javascript - 在没有表单的情况下从ajax向spring Controller 发送多部分请求？

Question

我正在从桌面上传图像，并将该图像转换为 JavaScript 中的基本代码。之后，我想通过多部分请求将此图像基本代码发送到 spring Controller 。但我没有使用表单。

HTML

 <input id="inputFileToLoad" type="file"  onchange="encodeImageFileAsURL()">   

JAVA脚本

 window.photoCakeUrl = '<c:url value="/media/image/upload"/>';
      function encodeImageFileAsURL() {
                var filesSelected = document.getElementById("inputFileToLoad").files;
                if (filesSelected.length > 0) {
                    var fileToLoad = filesSelected[0];
                    var fileReader = new FileReader();
                    fileReader.onload = function (fileLoadedEvent) {
                        var srcData = fileLoadedEvent.target.result; // <--- data: base64
                        var newImage = document.createElement('img');
                        var photoCake = srcData;
                        newImage.src = srcData;
                        document.getElementById("imgTest").innerHTML = newImage.outerHTML;
                        var ajax1 = $.ajax({
                            type: 'POST',
                            url: photoCakeUrl,
                            processData: false, // important
                            contentType: false, // important
                            dataType: 'json',
                            data: {photoCak: photoCake}
                        });

                      });

                    },
                            fileReader.readAsDataURL(fileToLoad);
                }
            }

Spring Controller :

@RequestMapping(value = "/media/image/upload", method = RequestMethod.POST)
    @ResponseBody
    public Map<String, String> productPictureUploadnew(MultipartHttpServletRequest request, HttpServletResponse response) {
        Map<String, String> resp = new HashMap<>();
        String photoCake = request.getParameter("photoCak");


        System.out.println("photoCake   " + photoCake);

        return resp;
    }

但是当我生成 AJAX 调用时，会出现 500 错误。如果我只是使用

public Map<String, String> productPictureUploadnew(HttpServletRequest
 request, HttpServletResponse response)

然后就可以了。意思是当我使用 MultipartHttpServletRequest 的地方 HttpServletRequest 请求那么它就不起作用。

Answer 1

我得到了解决方案，我们可以在javascript中使用formData，而无需在任何JSP中使用form来发送MultipartHttpServletRequest。

 window.photoCakeUrl = '<c:url value="/media/image/upload"/>';
      function encodeImageFileAsURL() {
                var filesSelected = document.getElementById("inputFileToLoad").files;
                if (filesSelected.length > 0) {
                    var fileToLoad = filesSelected[0];
                    var fileReader = new FileReader();
                    fileReader.onload = function (fileLoadedEvent) {
                        var srcData = fileLoadedEvent.target.result; // <--- data: base64
                        var newImage = document.createElement('img');
                        var photoCake = srcData;
                        newImage.src = srcData;
                        document.getElementById("imgTest").innerHTML = newImage.outerHTML;

                        var formData = new FormData();
                        formData.append("imgFile", document.getElementById("inputFileToLoad").files[0]);

                        var ajax1 = $.ajax({
                            type: 'POST',
                            url: photoCakeUrl,
                            dataType: 'json',
                            data: {photoCak: photoCake}
                        });

                      });

                    },
                            fileReader.readAsDataURL(fileToLoad);
                }
            }

var formData = new FormData();
formData.append("imgFile", document.getElementById("inputFileToLoad").files[0]);

Controller :

 @RequestMapping(value = "/media/image/upload", method = RequestMethod.POST)
        @ResponseBody
        public Map<String, String> productPictureUploadnew(MultipartHttpServletRequest request, HttpServletResponse response) {
            Map<String, String> resp = new HashMap<>();
            System.out.println("fsasasafsafsafsafsa");
            Iterator<String> itr = request.getFileNames();

            String photoCake = request.getParameter("photoCak");
            File file;
          ----------
           -------
          ----------              

            return resp;
        }

谢谢您，希望这对你们有帮助。