我正在使用一个路由来处理一个url。功能按预期启动。我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,但我想知道正确的节点方式是什么。

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>";}

最佳答案

基本上,发生的情况是在调用query函数之后将调用asyncresponse.end()处理程序函数。因此,任何写操作都将失败。
您需要从回调函数内部调用response.end(),并且在进入异步代码路径后注意不要调用response.end()。即return呼叫后立即。

07-24 09:38
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