JS获取url参数,JS发送json格式的POST请求方法
网络 2018-05-03 2092
1 | <script type="text/javascript"> |
一、获取url所有参数值
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | function US() { var name, value; var str = location.href; var num = str.indexOf("?"); str = str.substr(num + 1); var arr = str.split("&"); for (var i = 0; i < arr.length; i++) { num = arr[i].indexOf("="); if (num > 0) { name = arr[i].substring(0, num); value = arr[i].substr(num + 1); this[name] = value; } } } |
二、使用JS 发送JSON格式的POST请求
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | var us = new US(); var xhr = new XMLHttpRequest(); xhr.open("POST", "/searchguard/api/v1/auth/login", true); xhr.setRequestHeader("Content-type", "application/json"); xhr.setRequestHeader("kbn-version", "5.3.0"); xhr.onreadystatechange = function() { if (xhr.readyState == 4) { if (xhr.status == 200) { window.location.href = us.nextUrl; } } }; xhr.send(JSON.stringify({ "username" : us.u, "password" : us.p })); </script> |
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