十分な行が返されない

Question

私がここで理解しようとしているのは、必要な配列値にアクセスする方法です。私は次のクエリを持っていて、print_r（）が適用されたときに配列のためにこれを返します。何らかの理由で、dbテーブルの最初の行のみを実行します。それはまったく別の行を返す必要があります。あなたはちょうどあなたが結果にループが代わりに設定する必要があります最初の1、フェッチので

<?php 

session_start(); 

// Include the database page 
require ('../inc/dbconfig.php'); 
require ('../inc/global_functions.php'); 

//Login submitted 
if (isset($_POST['submit'])) { 

// Errors defined as not being any 
$errors = "no"; 

if((empty($_POST['answer1'])) || (trim($_POST['answer1'])=="") || ($_POST['answer1'] == NULL) || (!isset($_POST['answer1']))){$errors = "yes";} 
if((empty($_POST['answer2'])) || (trim($_POST['answer2'])=="") || ($_POST['answer2'] == NULL) || (!isset($_POST['answer2']))){$errors = "yes";} 

// Error checking, make sure all form fields have input 
if ($errors == "yes") { 

    // Not all fields were entered error 
    $message = "You must enter values to all of the form fields!"; 

    $output = array('errorsExist' => true, 'message' => $message); 

} else { 

    $userID = mysqli_real_escape_string($dbc,$_POST['userID']); 
    $answer1Post = mysqli_real_escape_string($dbc,$_POST['answer1']); 
    $answer2Post = mysqli_real_escape_string($dbc,$_POST['answer2']); 
    $question1 = mysqli_real_escape_string($dbc,$_POST['question1']); 
    $question2 = mysqli_real_escape_string($dbc,$_POST['question2']); 

    $query = "SELECT * FROM manager_users_secretAnswers WHERE userID = '".$userID."'"; 
    $result = mysqli_query($dbc,$query); 

    // Count number of returned results from query 
    if (mysqli_num_rows($result) > 0) { 

     while ($row = mysqli_fetch_array($result)) { 

      $answer = $row['answer']; 

      // Comparing the database password with the posted password 
      if ($answer == $answerPost) { 



      } else { 

       $errors = "yes"; 
       $message = "Your answers did not match the answers inside the database!"; 

       $output = array('errorsExist' => true, 'message' => $message); 

      } 


     } 

    } else { 

     $errors = "yes"; 
     $message = "We did not find any answers for your questions! Please consult the site administrator!"; 

     $output = array('errorsExist' => true, 'message' => $message); 

    } 

} 

} 

//Output the result 
$output = json_encode($output); 
echo $output; 

?> 

Answer 1

：ところで

$query = "SELECT * FROM manager_users_secretAnswers WHERE userID = '$userID'"; 
$result = mysqli_query($dbc,$query); 

if (mysqli_num_rows($result) > 0) { 
    while ($row = mysqli_fetch_array($result)) { 
     print_r($row); 
    } 
} 

を、あなたはSQL injectionを避けるためにprepared statementsを使用する必要があります。

Answer 2

ループ内でフェッチをラップする必要があります。例えば

if (mysqli_num_rows($result) > 0) 
{ 
    while (($row = mysqli_fetch_array($result)) !== false) 
    { 
    if ($row['answer'] == $answerPost) 
    { 
     // $row matches what we're looking for 
    } 
    else 
    { 
     $errors = "yes"; 
     $message = "Your answers did not match the answers inside the database!"; 
     $output = array('errorsExist' => true, 'message' => $message); 
    } 
    print_r($row); 
    } 
}