[英]get values from json array
i trying to get the sku
from a url www.mbsmfg.co/shop/coles-grey/?format=json-pretty (json) and show the users those values which are available under the variants for that particular product 我试图从网址www.mbsmfg.co/shop/coles-grey/?format=json-pretty(json )获取
sku
,并向用户显示在该特定产品的变体下可用的值
eg: when a user visits this page www.mbsmfg.co/shop/coles-grey/ the user must be able to see the SKU in the product body.
例如:当用户访问该页面www.mbsmfg.co/shop/coles-grey/时 ,用户必须能够在产品正文中看到SKU。
Im using YUI for this and so far 我到目前为止一直在使用YUI
function ajaxRequest() {
Y.io('//www.mbsmfg.co/shop/coles-grey/?format=json-pretty', {
on: {
success: function (x, o) {
var parsedResponse;
try {
d = Y.JSON.parse(o.responseText);
} catch (e) {
console.log("JSON Parse failed!");
return;
}
for (var i = 0; i < d.items.length; i++) {
var htmlString = '<h1>' + d.items[i].sku + d.items[i].price '</h1>' + d.items[i].body;
Y.one('.product-sharing').append(htmlString);
alert(htmlString);
}
}
}
});
}
Y.use('node', function() {
Y.on('domready', function() {
ajaxRequest();
});
});
It doesn't alert or append the values to the div. 它不会警告或将值附加到div。
Can someone please tell me what am i doing wrong here? 有人可以告诉我我在做什么错吗?
I think you are looking into the wrong place. 我认为您正在寻找错误的地方。 This should give you the sku value.
这应该为您提供sku值。
d.item.structuredContent.variants[0].sku
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