Question

学习PHP OOP方式：

将数据库类对象传递给其他类我收到此消息＆＃39;可捕获的致命错误：类PDO的对象无法转换为字符串＆＃39;同时，如果我对此行发表评论并且＃39; //$imageobj= new image($dbobj);＆＃39; ？

更新：在try {}之前，我在db.class.php中将此行$this->dbconn=null;添加到getConnection()函数。一切都很好!!!

的index.php

<?php
include "./classes/db.class.php";
include "./classes/image.class.php";

$dbobj= new db();
$imageobj= new image($dbobj);

$page_title="Shopping Center !";
include './template/header.php';

$dbobj->getConnection();

include './template/footer.php';

db.class.php

class db {
    private $host;
    private $dbname;
    private $username;
    private $password;
    private $dbconn;
    private $status;

    public function __construct() {
        // the require paramaters to start db connnection
        $this->host = '127.0.0.1';
        $this->dbname = 'shop_carta';
        $this->username = 'root';
        $this->password = '';
        $this->status = 0;
        //$this->dbconn = null;
    }

    public function getConnection() {
        try {
            $this->dbconn = new PDO("mysql:host=$this->host;dbname=$this->dbconn", $this->username, $this->password);
            if (!is_null($this->dbconn)) {

                $this->status = $this->dbconn->getAttribute(PDO::ATTR_CONNECTION_STATUS);
                $this->dbconn->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_ASSOC);
                $this->dbconn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

                return $this->dbconn;
            } else {
                echo "<div class='alert alert-danger'>"
                . " Our server Busy Now try again later.. </div></br>";
                return false;
            }
        } catch (PDOException $ex) {
              return false;
        }}}

image.class.php 这个类接收$ dbobject尝试$ dbobject-＆gt; getConncetion（）并设置retrieve

class image {
    private $table_name = 'product_images';
    private $dbconn;
    public $id;
    public $name;

    public function __construct($dbobj) {
         $this->dbconn = $dbobj->getConnection();
       }}

将dbconn指定为null solve： 捕获致命错误：类PDO的对象无法转换为字符串。有什么建议吗？

Answer 1

问题在于：

    new ExtractTextPlugin({
        filename: '[name].bundle.css'
    })

dbname应该是数据库名称。您提供的属性（在添加$ this-＆gt; dbconn = null之前）不存在，这就是您收到错误的原因。

添加$ this-＆gt; dbconn = null，确实可以防止非existant属性上的致命错误，但您仍然提供空数据库名称。

您应该通过将其替换为$ this-＆gt; dbname来修复它，如下所示：

"mysql:host=$this->host;dbname=$this->dbconn"

PDO无法转换为字符串php

1 个答案: