phonegap/msql（i）無法與數據庫建立連接

Question

我目前正在爲iOS開發一個Web應用程序，但該應用程序無法獲得連接權限。

當我試圖登錄時，我得到以下iOS彈出窗口：index.html錯誤。

而PHP文件是在網上，

這裏我的一些代碼：

connection.php

<?php //Make connection with database 
// Verbinden met MySQL Database 
$host = "localhost"; // Welke server : localhost 
$username = "*******"; // Gebruikersnaam 
$password = "****"; // Wachtwoord 
$dbnaam = "*****"; // Naam van de database 
$db_error1 = "<p>FOUT: verbinden met databaseserver is mislukt</p>"; // Foutmelding 1 
$db_error2 = "<p>FOUT: selecteren van database is mislukt</p>"; // Foutmelding 2 
$db_error3 = "<p>FOUT: sluiten van database is mislukt</p>"; // Foutmelding 3 

// Verbinden met Databaseserver 
$con=mysqli_connect($host, $username, $password, $dbnaam);// or die($db_error1); 
// verbinden met de database 
//mysql_select_db($dbnaam, $db) or die($db_error2); 
if (mysqli_connect_errno($con)) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 
?> 

demo.js

// define the URL of the server component 
//var url ="http://www.jorisgraaumans.nl/dutchmobile/"; 
//var url = "http://localhost:8888/Mobile/Festival/"; 
var url = "http://www.rikvandoorn.com/mobile/"; 

// afvangen van het standaard submit event 
// zie ook: http://api.jquery.com/submit/ 

$('#login').on('pageinit', function(event) { 
    $('#loginForm').submit(function() { 
     $.ajax({ 
      type: "POST", 
      url: "processLogin.php", 
      cache: "false", 
      dataType: "json", 
      data: { 
       email : $('#email').val(), 
       wachtwoord : $('#wachtwoord').val(), 
      }, 
      success: function(phpData){ 
       $("#return").data("login", phpData.login); 
       console.log("login is: "+ phpData.login); 
       if(phpData['error'] == true){ 
        $.mobile.changePage("#return", { 
         transition : "fade" 
        }) 
       } else { 
        $.mobile.changePage("#home", { 
         transition : "fade" 
        }) 
       } 
      }, 
      error: function(){ 
        alert('Error'); 
      }     
     }); 
     return false; // return false to prevent the default submit of the form to the server. 
    }); 
    // end: pageinit loginForm 
}); 
$('#return').on('pageshow', function(event) { 
    $("#error_message_log").empty(); 
    $("#error_message_log").prepend('<p>' + $(this).data('login') + '</p>'); 
}); 

processLogin.php

<?php //process login form 

include 'http://www.rikvandoorn.nl/mobile/connect.php'; //connection to database 
if(empty($_POST['email']) || empty($_POST['wachtwoord'])){ 
    $return['error'] = true; //return error 
    $return['login'] = 'Niet alle velden zijn ingevuld'; 
} else { 
    $email = $_POST['email']; 
    $wachtwoord = $_POST['wachtwoord']; 

    //Check if user exists 
    $sql = "SELECT * FROM gebruiker WHERE email = '$email'"; 
    $result = mysql_query($sql) or die(mysql_error()); 
    $num_rows = mysql_num_rows($result);  

    if($num_rows == 0){ 
     $return['error'] = true; 
     $return['login'] = 'Gebruiker bestaat niet'; 
    } else { 
     //Check if password is correct for user 
     $sql_2 = "SELECT * FROM gebruiker WHERE email = '$email' && wachtwoord = '$wachtwoord'"; 
     $query_2 = mysql_query($sql_2); 
     $num_rows_2 = mysql_num_rows($query_2); 
     if($num_rows_2 == 0){ 
      $return['error'] = true; 
      $return['login'] = 'Wachtwoord onjuist'; 
     } else { 
      $return['error'] = false; 
      $return['login']['email'] = $email; 
      $return['login']['wachtwoord'] = $wachtwoord; 
     } 
    } 
} 
echo json_encode($return); 

Answer 1

除非你已經充斥着錯別字的代碼，你將兩種完全獨立的MySQL庫：

$con=mysqli_connect($host, $username, $password, $dbnaam);// or die($db_error1); 
      ^---note the presence of an 'i' 

，並在其他腳本：

$result = mysql_query($sql) or die(mysql_error()); 
       ^---note the LACK of an 'i' 

mysql和mysqli的是NOT可互換，並建立在一個連接是完全無用的另一個。

mysql（沒有和i）庫已過時並已棄用。只能用mysqli（WITH）。