Question

我无法弄清楚我错在哪里我一直在尝试在SO上找到这么多的可能性和答案，但它并没有给我自定义属性的价值＆＃34; attr-pid＆＃ 34;

<div class="prod-col et_pb_column et_pb_column_4_4" style="">

<div attr-pid="2" class="bbb_prov product_select et_pb_column et_pb_column_2_4 model_container">
<div class="icon" attr-pid="2"><img src=" " style=" max-width: 100px;"</div></div>

<div attr-pid="4" class="bbb_prov product_select et_pb_column et_pb_column_2_4 model_container">
<div class="icon" attr-pid="4"><img src=" " style=" max-width: 100px;"></div></div>

</div>

我有这个

aProvider = $(".bbb_prov");

if ( me.savedJSON.providersToVisit == undefined ) 
{
    me.savedJSON.providersToVisit = [];
}

for( var y = 0 ; y < aProvider.length ; y++ ) 
{
    me.savedJSON.providersToVisit.push("&bxb_provider_id="+$(aProvider[y]).attr('attr-pid'));   
}

CEF_SaveDataJSON( me.savedJSON );   

console.log(JSON.stringify(me.savedJSON, null, 2));

这适用于我的另一个div，我只有一个类，但是这里有多个，console.log只显示我：

  "providersToVisit": []

Answer 1

小提琴http://jsbin.com/kutudogopo/edit?html,js,console

试试这个。既然你只是循环遍历它们，你可以使用JQuery的.each方法来遍历它们。

HTML：

  <div class="prod-col et_pb_column et_pb_column_4_4" style="">

<div attr-pid="2" class="bbb_prov product_select et_pb_column et_pb_column_2_4 model_container">
<div class="icon" attr-pid="2"><img src=" " style=" max-width: 100px;"</div></div>

<div attr-pid="4" class="bbb_prov product_select et_pb_column et_pb_column_2_4 model_container">
<div class="icon" attr-pid="4"><img src=" " style=" max-width: 100px;"></div></div>

JQuery的：

var aProvider = $(".bbb_prov");
console.log(aProvider.length);

$.each(aProvider, function(){
  var $this = $(this);
  console.log($this.attr('attr-pid'));

});

Answer 2

代码按提供的方式工作（请参阅此小提琴：https://jsfiddle.net/ta3sgh8e/）

HTML

<div class="prod-col et_pb_column et_pb_column_4_4" style="">
<div attr-pid="2" class="bbb_prov product_select et_pb_column et_pb_column_2_4 model_container">
<div class="icon" attr-pid="2"><img src=" " style=" max-width: 100px;"></div></div>
<div attr-pid="4" class="bbb_prov product_select et_pb_column et_pb_column_2_4 model_container">
<div class="icon" attr-pid="4"><img src=" " style=" max-width: 100px;"></div></div>
</div>

JS

function CEF_SaveDataJSON(data){
  //What happens in this function?
    //data.providersToVisit = [];
}
me = {savedJSON:{}};
aProvider = $(".bbb_prov");
if ( me.savedJSON.providersToVisit == undefined ) 
{
    me.savedJSON.providersToVisit = [];
}
for( var y = 0 ; y < aProvider.length ; y++ ) 
{
  me.savedJSON.providersToVisit.push("&bxb_provider_id="+$(aProvider[y]).attr('attr-pid'));   
}
CEF_SaveDataJSON(me.savedJSON);
console.log(JSON.stringify(me.savedJSON, null, 2));

我怀疑是在您的控制台日志之前在CEF_SaveDataJSON函数内修改了数据，除非有其他恶作剧，我们无法看到您从示例中排除了哪些内容。

Answer 3

是的它可以正常工作，所以我发现了我的错误而且我没有发布整个代码，我真的相信问题出在我发布的代码上。所以我刚刚删除了＆＃34;如果url ===＆＃34;

this.onPageLoad = function () 
{
    var url = window.location.href;
    var aModels;
    var aProvider;

    //iphone
    if ( url == "http://blablabla.com/apple/?menuf_id=1" ) //This is were i screwed up and had to remove it.
    {   
        aModels = $(".bbb_prov");


        if ( me.savedJSON.modelsToVisit == undefined ) {
            me.savedJSON.modelsToVisit = [];
        }

        if ( me.savedJSON.phoneModel == undefined ) {
            me.savedJSON.phoneModel = [];
        }

        for( var i = 0 ; i < aModels.length ; i++ ) 
        {
            me.savedJSON.modelsToVisit.push("http://blablabla.com/apple/?menuf_id=1&bxb_model_id="+$(aModels[i]).attr('attr-mid')); 
        }

        for( var x = 0 ; x < aModels.length ; x++ ) 
        {
            me.savedJSON.phoneModel.push($(aModels[x]).text());
        }

        CEF_SaveDataJSON( me.savedJSON );   


        me.blablabla_GotoNextModel();

        aProvider = $(".bbb_prov");

        if ( me.savedJSON.providersToVisit == undefined ) {
            me.savedJSON.providersToVisit = [];
        }


        for( var y = 0 ; y < aProvider.length ; y++ ) 
        {
            me.savedJSON.providersToVisit.push("&bxb_provider_id="+$(aProvider[y]).attr('attr-pid'));   
        }

        CEF_SaveDataJSON( me.savedJSON );   

        console.log(JSON.stringify(me.savedJSON, null, 3));
    }

}

javascript json jquery无法获取自定义属性值

3 个答案: