Question

我是php的新手，我已经创建了一个项目，它将数据库中的数据作为用户选择，当项目执行时，它在某个方法上发生了错误，而我现在还没有如何解决它。听到是PHP代码： -

     <?php

      $db = new PDO('sqlite:dinner.db');
      $meals = array('breakfast','lunch','dinner');

      if(in_array( $_POST['meal'], $meals)){

     $stmt = $db->prepare('SELECT dish,price FROM meals WHERE meal LIKE ?');
     $stmt->execute(array($_POST['meal']));
    $rows = $stmt->fetchAll();

      if(count($rows) == 0){
         print "No dishes available now";
      } else {

          print "<table><tr><th>dish</th><th>price</th></tr>";
          foreach($rows as $row) {
           print "<tr><td>$row[0]</td><td>$row[1]</td></tr>";
      }
          print "</table>";
      }
} else {
   print "Unknown meal";
  }

  ?>

那就是html页面代码：

          <html>
         <head>
         </head>
         <body>

      <form method="post" action="test.php">
      <lable>Select the meal to see all the available dishes:
      <input type="text" name="meal">


      </lable><input type="submit" value="Ok">
     </form>

   </body>
  </html>

，这是浏览器的输出： - enter image description here

Answer 1

根据PHP Docs execute，不应该有任何参数（void），它只会执行当前查询。

您应该通过bindValue或bindParam传递参数来填充查询。 Look at his example以正确的方式看待这种方式。

也许像这样的东西修复了它，$meal必须只有一个字符串而不是数组：

$stmt = $db->prepare('SELECT dish, price FROM meals WHERE meal LIKE ?');
$stmt->bindValue(1, $meal, SQLITE3_TEXT);

Answer 2

将你的参数传递给查询你需要使用bindparams而不是在这里执行如何：

<?php

  $db = new PDO('sqlite:dinner.db');
  $meals = array('breakfast','lunch','dinner');

  if(in_array( $_POST['meal'], $meals)){

 $stmt = $db->prepare('SELECT dish,price FROM meals WHERE meal LIKE ?');
 $stmt->bindParam(1, $_POST['meal']);
 $stmt->execute();
$rows = $stmt->fetchAll();

  if(count($rows) == 0){
     print "No dishes available now";
  } else {

      print "<table><tr><th>dish</th><th>price</th></tr>";
      foreach($rows as $row) {
       print "<tr><td>$row[0]</td><td>$row[1]</td></tr>";
  }
      print "</table>";
  }
} else { print "Unknown meal";}  ?>

PHP，数据库方法的错误

2 个答案: