0

我正在从 java 将数据发布到一个 URL，该 URL 用于从 Web 服务器上的 MySQL 字段PHP返回一个json_encode(d)blob。

现在，当我使用返回的数据（使用InputStreamor getBytes）创建 abitmap并将 a 设置ImageView为此位图setImageBitmap时，我得到空的 imageview。即，decodeByteArray或decodeStream返回NULL。

我在整个互联网上搜索了这个，发现许多开发人员都面临这个问题。请尽快帮忙。

谢谢你。

我检索文本细节没有问题，检索图像是个问题。这是Java和PHP的代码片段..

这是我的 java 的 onCreate 方法。

public void onCreate(Bundle icicle) {

super.onCreate(icicle);

setContentView(R.layout.loginlayout);
TextView tv1, tv2, tv3, tv4, tv5;
imageview = (ImageView) findViewById(R.id.imageView1);

button  = (Button) findViewById(R.id.button1);
tv1 = (TextView) findViewById(R.id.dispuser);
tv2 = (TextView) findViewById(R.id.tv2);
tv3 = (TextView) findViewById(R.id.tv3);
tv4 = (TextView) findViewById(R.id.tv4);
tv5 = (TextView) findViewById(R.id.tv5);


SharedPreferences userDetails = this.getSharedPreferences("logindetails",      MODE_PRIVATE);
String username = userDetails.getString("username", "");
String password = userDetails.getString("password", "");

button.setOnClickListener(this);


httpclient = new DefaultHttpClient();

httppost = new HttpPost("myurl");


try{

nameValuePairs = new ArrayList<NameValuePair>();


nameValuePairs.add(new BasicNameValuePair("username", username));
nameValuePairs.add(new BasicNameValuePair("password", password));



httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

response = httpclient.execute(httppost);

if(response.getStatusLine().getStatusCode() == 200){

    entity = response.getEntity();


    if(entity != null) {

        InputStream instream = entity.getContent();

        JSONObject jsonResponse = new JSONObject(convertStreamToString(instream));




String sex = jsonResponse.getString("Sex");
String age = jsonResponse.getString("Age");
String name = jsonResponse.getString("Name");
String phone = jsonResponse.getString("Phone");
String pic = jsonResponse.getString("picture");

byte[] image = Base64.encodeBytesToBytes(pic.getBytes("UTF-8"));

imageview.setImageBitmap(BitmapFactory.decodeByteArray(image, 0, image.length));


 tv1.setText("Welcome " + username);
 tv2.setText("Your  is Name: " + name);
 tv3.setText("Your phone " + phone);
 tv4.setText("Your a  " + sex );
 tv5.setText("Your age is " + age);


}
} 

}
catch (Exception e) {

e.printStackTrace();



}
finally {
httppost.abort();
}
}

在PHP方面..

    <?
    include 'connect.php';

    $username = $_POST['username'];
    $password = $_POST['password'];

    $query = mysql_query("SELECT * FROM applogin WHERE username = '$username' AND password = '$password'");

    $num = mysql_num_rows($query);

    if($num == 1)
    {
    while($list = mysql_fetch_assoc($query))
    {
    $output = $list;
    }
    echo json_encode($output);
    mysql_close();
    }
    ?>
4

1 回答 1

0 
String input;//your pic string 
byte[] decodedByte = Base64.decode(input, 0);
BitmapFactory.decodeByteArray(decodedByte, 0, decodedByte.length);
于 2013-01-25T07:41:20.607 回答