Android OkHttp3 上传多张图片

来源：互联网发布：书生电子图书数据库编辑：程序博客网时间：2024/06/05 06:33

经过实践，android与php交互，已经成功搞定！

一、Android 端

 /**     * 上传文件及参数     */    private void sendMultipart(){        File sdcache = getExternalCacheDir();        int cacheSize = 10 * 1024 * 1024;        //设置超时时间及缓存        OkHttpClient.Builder builder = new OkHttpClient.Builder()                .connectTimeout(15, TimeUnit.SECONDS)                .writeTimeout(20, TimeUnit.SECONDS)                .readTimeout(20, TimeUnit.SECONDS)                .cache(new Cache(sdcache.getAbsoluteFile(), cacheSize));        OkHttpClient mOkHttpClient=builder.build();        MultipartBody.Builder mbody=new MultipartBody.Builder().setType(MultipartBody.FORM);        List<File> fileList=new ArrayList<File>();        File img1=new File("/sdcard/wangshu.jpg");        fileList.add(img1);        File img2=new File("/sdcard/123.jpg");        fileList.add(img2);        int i=0;        for(File file:fileList){            if(file.exists()){                Log.i("imageName:",file.getName());//经过测试，此处的名称不能相同，如果相同，只能保存最后一个图片，不知道那些同名的大神是怎么成功保存图片的。                mbody.addFormDataPart("image"+i,file.getName(),RequestBody.create(MEDIA_TYPE_PNG,file));                i++;            }        }        RequestBody requestBody =mbody.build();        Request request = new Request.Builder()            .header("Authorization", "Client-ID " + "...")            .url("http://192.168.1.105/interface/index.php?action=sendMultipart")            .post(requestBody)            .build();        mOkHttpClient.newCall(request).enqueue(new Callback() {            @Override            public void onFailure(Call call, IOException e) {            }            @Override            public void onResponse(Call call, Response response) throws IOException {                Log.i("InfoMSG", response.body().string());            }        });    }

二、Php服务端

if ($act == "sendMultipart") {$result="";try {foreach ($_FILES as $key => $val) {$imgName = time().rand(1000, 9999);//随机数$folder = "images";//接口文件同目录下创建此文件夹，当然也可以通过代码的形式判断/创建$file_dir = $folder . "/" . $imgName . ".jpg";if (move_uploaded_file($val["tmp_name"], $file_dir)) {$result .= $val["name"] . "保存成功";} else {$result .= "在服务器中保存失败" . $val["tmp_name"] . "--------";}echo "返回信息：" . $result."<br />\n";}} catch(exception $ex) {echo $ex;}}

没什么可总结的了，都在注释里了。

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