c# - 在 XAMARIN C# 中使用 FILE 发送 POST 请求

问题描述

在网上看，我看到了一点如何发布“文件”，因此，我以这种方式编写了我的代码：

        var upfilebytes = File.ReadAllBytes((string)fPath);

        //create new HttpClient and MultipartFormDataContent and add our file, and StudentId
        HttpClient client = new HttpClient();

        MultipartFormDataContent content = new MultipartFormDataContent();
        ByteArrayContent baContent = new ByteArrayContent(upfilebytes);
        content.Add(baContent, "img", fPath);

        StringContent emailText = new StringContent(lbl_email.Text);
        content.Add(emailText, "email");

        string url = "http://192.168.178.77/TestLoginURL/api/updateUserImage.php";
        //upload MultipartFormDataContent content async and store response in response var
        var response =
            await client.PostAsync(url, content);
        //read response result as a string async into json var
        var responsestr = response.Content.ReadAsStringAsync().Result;

现在问题是另一个......这是我的 API 代码：

<?php
$response = array();

if($_SERVER['REQUEST_METHOD']=='POST'){

    //getting values
    $img = $_FILES['img']['name'];
    $email = $_POST['email'];

    //including the db operation file
    require_once '../includes/DbOperation.php';

    $db = new DbOperation();

    $target = "/Applications/XAMPP/xamppfiles/htdocs/TestLoginURL/images/".basename($img);

    //inserting values
    if($db->updateImage((String)basename($img),$email)){
        $response['error']=false;
        $response['message']='Image added successfully - test fileName = '.(String)basename($img);
    }else{
        $response['error']=true;
        $response['message']='Could not add image';
    }
    if(move_uploaded_file($_FILES['img']['tmp_name'], $target)) {
        /*$response['error']=false;
        $response = "Image uploaded successfully";*/
    }else{
        $response['error']=true;
        $response = "Failed to upload image";
    }
}else{
    $response['error']=true;
    $response['message']='You are not authorized';
}
echo json_encode($response);

使用 Postman 它可以完美运行，更新数据库中图像的名称并将图像物理插入指定路径，响应消息例如是这样的：{“error”：false，“message”：“图像添加成功 - 测试文件名 = trollolollo.png"}

现在，应用程序将文件保存在正确的存储库中，但不会更新数据库中“图像”的名称……但奇怪的是，调试器中的“响应”消息也正确显示了文件的名称……所以我只是不明白我错在哪里......有人可以帮我写代码吗？谢谢

OFF_TOPIC：通常，当我只需要发送字符串时，我会发送一个以这种方式编写的 post 请求

            string url = "http://192.168.178.77/TestLoginURL/api/insertUser.php";

            FormUrlEncodedContent formContent = new FormUrlEncodedContent(new[]
            {
              new KeyValuePair<string, string>("nickname", nickname),
              new KeyValuePair<string, string>("password", password1),
              new KeyValuePair<string, string>("email", email)
            });

            var httpClient = new HttpClient();
            httpClient.DefaultRequestHeaders.Add("Accept", "application/json");
            try
            {
                CancellationTokenSource cts = new CancellationTokenSource();
                var responseMessage = httpClient.PostAsync(url, formContent).Result;
            }
            catch (Exception ex)
            {
                ex.Message.ToString();
                throw;
            }

但是现在，作为一个文件并且没有经验，我遇到了困难...再次感谢，我希望有人可以帮助我编写代码

标签： c#filexamarinposthttpclient

c# - 在 XAMARIN C# 中使用 FILE 发送 POST 请求

问题描述

解决方案

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