2013-05-14 62 views
0

我在我的應用程序的一個類中工作正常,下面的代碼。以下代碼段從遠程服務器收到一個字符串「Okay」。從sql server接收數據時出錯

try { 
        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2); 

        nameValuePairs.add(new BasicNameValuePair("userid", et1.getText().toString())); 
        nameValuePairs.add(new BasicNameValuePair("pass", et2.getText().toString()));      
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 
        client.execute(httppost); 
        HttpResponse response = client.execute(httppost); 
        HttpEntity respEntity = response.getEntity(); 

        if (respEntity != null) { 
         // EntityUtils to get the response content 
         content = EntityUtils.toString(respEntity); 

        Log.d("valueeeeeeeeeeee", content); 
        } 

的PHP代碼是在以下幾點:

<?php 

$host = "localhost"; 
$user = "admin"; 
$pass = "123"; 
$connection = mysql_connect($host,$user,$pass) or die (mysql_errno().": ".mysql_error()."<BR>"); 

$userid = mysql_real_escape_string($_POST['userid']); 
    $pass = mysql_real_escape_string($_POST['pass']); 

$db_select=mysql_select_db("mydb"); 
if(!$db_select){ 
    die(mysql_error()); 
    echo "error"; 
} 

    $query = "select count(1) as count_users from wp_users where user_login = '".$userid."' and user_pass='".$pass."'"; 

    $result = mysql_query($query); 
    $row = mysql_fetch_assoc($result); 

    if($row['count_users']>0) 
    { 
     echo "Okay"; 
    } 
    else 
    { 
     echo "Not found"; 
    } 

if($medo=mysql_query($query)){ 
    header("localhost/filename"); 
    exit; 
}else{ 
    echo"<p> Error</p>"; 
    die(mysql_error()); 
} 

而當我試圖在PHP運行下面的代碼它給我同樣採用Android代碼中的錯誤。

<?php 

$host = "localhost"; 
$user = "admin"; 
$pass = "123"; 
$connection = mysql_connect($host,$user,$pass) or die (mysql_errno().": ".mysql_error()."<BR>"); 

$userid = mysql_real_escape_string($_POST['userid']); 
    $pass = mysql_real_escape_string($_POST['pass']); 

$db_select=mysql_select_db("mydb"); 
if(!$db_select){ 
    die(mysql_error()); 
    echo "error"; 
} 

    $query= "SELECT * FROM wp_users"; 
    $result = mysql_query($query); 
    $num_rows = mysql_num_rows($result); 

    if("$num_rows" > 0) 
    { 
     echo "$num_rows"; 
    } 
    else 
    { 
     echo "Not found"; 
    } 


    if($medo=mysql_query($query)){ 
    header("localhost/filename"); 
    exit; 
}else{ 
    echo"<p> Error</p>"; 
    die(mysql_error()); 
} 

是否有不同的方式接收Android行末尾的行或列中的行數或數據?

回答

0

首先改變的

if("$num_rows" > 0) 
{ 
    echo "$num_rows"; 
} 

if ($num_rows > 0) 
{ 
    echo $num_rows; 
} 

其次,如果你只是想獲得的行數使用查詢

$query= "SELECT COUNT(*) FROM wp_users"; 
+0

它仍然給了我「NumberFormatException異常:無效的詮釋「error – 2013-05-14 03:32:01

+0

$ query =」SELECT COUNT(*)FROM wp_users「; <------並沒有真正的幫助,因爲它給了我不正確的行數。 – 2013-05-14 03:36:37

+0

@MohammadSohaib您的查詢是'$ query =「SELECT * FROM wp_users」;'因此'SELECT COUNT(*)FROM wp_users'給出了您在問題中提出的問題 - 結果集中的行數。要麼改變你的查詢,要麼清楚地解釋你試圖得到的數字。 – peterm 2013-05-14 03:40:55

相關問題