" }, { data: [[1,41], [2,87], [3,30], [4,29], ], label: "All Returns" } ], {series: {lines: { show: true },points: { show: true }}, grid: { hoverable: true, clickable: true }, yaxis: { min: 0, max: 100 }, xaxis: { ticks: [[1,"January"],[2,"February"],[3,"March"],[4,"April"],[5,"May"],[6,"June"],[7,"July"],[8,"August"],[9,"September"],[10,"October"],[11,"November"],[12,"December"]] }} 

回答

1

务必将您的dataType: "json"选项也是如此。顺便说一句,你能做到这一点的success回调函数以及不锁定用户界面,同时等待响应,像这样:

$.ajax({ 
    type: "POST", 
    dataType: "json", 
    url: "../api/?key=xxx&api=report&crud=return_months&format=json", 
    success: function(total) { 
    var plot = $.plot($("#placeholder"),total); 
    //do more work if needed 
    } 
}); 

或者,使用$.post()做同样在较短的形式,就像这样:

$.post("../api/?key=xxx&api=report&crud=return_months&format=json", 
    function(total) { 
    var plot = $.plot($("#placeholder"),total); 
    //do work 
    }, "json"); 
+0

唉太多的常见错误。 谢谢尼克:-) – 2010-04-27 02:06:28

+0

其实我只是尝试过两种解决方案,都没有工作 – 2010-04-27 02:32:33

+0

@亚当 - 澄清一下?不成功的'成功'回调没有运行,这是错误的,其他? – 2010-04-27 02:35:10