Question

在UBUNTU 10.04上使用Eclipse 3.5.2（PDT，WEB工具，APTANA 2和使用XDEBUG进行调试）...

我无法弄清楚这个代码有什么问题。我作为.NET开发人员已经工作多年了，所以我想我知道如何阅读文档和配置基本应用程序，尽管我是LAMP的新手。

以下页面仅显示“PHP正在运行”以及“我”，“您”和“她”的第一个列表。我通过登录PhpMyAdmin双重检查了数据库凭据。

调试时，我进入core_db.php，变量全部列为<Uninitialized>，一切都停在$stmt->execute();行

list_users.php

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<?php include 'db/chore_db.php' ?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=encoding">
<title>Insert title here</title>
</head>
<body>
<?php echo 'Php is running'; ?>
<ul>
<?php $ar = array('me','you','her');
foreach ($ar as $user) {
    ?>
    <li><?php   echo $user;}?></li>
</ul>

<div>

<?php 
$db = new chore_db();
$myusers = $db->get_users(null,null);

foreach ($myusers as $u) {
    ?>


<ul>
    <li><?php echo $u;  }?></li>
</ul>
</div>
</body>
</html>

chore_db.php

<?php    
class chore_db{
    /**
     * @param int $id
     * @param string $user_name
     * @return Ambigous <multitype:, usersTbl>
     */
    public function  get_users($id, $user_name){
        $users = array();
        $sql = '';
        $con = null;


        $sql = 'SELECT id, chore_type, name, description, created_dt, created_by, modified_dt, modified_by FROM chores';// where id = IFNULL(?,id) and name = IFNULL(?,name)";
        $con = new mysqli('localhost','chore_admin','!chore_admin','chores');

        $stmt = $con->prepare($sql);

        //bind parameters
        //$stmt->bind_param('ss', $id_param, $name_param);
        //$id_param = $id;
        //$name_param = $user_name;
        //execute the sql

        $stmt->execute();

        //now bind the results to variables
        $stmt->bind_result($rs_id, $rs_chore_type, $rs_name, $rs_description, $rs_created_dt, $rs_created_by, $rs_modified_dt, $rs_modified_by);

        while($stmt->fetch()){
            $users[] = new usersTbl($rs_id, $rs_chore_type, $rs_name, $rs_description, $rs_created_dt, $rs_created_by, $rs_modified_dt, $rs_modified_by);

        }

        return $users;
    }
}



class usersTbl{
    public $id;
    public $user_name;
    public $first_name;
    public $last_name;
    public $created_dt;
    public $created_by;
    public $modified_dt;
    public $modified_by;

    public $user_chore_list;
    public $user_chore_date_list;

    //public function __construct(){}
    public function __construct($userId, $uName, $fName, $lName, $createdDT, $createdDT, $createdBY, $modifiedDT, $modifiedBy){
        $this->id = $userId;
        $this->user_name = $uName;
        $this->first_name = $fName;
        $this->last_name = $lName;
        $this->created_by = $createdBY;
        $this->created_dt = $createdDT;
        $this->modified_by = $modifiedBy;
        $this->modified_dt = $modifiedDT;
    }
}

Answer 1

问题是您正在尝试回显对象，而没有说明应如何格式化其内容。你应该仍然得到一些输出，所以显然还有其他东西在发生，但这里有一个版本已修复很多问题：

<强> list_users.php

<?php include 'db/chore_db.php' ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
  <head>
    <meta http-equiv="Content-Type" content="text/html; charset=encoding">
    <title>Insert title here</title>
  </head>
  <body>
    <?php echo 'Php is running'; ?>
    <ul>
      <?php 
        $ar = array('me','you','her');
        foreach ($ar as $user) {
      ?>
      <li><?php echo $user; ?></li>
    <?php } ?>
    </ul>
    <div>
      <ul>
        <?php 
          $db = new chore_db();
          $myusers = $db->get_users(null,null);
          foreach ($myusers as $u) {
        ?>
        <li><?php echo $u->user_name; ?></li> <!-- You need to echo a property of the object (or several, rather than the object itself... -->
        <?php } ?>
      </ul>
    </div>
  </body>
</html>

Answer 2

<ul>
<?php $ar = array('me','you','her');
foreach ($ar as $user) {
    ?>
    <li><?php   echo $user;}?></li>  <<< this is at least one of the problems
</ul>

简单的php页面：我在这里做错了什么？

2 个答案: