Question

你能帮我理解为什么这会失败吗？ Response.data返回true，我将其分配给变量＆＃39; valid＆＃39;但是当我最后归还它时，这是错误的。

var  valid = false;

factory.validate = function(id)
{

            data ={ 'id' : id };
            $http.post('php/validate.php', data).then
            (
                function(response)
                {
                    valid = response.data;
                    // Prints true here
                    console.log(response.data);
                    console.log(valid);
                },
                function(error)
                {
                    console.log(error);
                }
            );

            // Returns and prints false here
            console.log(valid);
            return valid;
        };

提前谢谢。

Answer 1

该函数返回false，因为它是异步的，并且它还没有从您的post请求收到服务器响应。

如果您希望在表单验证/服务器响应后使用函数返回值来执行某些操作，那么您可以使用angular＆＃39; s $ q

this.validate = function() {
    var deferred = $q.defer(); 
    setTimeout(function() {//or do your post request that takes some time
        deferred.resolve();//validate the response etc and resolve as true.
        //or if something went wrong or didnt validate, reject.
        deferred.reject("Your form contained some errors"); 
    },1000);
    return deferred.promise;
}

如果是您的示例，您只需返回$ http.post（）的返回值，然后使用它（$ http请求返回一个promise对象）。

//your code with minor edits
factory.validate = function(id) {
    var deferred = $q.defer();
    var data ={ 'id' : id };
    $http.post('php/validate.php', data).success(function(response) {
        if ( response.is_valid ) {deferred.resolve("all OK");}
        else {deferred.reject("Error: your data is invalid");}
    }).error(function() {
        deferred.reject("Error:could not contact the server");
    });
    return deferred.promise;
};

//then use the validation code like this
validate(112).then(function() {
    //things validated and all is OK, give user some feedback
    alert("your stuff is valid");
},function(message) {
    alert(message);
});

在ajax调用之后返回打印false

1 个答案: