Question

请求网址：$scope.sid="scpid"; $scope.sid = [{todoText:'Clean House', done:false}]; $scope.todoAdd = function() { var todoInput =""; $scope.sid.push({todoText:$scope.todoInput, done:false}); $scope.todoInput = ""; console.log(todoInput); }; $scope.remove = function() { var oldList = $scope.sid; $scope.sid = []; angular.forEach(oldList, function(x) { if (!x.done) $scope.sid.push(x); }); };
请求方法：POST
状态代码：200 OK
我在我的网络上遇到此错误，有人可以帮我解决这个问题这是我的功能

function filter2ndSearch(sid, bname, bcode, gnder, ptype, pname, size_oz, size_ml, size_g, size_pcs){
    return $http({
        method: 'post',
        url: rootURL + '/2ndlayer/search/' + sid + '/' + bname + '/' + bcode + '/' + gnder + '/' + ptype + '/' + pname ,        
        headers: {
            'Content-Type': 'application/x-www-form-urlencoded'
        }
    });
}

我的服务

<application
        android:allowBackup="true"
        android:icon="@mipmap/ic_launcher"
        android:label="@string/app_name"
        android:supportsRtl="true"
        android:theme="@style/AppTheme">

Answer 1

我知道你的sid对象结构，并在你的URL中添加它，bname的情况也是如此，这里bname的属性是什么？

 url: rootURL +    rootURL + '/2ndlayer/search/' + sid.todoInput + '/' + bname + '/' + bcode + '/' + gnder + '/' + ptype + '/' + pname ,

Answer 2

Angular将服务作为httpParamSerializerJQLike提供，因为1.4发送了内容类型＆＃39;应用程序/ x-www-form-urlencoded＆＃39;的数据。你可以这样做 -

$http({
url: 'url',
method: 'POST',
data: $httpParamSerializerJQLike(data), // Make sure to inject the 
service in to the controller
headers: {
'Content-Type': 'application/x-www-form-urlencoded' // header
}
}).then(function(response) { /* do something here */ });

要详细了解它，您可以参考 - URL-Encoding

请求[object％20Object]

2 个答案: