java中ajax post调用的等效方法答案

【问题标题】：Equivalent method to ajax post call in javajava中ajax post调用的等效方法
【发布时间】：2019-01-16 22:34:25
【问题描述】：

我在 javascript 中发布了一个带有表单数据的 ajax 调用，并且表单数据也包含一个文件。我想对 java 做同样的事情。

我尝试过使用spring rest 模板，但没有成功，发送文件导致出现异常问题。请建议我如何使用 http 客户端通过 post call 发送文件，就像我在 javascript 代码中所做的一样。

const uploadFormData = new FormData();
    uploadFormData.append('file', this.state.uploadFile);
    uploadFormData.set('folderId' , this.state.uploadFoldarId);
    uploadFormData.set('repositoryId' , this.state.uploadRepositoryId);
    uploadFormData.set('orgName' , this.state.uploadOrgName);

    uploadFormData.set('sourceFileName','document_forTPAPI.txt');
    uploadFormData.set('title','from_React');
    uploadFormData.set('description','test');
    uploadFormData.set('changeLog','no');
    uploadFormData.set('mimeType','application\\txt');
    uploadFormData.set('serviceContext ','{}');
    $.ajax({
            url: 'https://tst.com/api/jsonws/dlapp/add-file-entry',
            type: 'POST',
            data: uploadFormData,
            async: false,
            cache: false,
            contentType: false,
            enctype: 'multipart/form-data',
            processData: false,

            success: function (response) {
                //alert(response);
                console.log(response);
            }
        });

【问题讨论】：

尝试使用 HTTP Post 请求，这里有一个例子：stackoverflow.com/questions/10604001/…
为什么不使用 Apache Common HttpClient（包含在 liferay 中）？请参阅 stackoverflow.com/a/6917303/379902
+1 对@DanieleBaggio 的评论，我很想在重复投票中建议这个答案。最重要的是，请比“不起作用”更具体：您到底做了什么，发生了什么，您期望什么？
@AS_Tomar 请编辑您的问题以指定与上述 cmets 中链接答案之一的差异：您需要什么但那里没有回答？否则我希望这是一个重复的问题。
嗨 olaf kock...请找到我的工作代码...我按照 Manuel 和 Daniele 的建议做了。

标签： javascript java file http liferay

【解决方案1】：

 uploadHttpClient = HttpClientBuilder.create().build();
    String userCredentialsBasicAuth =
        "Basic " + new String(Base64.encodeBase64(userCredentials.getBytes()));
    HttpPost post = new HttpPost(url + "/add-file-entry");
    post.setHeader(AUTHORIZATION, userCredentialsBasicAuth);
    BasicHttpContext ctx = new BasicHttpContext();

    FileBody filebody = new FileBody(getFileFromMultipartFile(file));
    StringBody repositoryIdBody = new StringBody(repositoryId, ContentType.TEXT_PLAIN);
    StringBody folderIdBody = new StringBody(folderId, ContentType.TEXT_PLAIN);
    StringBody sourceFileName =
        new StringBody(file.getOriginalFilename(), ContentType.TEXT_PLAIN);
    StringBody mimeType = new StringBody(file.getContentType(), ContentType.TEXT_PLAIN);
    StringBody title = new StringBody(file.getOriginalFilename(), ContentType.TEXT_PLAIN);
    StringBody description = new StringBody(StringPool.BLANK, ContentType.TEXT_PLAIN);
    StringBody changeLog = new StringBody(StringPool.BLANK, ContentType.TEXT_PLAIN);
    StringBody serviceContext = new StringBody("{}", ContentType.TEXT_PLAIN);

    org.apache.http.HttpEntity entity =
        MultipartEntityBuilder.create()
            .addPart("file", filebody)
            .addPart("repositoryId", repositoryIdBody)
            .addPart("folderId", folderIdBody)
            .addPart("sourceFileName", sourceFileName)
            .addPart("mimeType", mimeType)
            .addPart("title", title)
            .addPart("description", description)
            .addPart("changeLog", changeLog)
            .addPart("serviceContext", serviceContext)
            .build();

    post.setEntity(entity);
    HttpResponse resp = uploadHttpClient.execute(post, ctx);

【讨论】：

感谢您的帮助。