Question

我想从我的Android应用程序向我的网络托管数据库中插入一个值。它会触发我的create.php脚本，但它不会将值插入我从编辑文本中键入的标题列。

这是我插入的方式

    try {
                URL url = new URL("mydburl/create.php");
                conn = (HttpURLConnection) url.openConnection();
                conn.setRequestMethod("POST");
                conn.setDoOutput(true);
                //conn.setDoInput(true);
                conn.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
                //conn.setRequestProperty("Accept", "application/json");
                //conn.connect();

                //Log.d("In Async param",params[0]);

                String title = params[0];
               // Log.d("In Async title",title);

                jsonObj = new JSONObject();
                jsonObj.put("title", title);
                //Log.d("In Async Json toString",jsonObj.toString());
            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (ProtocolException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            } catch (JSONException e) {
                e.printStackTrace();
            }
            try {

                os = (conn.getOutputStream());
                os.write(jsonObj.toString().getBytes("UTF-8"));
                //os.flush();
                os.close();
            } catch (IOException e) {
                e.printStackTrace();
            }

            try {
                Log.e("connection status",conn.getResponseMessage());
                InputStream in = new BufferedInputStream(conn.getInputStream());
                BufferedReader reader = new BufferedReader(new InputStreamReader(in));
                strBuilder = new StringBuilder();
                String line;
                while ((line = reader.readLine()) != null) {
                    strBuilder.append(line);
                }
                os.close();
                reader.close();
                conn.disconnect();

            } catch (IOException e) {
                e.printStackTrace();
            }

在我的android studio log cat上，反映出连接状态正常。

我也在异步任务方法中做了一些调试。（fffg是我在编辑文本中输入的内容）

Log.d("In Async param",params[0]); 
Log.d("In Async Json toString",jsonObj.toString());

这是我的php脚本（create.php）。

<?php
    $response = array();
    $servername = "myservername";
    $username = "user";
    $password = "pwd";
    $dbname = "myDB";

    $con=mysql_connect("$servername","$username","$password");
    mysql_select_db("$dbname", $con);
    $title=$_POST['title'];
    mysql_query("INSERT INTO Test(title) VALUES ('$title')");
    header('Content-type: application/json; charset=UTF-8');
?>

请帮忙。

Answer 1

确保标题不为空白。

<?php
$response = array();
$servername = "myservername";
$username = "user";
$password = "pwd";
$dbname = "myDB";

$con=mysql_connect("$servername","$username","$password");
mysql_select_db("$dbname", $con);

 if (isset($_POST['title'])) {
    $title=$_POST['title'];
    mysql_query("INSERT INTO Test(title) VALUES ('$title')");
    header('Content-type: application/json; charset=UTF-8');
}
else {
    echo "Required field is missing";
}
?>

Answer 2

我认为原因是，当服务器读取为JsonObject时，您正在发送key-value-pair。尝试这样设置。

Uri.Builder builder = new Uri.Builder()
        .appendQueryParameter("title", title);
String query = builder.build().getEncodedQuery();

OutputStream os = conn.getOutputStream();
BufferedWriter writer = new BufferedWriter(
            new OutputStreamWriter(os, "UTF-8"));
writer.write(query);
writer.flush();
writer.close();
os.close();    
conn.connect();

Answer 3

管理解决它

$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$json = file_get_contents('php://input');
$obj = json_decode($json);
$title = $obj->{'title'};
$sql = "INSERT INTO Test (title) VALUES ('$title')";

if ($conn->query($sql) === TRUE) { 
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}
    $conn->close();

数据库插入空值

3 个答案: