Question

我正在努力解决这两个输入INSERT查询问题。它不会向db插入任何内容。首先，它应该接受用户输入和下拉菜单选择，这是来自另一个表的id因此SELECT查询。修复此代码后，我将通过'pubid'设置两个表之间的关系。

<?php

   error_reporting(E_ALL); ini_set('display_errors', 1);

   require_once("dbconn.php");

   try {
      $conn = new PDO("mysql:host=$dbhost;dbname=$dbname;charset=utf8",$dbuser,$dbpass,$dbo);
   } catch (PDOException $e) {
      echo $e->getMessage();
      exit;
   }

   if(isset($_POST['addBtn'])){
      $q = $conn->prepare("INSERT INTO series (title, pubid) VALUES (:title, :pubid)");
      $q->bindParam(':title',$title,PDO::PARAM_STR);
      $q->bindParam(':pubid',$pubid,PDO::PARAM_INT);
      $q->execute();
   }

   $sql = 'SELECT pubid, name FROM publisher ORDER BY name';
   $q = $conn->query($sql);
   $q->setFetchMode(PDO::FETCH_ASSOC);

?>
<!DOCTYPE html>
<html lang="eng">
   <head>
      <meta charset="UTF-8">
      <title>
         Comics DB > Add Series
      </title>
      <link rel="stylesheet" type="text/css" href="comicsdb.css">
   </head>
 <body>
    <div class="main">
       <menu>
          <ul>
             <li><a href="index.php">Home</a></li>
             <li><a href="add.php">Add</a></li>
             <li><a href="edit.php">Edit</a></li>
             <li><a href="delete.php">Delete</a></li>
             <li><a href="list.php">List</a></li>
             <li><a href="search.php">Search</a></li>
          </ul>
      </menu>
    <div class="pub_menu">
       <form action="addseries.php" method="POST">
          <p>
            Series: 
            <input type="text" name="title" id="title" size="40" /><br />
            <select>
                <option>Select</option>
                <?php while ($row = $q->fetch()){ ?>
                <option name="pubid" id="pubid" value="<?php echo $row['pubid']; ?>"><?php echo $row['name']; ?></option>
                <?php } ?>
            </select>
            <input type="submit" value="Add Series" name="addBtn" />
         </p>
       </form>
     </div>
   </body>
</html>

dbconn.php

<?php

$dbhost         = '127.0.0.1';
$dbname     = 'comicsdb';
$dbuser         = 'root';
$dbpass     = 'FuckYou';
$dbport         = '3306';
$charset    =   'utf8';
$dbo            = array(
// important! use actual prepared statements (default: emulate prepared statements)
PDO::ATTR_EMULATE_PREPARES => false
// throw exceptions in case of errors (default: stay silent)
, PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION
// fetch associative arrays (default: mixed arrays)
, PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC
);

[编辑]

<?php

   error_reporting(E_ALL); ini_set('display_errors', 1);

   require_once("dbconn.php");

   try {
      $conn = new PDO("mysql:host=$dbhost;dbname=$dbname;charset=utf8",$dbuser,$dbpass,$dbo);
   } catch (PDOException $e) {
      echo $e->getMessage();
      exit;
   }

   if(isset($_POST['addBtn'])){
      $title = $_POST['title'];
      $pubid = $_POST['pubid'];
      $q = $conn->prepare("INSERT INTO series (title, pubid) VALUES (:title, :pubid)");
      $q->bindParam(':title',$title,PDO::PARAM_STR);
      $q->bindParam(':pubid',$pubid,PDO::PARAM_INT);
      $q->execute();
  }

  $sql = 'SELECT pubid, name FROM publisher ORDER BY name';
  $q = $conn->query($sql);
  $q->setFetchMode(PDO::FETCH_ASSOC);

?>
<!DOCTYPE html>
<html lang="eng">
   <head>
      <meta charset="UTF-8">
      <title>
         Comics DB > Add Series
      </title>
      <link rel="stylesheet" type="text/css" href="comicsdb.css">
   </head>
   <body>
      <div class="main">
         <menu>
            <ul>
               <li><a href="index.php">Home</a></li>
               <li><a href="add.php">Add</a></li>
               <li><a href="edit.php">Edit</a></li>
               <li><a href="delete.php">Delete</a></li>
               <li><a href="list.php">List</a></li>
               <li><a href="search.php">Search</a></li>
            </ul>
         </menu>
         <div class="pub_menu">
            <form action="addseries.php" method="POST">
               <p>
                  Series: 
                  <input type="text" name="title" id="title" size="40" /><br />
                  <select>
                     <option name="pubid" id="pubid">Select</option>
                     <?php while ($row = $q->fetch()){ ?>
                     <option  value="<?php echo $row['pubid']; ?>"><?php echo $row['name']; ?></option>
                     <?php } ?>
                  </select>
                  <input type="submit" value="Add Series" name="addBtn" />
              </p>
           </form>
        </div>
    </body>
</html>

Answer 1

您没有为变量分配任何内容。

将它们分配给POST阵列：

if(isset($_POST['title'],$_POST['pubid'])){

$title = $_POST['title'];
$pubid = $_POST['pubid'];

      $q = $conn->prepare("INSERT INTO publisher (title, pubid) VALUES (:title, :pubid)");
      $q->bindParam(':title',$title,PDO::PARAM_STR);
      $q->bindParam(':pubid',$pubid,PDO::PARAM_INT);
      $q->execute();
   }

感谢幽灵抓住这个。

<select>带有name属性，而不是<option>。

将您的<select>更改为<select name="pubid">，然后从name="pubid"中移除<option>。

修改

根据您的编辑，您仍然拥有选择的未命名属性：

<select> <option name="pubid" id="pubid">Select</option>

需要阅读：

<select name="pubid"> <option>Select</option>

这就是为什么你仍然得到一个未定义的索引通知。

另外，你可以在里面移动你的身份：

<select name="pubid" id="pubid">

PHP; PDO;带有下拉列表的MySQL INSERT查询

1 个答案: