想找出办法希望这里有人能帮上忙。尝试创建一个JavaScipt,它可以捕获'promoCode‘参数,而不需要参数名和后面的所有垃圾(参数值之后的其他值),然后将结果存储在sessionStorage中。
下面是一个URL示例:
https://www.constellation.com/solutions/for-your-home/residential-signup.html?zip=77493&promoCode=rockstarcontent/constellation/en/campaigns/DisplayNew2020.html%3futm_source=criteo这就是我想要的结果:“摇滚明星”
注意,我不想要这个部分的'content/constellation/en/campaigns/DisplayNew2020.html%3futm_source=criteo‘,而且在'promoCode’值之前也不需要任何东西。
(非常感谢:)
发布于 2022-10-12 17:31:52
没有regex。如果您知道它总是位于相同的位置(本例中的第一个位置),则可以使用以下方法:
const url = 'https://www.constellation.com/solutions/for-your-home/residential-signup.html?zip=77493&promoCode=rockstarcontent/constellation/en/campaigns/DisplayNew2020.html%3futm_source=criteo'
const n = new URLSearchParams(url);
const o = Object.fromEntries(n.entries());
const w = o.promoCode.split("/")[0]
const s = w.replace("content","");
console.log(s)
发布于 2022-10-12 19:04:48
var url = window.location.search;
var n = new URLSearchParams(url);
var o = Object.fromEntries(n.entries());
var w = o.promoCode.split("/")[0]
var s = w.replace("content","");
var s = w.replace("state","");
sessionStorage.setItem("promo_code", (s));
console.log(s)https://stackoverflow.com/questions/74045752
复制相似问题