Question

请参阅以下代码。当我在""中使用php并被发送到android时，可以运行以下条件：

   // Show response on activity
        String text2 = "";
        boolean check = text.equals(text2);
        if(check == true)
        { String display;
            display = (String)getText(R.string.display);
            Toast.makeText(getApplicationContext(), display, Toast.LENGTH_LONG).show();}
else {textView.setText(text);}

和

<?php
    if($_SERVER["REQUEST_METHOD"] == "POST") {
    $username = urldecode($_POST['username']);
    $login = urldecode($_POST['login']);
    if(empty($username) || empty($login)) {
    $display = "";
    echo $display;
    }}

在上面的代码中，当我使用$display = "";时，将空格字符发送到text对象，因为text2 = "" Toast类也会显示R.string.display。

但是当我使用text之类的字符或其他不像空格字符（""）的字符时，$username或$login是为空，在以下代码中，Toast类无法工作，而else {textView.setText(text);}可以正常工作。

   // Show response on activity
        String text2 = "text";
        boolean check = text.equals(text2);
        if(check == true)
        { String display;
            display = (String)getText(R.string.display);
            Toast.makeText(getApplicationContext(), display, Toast.LENGTH_LONG).show();}
else {textView.setText(text);}

和

<?php
if($_SERVER["REQUEST_METHOD"] == "POST") {
$username = urldecode($_POST['username']);
$login = urldecode($_POST['login']);
if(empty($username) || empty($login)) {
$is = "text";
echo $is;
}}
 ?>

全功能表格

   public  void  GetText()  throws UnsupportedEncodingException
    {
        // Get user defined values
        name = username.getText().toString();
        pass   = password.getText().toString();


        // Create data variable for sent values to server

        String data = URLEncoder.encode("username", "UTF-8")
                + "=" + URLEncoder.encode(name, "UTF-8");

        data += "&" + URLEncoder.encode("login", "UTF-8") + "="
                + URLEncoder.encode(pass, "UTF-8");

        String text = "";
        BufferedReader reader=null;

        // Send data
        try
        {

            // Defined URL  where to send data
            URL url = new URL("http://127.0.0.1:8080/apps/example.php");

            // Send POST data request

            URLConnection conn = url.openConnection();
            conn.setDoOutput(true);
            OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
            wr.write( data );
            wr.flush();

            // Get the server response

            reader = new BufferedReader(new InputStreamReader(conn.getInputStream()));
            StringBuilder sb = new StringBuilder();
            String line = null;

            // Read Server Response
            while((line = reader.readLine()) != null)
            {
                // Append server response in string
                sb.append(line + "\n");
            }


            text = sb.toString();
        }
        catch(Exception ex)
        {

        }
        finally
        {
            try
            {

                reader.close();
            }

            catch(Exception ex) {}
        }

        // Show response on activity
        String text2 = "text";
        boolean check = text.equals(text2);
        if(check == true)
        { String display;
            display = (String)getText(R.string.display);
            Toast.makeText(getApplicationContext(), display, Toast.LENGTH_LONG).show();}
else {textView.setText(text);}
    }}

Answer 1

在你的while循环中，你将\n附加到每一行。这意味着，当您阅读字符串"text"时，它实际上会在"text\n"中读取。只需将您要比较的字符串更改为"text\n"，它就可以正常工作。

reader = new BufferedReader(new InputStreamReader(conn.getInputStream()));
StringBuilder sb = new StringBuilder();
String line = null;

// Read Server Response
while((line = reader.readLine()) != null)
{
    // Append server response in string
    sb.append(line + "\n");
}

...

// Show response on activity
String text2 = "text\n";

我只能得到从php到android的空格字符

1 个答案: