我正在尝试将文件上传到 php 我的 server.file 和通过多部分/form-data 上传数据，在 php 服务器上收到的文件和数据，但在我的 php 服务器中返回 json 响应。请帮助我如何读取 json 响应在我的网页中，如果它的成功(代码=0)意味着它重定向另一个页面。php 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"}

<div style="height:0px;overflow:hidden">
    <form id="myForm" action="http://192.168.2.4/digiid/api/addid" 
        method="post" enctype="multipart/form-data" runat="server">

        <input type="file" name="file" id="file" onchange="showMyImage(this)" />
        <input type="hidden" name="userid" value="<?php echo $_SESSION["userid"]?>">
        <input type="hidden" id="inputfilename" name="filename" value="here">
    </form>
</div>

<a class="button1" id="browseButton" onclick=""  style="width:12%;height: 30px; text-decoration:none;"><font color="white" size="5px">Select ID</font></a>
<br/>

<div>

            <img src='images/capture_picture_size.png' id='imgscreen'  width='200' height='200'>

<br/>


<p id="filename" style="color: #ffffff; font-size: 20px" >
    Title of the ID<br/></p>

<a class="button1"onclick="myFunction()" style= " width:12%;height: 30px; text-decoration:none;"><font color="white" size="5px">Save ID</font></a></form>

</div>  

<script>
    function myFunction() {
       document.getElementById("myForm").submit();
    }
</script>

<script>
    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);

 }</script>

最佳答案

我认为它应该适合您。像我一样使用 AJAX

     //Your php code
        $arrToJSON = array(
        "dataPHPtoJs"=>"yourData",
        "asYouWant"=>"<div class=\".class1\">soemting</div>"    
        );  
        return json_encode(array($arrToJSON));




    //Your javaScript code
    $(document).on("event", "#idElement", function(){
        //Data you want to send to 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);
            }); 
    }

关于javascript - 如何以 html 形式上传文件并从 php 服务器读取 json 响应，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/26052722/