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PHP PDO fetchObject返回2个实例

fetch pdo php

Paul Slater · 技术社区 · 6 月前

我正在学习PHP和MySQL。我有一个数据库,正在进行一些检索,但当我使用fetchObject()时,我返回了2个实例。第一次尝试:

class Food {
  // Properties
  public $id;
  public $version;
  public $name;
  public $solid;
  public $servingtext;
  public $servingmeasure;
  public $servingamount;

  private function __construct__() {}

  public static function get_with_id ($dbconn,$id, $version) {
      $stmt = $dbconn->prepare('SELECT * FROM Foods WHERE Id = :Id and Version = :Version;');
      $stmt->execute([':Id' => $id, ':Version' => $version]);
      return  $stmt->fetchObject(__CLASS__);
    //   return $answer;
   }

  // Methods
  function set_id($id){
    $this->id = $id;
  }
   etc. with sets and gets

打电话给

require "./templates/header.php";

try 
{
    include __DIR__ . '../../connection.php';
    $conobj = new Connection();
    $dbconn = $conobj->get_connection();
} catch (Throwable $t) {
    error_log(__FILE__ . ':' . $t->getMessage());
    echo 'Connection Error ' . $t->getMessage();
}
require "./models/food.php";
$food = Food::get_with_id ($dbconn,2,0);
var_dump($food);

结果:

object(Food)#4 (14) { ["id"]=> NULL ["version"]=> NULL ["name"]=> NULL ["solid"]=> NULL ["servingtext"]=> NULL ["servingmeasure"]=> NULL ["servingamount"]=> NULL ["Id"]=> int(2) ["Version"]=> int(0) ["Name"]=> string(35) "Sanitarium Weet-Bix Blends Hi-Bran+" ["Solid"]=> int(1) ["ServingText"]=> string(28) "1 serving = 40g (2 biscuits)" ["ServingMeasure"]=> string(1) "g" ["ServingAmount"]=> int(40) }

应该只有7个字段,但第一组都是NULL,然后就是我想要的对象。

我试过了。。

<?php
class Food {
  // Properties
  public $id;
  public $version;
  public $name;
  public $solid;
  public $servingtext;
  public $servingmeasure;
  public $servingamount;

  private function __construct__() {}

  // Methods
  function set_id($id){
    $this->id = $id;
  }

具有

<?php
require "./templates/header.php";

try 
{
    include __DIR__ . '../../connection.php';
    $conobj = new Connection();
    $dbconn = $conobj->get_connection();
} catch (Throwable $t) {
    error_log(__FILE__ . ':' . $t->getMessage());
    echo 'Connection Error ' . $t->getMessage();
}
require "./models/food2.php";
$id = 2;
$version = 0;
$stmt = $dbconn->prepare('SELECT * FROM Foods WHERE Id = :Id and Version = :Version;');
$stmt->execute([':Id' => $id, ':Version' => $version]);
$food = $stmt->fetchObject('Food');
var_dump($food);

同样的结果。然后用同样的食物类尝试:

<?php
require "./templates/header.php";

try 
{
    include __DIR__ . '../../connection.php';
    $conobj = new Connection();
    $dbconn = $conobj->get_connection();
} catch (Throwable $t) {
    error_log(__FILE__ . ':' . $t->getMessage());
    echo 'Connection Error ' . $t->getMessage();
}
require "./models/food2.php";
$id = 2;
$version = 0;
$food = $dbconn->query('SELECT * FROM Foods WHERE Id = ' . $id . ' AND Version = ' . $version . ';')->fetchObject('Food');
var_dump($food);

再次,结果有2个条目。

有人能帮我解释一下为什么我要先得到NULLs版本吗。我不想要也不需要它。很高兴被告知这是一个新手错误!我在检索数组方面取得了很多成功,但我想走OO的道路。

谢谢你,保罗

Docker容器中的PHP 8.1.30、MySQL 9.1.0和NGINX 1.27.2

1 回复 | 直到 6 月前

lemon8de 6 月前

fetchObject() 生成类,并在您将其添加到类中时为您提供前7列null Food ,它们默认为 NULL

// Your properties, they all get NULL
public $id;
public $version;
public $name;
public $solid;
public $servingtext;
public $servingmeasure;
public $servingamount;

然后fetchObject()注入更多变量:结果查询的列也变成类属性,使其看起来像是重复的,但我们可以从vardump()的结果中跟踪它。

id 重复与 Id

version 重复与 Version

name 重复与 Name

...

使公共属性变量与查询中获得的列名匹配。从 身份证件 到 身份证件 , servingtext 到 ServingText 等等。

或者找到一种在不使用的情况下生成类的方法 fetchObject() .

https://www.php.net/manual/en/pdostatement.fetchobject.php