javascript – 如何html形式post – 文件上传和从php服务器读取json响应

我正在尝试将文件上传到PHP我的server.file和数据上传通过多部分/表格数据,在PHP服务器上收到的文件和数据但在我的PHP服务器上返回json响应.请帮助我如何在我的网页中读取json响应如果它的成功(代码= 0)意味着它重定向另一个页面.phroid服务器对于android和网页都是常见的.json响应看起来像{“code”：0,“message”：“success”}

PHP echo $_SESSION["userid"]?>">
        browseButton" onclick=""  style="width:12%;height: 30px; text-decoration:none;">decoration:none;">browseButton.onclick=function chooseFile() {
        document.getElementById("file").click(); 
    };

    function showMyImage(fileInput) {

        var files = fileInput.files;

        var file = files[0];
        var imageType = /image.*/;

        var img=document.getElementById("imgscreen");
        var reader = new FileReader();
        reader.onload = (function(aimg) {
            return function(e) {
            //x=e.target.result

            img.src = e.target.result;
            var extfilename=file.name;
            document.getElementById("filename").innerHTML=extfilename.slice(0,-5) ;

            document.getElementById("inputfilename").value=extfilename.slice(0,-5);
     };
 })(img);

 reader.readAsDataURL(file);

 }

最佳答案

我认为它应该适合你.像我一样使用AJAX

     //Your PHP code
        $arrToJSON = array(
        "dataPHPtoJs"=>"yourData","asYouWant"=>"
PHP evaluate
         var dt={ 
                  ObjEvn:"btn_Login",dataJsToPHP: $("#txt_EmailLogin").val()
                };

        //Ajax      
         var request =$.ajax({//http://api.jquery.com/jQuery.ajax/
                                url: "yourServer.PHP",type: "POST",data: dt,dataType: "json"
                            });

        //Ajax Done catch JSON from PHP 
            request.done(function(dataset){
                for (var index in dataset){ 
                     dataPHPtoJsJS=dataset[index].dataPHPtoJs;
                     asManyasYouWantJS=dataset[index].asYouWant;
                 }

                 //JavaScript conditions. Here you can control the behaivior of your html object,based on your PHP response
                 if(dataPHPtoJsJS){
                    $( "#idYourHtmlElement" ).removeClass( "class1" )
                    $( "#idYourHtmlElement" ).addClass( "class2" )
                 }


         }); 

        //Ajax Fail 
            request.fail(function(jqXHR,textStatus) {
                alert("Request Failed: " + textStatus);
            }); 
    }