Question

我正在创建一个应用程序，其中我使用php从相应的sql表中获取用户数据。

<?php 

$email = $_POST["email"];

@mysql_connect("localhost","root","root") or die(@mysql_error());
@mysql_select_db("dtbse") or die(@mysql_error());

$x = mysql_query("select * from dtbse where email = '$email' ") or die(@mysql_error());
$result = array();

while ($y=mysql_fetch_array($x)) {
            echo $y["uname"]."<br>";
            echo $y["gender"]."<br>";
            echo $y["pass"]."<br>";
            echo $y["address"]."<br>";
            echo $y["email"]."<br>";

        }

?>

任何帮助都会非常感激。我知道这个问题很多次，但我不认为有一些东西可以复制这个问题。感谢。

以下是负责获取和解析的代码段。 final ArrayList arr = new ArrayList（）; arr.add（new BasicNameValuePair（“email”，uname））;

            try {

                DefaultHttpClient httpclient = new DefaultHttpClient(new BasicHttpParams());
                HttpPost httppost = new HttpPost("http://xxyoxx.esy.es/getDetails.php");
                httppost.setEntity(new UrlEncodedFormEntity(arr));
                HttpResponse hr = httpclient.execute(httppost);
                HttpEntity ent = hr.getEntity();
                is = ent.getContent();
                Toast.makeText(getApplicationContext(),"1 wrk ",Toast.LENGTH_LONG).show();


            } catch (Exception fl) {
                Toast.makeText(getApplicationContext(),"First Try error "+fl,Toast.LENGTH_LONG).show();

            }
            /*// Depends on your web service
            httppost.setHeader("Content-type", "application/json");*/

            String result=null;
            try {
                // json is UTF-8 by default
                BufferedReader reader = new BufferedReader(new InputStreamReader(is, "UTF-8"), 8);
                StringBuilder sb = new StringBuilder();
                String line;
                while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
                }
                result = sb.toString();
                Toast.makeText(getApplicationContext(),"2 str\n "+result,Toast.LENGTH_LONG).show();
            } catch (Exception sl) {
                sl.printStackTrace();

                Toast.makeText(getApplicationContext(),"2 err\n "+sl,Toast.LENGTH_LONG).show();
            }

            try{
                String aa = "", b = "", c = "";
                JSONArray ar = new JSONArray(result);
                for (int i = 0; i < ar.length(); i++) {
                    JSONObject jo = ar.getJSONObject(i);
                    aa = jo.getString("uname");
                    b = jo.getString("address");
                    c = jo.getString("email");
                }
                nm.setText(aa);
                addr.setText(b);
                mail.setText(c);
                Toast.makeText(getApplicationContext(),"3 wrk"+result,Toast.LENGTH_LONG).show();
            }
            catch (Exception tl){
                Toast.makeText(getApplicationContext(),"3 err "+tl,Toast.LENGTH_LONG).show();
            }

Answer 1

由<br>分隔的字符串不是有效的JSON数组。 PHP可以使用json_encode

创建JSON字符串

如果您需要在Android中读取JSON数组，则需要从PHP回显JSON数组：

<?php 

$email = $_POST["email"];

mysql_connect("localhost","root","root") or die(mysql_error());
mysql_select_db("dtbse") or die(mysql_error());

$x = mysql_query("select * from dtbse where email = '$email' ") or die(mysql_error());
$result = array();

$res=[];    
while ($y=mysql_fetch_array($x)) {
     $res[] = [ 
              $y["uname"], 
              $y["gender"],                   
              $y["pass"], 
              $y["address"],
              $y["email"] 
     ];    
 }
 echo json_encode($res); //Make PHP return a valid JSON response

此外，错误抑制操作符可能会隐藏有价值的调试信息，这可能有助于您诊断其他问题。

如果您更喜欢将JSON对象传递给Java，那么您可以执行以下（更简单）的操作。

<?php 

$email = $_POST["email"];

mysql_connect("localhost","root","root") or die(mysql_error());
mysql_select_db("dtbse") or die(mysql_error());

$x = mysql_query("select * from dtbse where email = '$email' ") or die(mysql_error());
$result = array();

$res=[];    
while ($y=mysql_fetch_array($x)) {
     $res[] = $y;
 }
 echo json_encode($res); //Make PHP return a valid JSON response

值[string] <br/> [string] <br/>类型java.lang.String无法转换为JSONArray

1 个答案: