Question

我如何填充第二个选择元素？我已经弄清楚如何做第一个。但是我怎么能在第二次做同样的事情，这取决于选择哪个“Make”？我试图通过它自己说话，同时采取一些小步骤，但我认为这对我来说可能太先进了。

var cars = '{"USED":[{"name":"Acura","value":"20001","models":[{"name":"CL","value":"20773"},{"name":"ILX","value":"47843"},{"name":"ILX Hybrid","value":"48964"},{"name":"Integra","value":"21266"},{"name":"Legend","value":"21380"},{"name":"MDX","value":"21422"},{"name":"NSX","value":"21685"},{"name":"RDX","value":"21831"},{"name":"RL","value":"21782"},{"name":"RSX","value":"21784"},{"name":"SLX","value":"21879"},{"name":"TL","value":"22237"},{"name":"TSX","value":"22248"},{"name":"Vigor","value":"22362"},{"name":"ZDX","value":"32888"}]},{"name":"Alfa Romeo","value":"20047","models":[{"name":"164","value":"20325"},{"name":"8c Competizione","value":"34963"},{"name":"Spider","value":"22172"}]}';
var carobj = eval ("(" + cars + ")");
var select = document.getElementsByTagName('select')[0];

//print array elements out
for (var i = 0; i < carobj.USED.length; i++) {
    var d = carobj.USED[i];
    select.options.add(new Option(d.name, i))
};

Answer 1

如果我正确地阅读了您的问题，您希望在第一个选择中使用make的模型填充第二个选择。请参阅下面的纯JS方法（使用jsfiddle）。如果可能的话，我建议调查jQuery，因为我更喜欢jQuery解决方案。

http://jsfiddle.net/m5U8r/1/

var carobj;

window.onload = function () {
    var cars = '{"USED":[{"name":"Acura","value":"20001","models":[{"name":"CL","value":"20773"},{"name":"ILX","value":"47843"},{"name":"ILX Hybrid","value":"48964"},{"name":"Integra","value":"21266"},{"name":"Legend","value":"21380"},{"name":"MDX","value":"21422"},{"name":"NSX","value":"21685"},{"name":"RDX","value":"21831"},{"name":"RL","value":"21782"},{"name":"RSX","value":"21784"},{"name":"SLX","value":"21879"},{"name":"TL","value":"22237"},{"name":"TSX","value":"22248"},{"name":"Vigor","value":"22362"},{"name":"ZDX","value":"32888"}]},{"name":"Alfa Romeo","value":"20047","models":[{"name":"164","value":"20325"},{"name":"8c Competizione","value":"34963"},    {"name":"Spider","value":"22172"}]}]}';

    carobj = eval ("(" + cars + ")");
    var makes = document.getElementById('make');

    for (var i = 0; i < carobj.USED.length; i++) {
        var d = carobj.USED[i];
        makes.options.add(new Option(d.name, i));
    }

    makes.onchange = getModels;
    getModels();
}

// add models based on make
function getModels () {
    var makes = document.getElementById('make');
    var make = makes.options[makes.selectedIndex].text;
    for (var i = 0; i < carobj.USED.length; i++) {
        if (carobj.USED[i].name == make) {
            var models = document.getElementById('model');
            models.options.length = 0;
            for (var j= 0; j < carobj.USED[i].models.length; j++) {
                var model = carobj.USED[i].models[j];                
                models.options.add(new Option(model.name, j));
            }
            break;            
        }
    }
}

我还建议您研究更安全的JSON解析。如果在任何用户输入上运行，则使用eval存在安全风险。您可以查看JSON.org及其json2.js。或者如果你想使用jQuery：parseJSON。以下是jQuery版本：

jQuery.parseJSON(jsonString);

来自Safely turning a JSON string into an object的JSON解析提示。

基于json填充select元素

1 个答案: