我正在尝试使用 PHP 连接到 MySQL,从 Android 传递数据库连接参数。我不想硬编码连接参数,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,我的代码运行良好,但现在我尝试使用如下名称值对将它们从 Java 传递到 PHP 时不起作用。
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,所以我怀疑存在一些格式或 REGEX 问题,但找不到任何解决方案。非常感谢任何帮助!
在 VolkerK 的帮助下解决的问题。请参阅原始 PHP 代码并在下面进行更新。
原始 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)); 
摘自我的 Android 代码(更改了细节以保护无辜者!):
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