PHP Prepared Statement - 回显值

Question

我最近使用准备好的语句修复了我的旧的且非常不安全的 MySQL 查询。但是，当我运行这个新查询时，除了使用 print_r 进行调试外，不会返回任何值。如果我只是简单地回显该值（即回显 $eventid），我将收到一个错误，即我无法将数组转换为字符串。我是否需要为此查询添加某种 foreach 循环？非常感谢大家继续帮助我学习！ :)

   <?php
echo "<div class='container target'>

        <div class='container'>    
    <div id='quick-access'>
      <form class='form-inline quick-search-form' role='form'>
        <div class='form-group'>
                <input type='text' id='name' name='name' class='form-control' placeholder='Driver name'>
        </div>
        <button type='submit' id='quick-search' class='btn btn-custom'><span class='glyphicon glyphicon-search custom-glyph-color'></span></button>
      </form>
    </div>
</div>";
?>


<?php

$conn = new mysqli('localhost', 'Username', 'password', 'Database');

if($conn->connect_errno > 0)
{
    die('Unable to connect to database [' . $conn->connect_error . ']');
}

# SQL WITH QMARK PLACEHOLDER (USING TABLE ALIASES)
$sql = "SELECT e.personaId, e.ID AS event_id, e.EVENTID, e.rank, e.carId, 
               e.alternateEventDurationInMilliseconds, 
               p.iconIndex, p.cash, p.level, p.created, p.score, p.motto, 
               p.repAtCurrentLevel, p.rep, p.name AS p_name
        FROM EVENT_DATA e
        INNER JOIN PERSONA p ON e.personaId = p.ID
        WHERE (p.name = ? AND e.EVENTID = '43' 
          AND  e.alternateEventDurationInMilliseconds > '0')";

# INITIALIZE ARRAYS - TWO METHODS: array() or []
$name = array(); $avatarimg = []; $cash = array(); $level = array(); $createddate = [];
$driverscore = array(); $motto = []; $repcurrent = array(); $reptotal = []; 
$personaid = array(); $eventid = []; $milliseconds = array();

# PREPARED STATEMENT
$stmt = mysqli_prepare($conn, $sql) or die(mysqli_error($conn));

# BIND PARAM 
$param=mysqli_real_escape_string($conn, $_GET['name']);
mysqli_stmt_bind_param($stmt, "s", $param);


# EXECUTE STATEMENT
if (mysqli_stmt_execute($stmt)){
    $result = $stmt->get_result();
    # CHECK ROWS 
    if(mysqli_num_rows($result) > 0)  { 
       # ITERATE THROUGH ROWS
       while ($row = mysqli_fetch_array($result)){
            # APPEND TO ARRAYS
            $name[] = mysqli_real_escape_string($conn, $row['p_name']);
            $avatarimg[] = mysqli_real_escape_string($conn, $row['iconIndex']);
            $cash[] = mysqli_real_escape_string($conn, $row['cash']);
            $level[] = mysqli_real_escape_string($conn, $row['level']);
            $createddate[] = mysqli_real_escape_string($conn, $row['created']);
            $driverscore[] = mysqli_real_escape_string($conn, $row['score']);
            $motto[] = mysqli_real_escape_string($conn, $row['motto']);
            $repcurrent[] = mysqli_real_escape_string($conn, $row['repAtCurrentLevel']);
            $reptotal[] = mysqli_real_escape_string($conn, $row['rep']);
            $personaid[] = $row['personaId'];
            $eventid[] = $row['EVENTID'];
            $milliseconds[] = $row['alternateEventDurationInMilliseconds'];    
       }    
    }    
}



 print_r($name);
  print_r($eventid);


?>

Answer 1

当您使用$name[] = mysqli_real_escape_string($conn, $row['p_name']); 时，它会将您的选择查询的每一个结果添加到一个数组中。这就是为什么当您尝试回显时它不起作用的原因。

这是查看它的简单方法。

foreach ($name as $n) {
    echo $n . '<br />';
}