Question

我正在使用一个路由来处理网址。功能按预期启动。我想要做的是在回调中使用response.write（）。我知道这不起作用，因为回调不能访问与调用它的函数相同的变量，但我想知道正确的NODE方式是做什么的。

route.post('/{type}subject/{method}', function (request,response) {
var post = "";
request.on('data', function (chunk){
    post += chunk;
});
request.on('end', function (){
    postData = qs.parse(post);
    response.writeHead(200);
    switch(request.params['method'].toLowerCase())
    {
        case "registerobserver":
            if (postData['uri']){
                registerObserver (request.params['type'], postData['uri']);
                response.write(success("registerobserver"));
            }
            else
                response.write(failure("1","uri undefined"));

            break;
        case "unregisterobserver":
            if (postData['uri']){
                client.query ('DELETE observers FROM observers INNER JOIN type ON (observers.TypeID = type.TypeID) WHERE observers.uri ="'+postData['uri']+'" AND type.name = "'+request.params['type']+'"', 
                function(err, info) 
                {
                    if(err){
                        response.write(failure("2", "general database failure"));}
                    else{   
                    if(info.affectedRows != 0)
                        response.write(success("unregisterobserver")); //this code does not trigger a response due to namespace

                    else
                        response.write(failure("1", "uri not an observer"));//this code does not trigger a response
                        console.log("uri not observer");
                    }

                    console.log("done");
                })

            }
            else
                response.write(failure("1","uri required"));

            break;
        default:

    }

    response.end();
})
//response.write("type: "+request.params['type']+"<br/>method="+request.params['method']);

});

function success(method){return "<?xml version=\"1.0\"?>\n<response stat=\"ok\">\n\t<method>"+method+"</method>\n</response>";}
function failure(code, message){return "<?xml version=\"1.0\"?>\n<response stat=\"fail\">\n\t<err code=\""+code+"\" msg = \""+message+"\" />\n</response>";}

Answer 1

基本上发生的事情是，在 query函数调用之后，异步response.end()处理函数将被称为。因此，任何写入都将失败。

您需要从回调内部调用response.end()，并且在处于异步代码路径时请注意不要再调用response.end()。 IE浏览器。 return致电后立即client.query()。

NodeJS response.write里面的回调

1 个答案: