如何将namevaluepairs中的MySQL数据库连接数据从Android传递到PHP

Question

我正尝试使用PHP连接到MySQL，从Android传递数据库连接参数。我不想对连接参数进行硬编码，也不想将它们存储在单独的文件中。当我在PHP中有数据库参数时，我的代码工作正常，但现在不能工作，因为我试图通过名称值对将它们从Java传递到PHP，如下所示。

除了使用传递变量而不是硬编码的PHP连接外没有任何变化，所以我怀疑有些格式化或REGEX问题，但找不到任何解决方案。任何帮助非常感谢！

根据VolkerK的协助解决问题。查看原始的PHP代码并在下面更新。

ORIGINAL SQLQuery.php：

<?php 
mysql_connect($_REQUEST['url'],$_REQUEST['username'],$_REQUEST['password']); 
mysql_select_db($_REQUEST['database']); 
$q=mysql_query($_REQUEST['SQL']); 
while($e=mysql_fetch_assoc($q)) 
     $output[]=$e; 
print(json_encode($output)); 
mysql_close(); 
?> 

工作SQLQuery.php：

<?php 
if (get_magic_quotes_gpc()) { 
    $process = array(&$_GET, &$_POST, &$_COOKIE, &$_REQUEST); 
    while (list($key, $val) = each($process)) { 
     foreach ($val as $k => $v) { 
      unset($process[$key][$k]); 
      if (is_array($v)) { 
       $process[$key][stripslashes($k)] = $v; 
       $process[] = &$process[$key][stripslashes($k)]; 
      } else { 
       $process[$key][stripslashes($k)] = stripslashes($v); 
      } 
     } 
    } 
     unset($process); 
} 

define('DEBUGLOG', true); 
$output = array(); 

$mysql = mysql_connect($_REQUEST['url'],$_REQUEST['username'],$_REQUEST['password']); 
if (!$mysql) { 
    $output['status']='Error'; 
    $output['errormsg']='MySQL connect error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array( 
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'username'=>$_REQUEST['username'], 
      'password'=>$_REQUEST['password'] 
     ); 
    } 
} 
else if (!mysql_select_db($_REQUEST['database'])) { 
    $output['status']='Error'; 
    $output['errormsg']='Database select error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array( 
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'database'=>$_REQUEST['database'] 
     ); 
    } 
} 
else if (false===($q=mysql_query($_REQUEST['SQL']))) { 
    $output['status']='Error'; 
    $output['errormsg']='Query error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array( 
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'SQL'=>$_REQUEST['SQL'] 
     ); 
    } 
} 
else { 
    while($e=mysql_fetch_assoc($q)) { 
     $output[]=$e; 
    } 
} 

print(json_encode($output)); 

从我的Android代码（！细节改变，以保护无辜）提取物：

String phpDBURL = "mysqlserver.blah.com:3306"; 
String phpURL = "http://www.blah.com/php/"; 
String dbname ="dbref_Evaluate"; 
String username = "dbref_admin"; 
String password = "password"; 
String SQL = "SELECT ID, ShortDesc FROM User WHERE Account = 'myname@gmail.com'"; 
//the query to send 
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(); 
nameValuePairs.add(new BasicNameValuePair("url",phpDBURL)); 
nameValuePairs.add(new BasicNameValuePair("username",username)); 
nameValuePairs.add(new BasicNameValuePair("password",password)); 
nameValuePairs.add(new BasicNameValuePair("database",dbname)); 
nameValuePairs.add(new BasicNameValuePair("SQL",SQL)); 
Log.v("Common.SQLQuery", "Passing parameters: " + nameValuePairs.toString()); 
//http post 
try{ 
     HttpClient httpclient = new DefaultHttpClient(); 
     HttpPost httppost = new HttpPost(phpURL + "SQLQuery.php"); 
     httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 
     HttpResponse response = httpclient.execute(httppost); 
     HttpEntity entity = response.getEntity(); 
     InputStream is = entity.getContent(); 
//convert response to string 

等等。

Answer 1

您需要 更多 错误处理。 任何mysql_ *函数都可能失败，您的代码必须对此作出反应。

原油例如：

<?php 
define('DEBUGLOG', true); 
$output = array(); 

$mysql = mysql_connect($_REQUEST['url'],$_REQUEST['username'],$_REQUEST['password']); 
if (!$mysql) { 
    $output['status']='error'; 
    $output['errormsg']='database error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array(
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'username'=>$_REQUEST['username'], 
      'password'=>$_REQUEST['password'] 
     ); 
    } 
} 
else if (!mysql_select_db($_REQUEST['database'])) { 
    $output['status']='error'; 
    $output['errormsg']='database error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array(
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'database'=>$_REQUEST['database'] 
     ); 
    } 
} 
else if (false===($q=mysql_query($_REQUEST['SQL']))) { 
    $output['status']='error'; 
    $output['errormsg']='database error'; 
    if (defined('DEBUGLOG') && DEBUGLOG) { 
     $output['errordetails'] = array(
      'msg'=>mysql_error(), 
      'url'=>$_REQUEST['url'], 
      'SQL'=>$_REQUEST['SQL'] 
     ); 
    } 
} 
else { 
    while($e=mysql_fetch_assoc($q)) { 
     $output[]=$e; 
    } 
} 

print(json_encode($output)); 

Answer 2

使用POST而不是PHP文件中的REQUEST