Question

我不知道是否有人可以阅读所有这些并且无论如何都要花时间我试图使用Android应用程序将图像保存到mysql数据库，但它失败了，没有添加任何内容，所以我试图显示一些我用来保存图像的变量，以便查看是否有任何错误因此使用Log.d方法我显示两个变量，请查看它们的内容并帮助我找出问题所在：

MainActivity

<br />

String uploadImage ：

9J / 4AAQSkZJRgABAQAAAQABAAD / 2wBDAAEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEBAQEB ... ECT

字符串结果：此变量显示balise html br：

＆＃13;

public String sendPostRequest(String requestURL,HashMap<String, String> postDataParams) {
    URL url;
    String response = "";
    try {
        url = new URL(requestURL);

        HttpURLConnection conn = (HttpURLConnection) url.openConnection();
        conn.setReadTimeout(15000);
        conn.setConnectTimeout(15000);
        conn.setRequestMethod("POST");
        conn.setDoInput(true);
        conn.setDoOutput(true);

        OutputStream os = conn.getOutputStream();
        BufferedWriter writer = new BufferedWriter(
                new OutputStreamWriter(os, "UTF-8"));
        writer.write(getPostDataString(postDataParams));

        writer.flush();
        writer.close();
        os.close();
        int responseCode = conn.getResponseCode();

        if (responseCode == HttpsURLConnection.HTTP_OK) {
            BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream()));
            response = br.readLine();
        } else {
            response = "Error Registering";
        }
    } catch (Exception e) {
        e.printStackTrace();
    }

    return response;
}

private String getPostDataString(HashMap<String, String> params) throws UnsupportedEncodingException {
    StringBuilder result = new StringBuilder();
    boolean first = true;
    for (Map.Entry<String, String> entry : params.entrySet()) {
        if (first)
            first = false;
        else
            result.append("&");

        result.append(URLEncoder.encode(entry.getKey(), "UTF-8"));
        result.append("=");
        result.append(URLEncoder.encode(entry.getValue(), "UTF-8"));
    }

    return result.toString();
}

＆＃13;

RequestHandler：

<?php
if($_SERVER['REQUEST_METHOD']=='POST'){
		
		$image = $_POST['image'];
		
		require_once('dbConnect.php');
		
		$sql = "INSERT INTO images (image) VALUES (?)";

		$stmt = mysqli_prepare($con,$sql);
		
		mysqli_stmt_bind_param($stmt,"s",$image);
		mysqli_stmt_execute($stmt);
		
		$check = mysqli_stmt_affected_rows($stmt);
		
		if($check == 1){
			echo "Image Uploaded Successfully";
		}else{
			echo "Error Uploading Image";
		}
		mysqli_close($con);
	}else{
		echo "Error";
	}
	?>

问题是一个sendPostRequest Methode，但我找不到有人能找到那里的问题？

这是我使用的脚本php：

＆＃13;

.htaccess

＆＃13;

Answer 1

1-比较图像的BLOB类型的数据大小和数据大小

  $image = $_POST['image'];

  $image = base64_decode($image );

并且由于您在将图像发送到db之前对其进行了编码，因此您需要在存储之前对其进行解码。

Android无法将图像保存到mysql数据库

1 个答案: