Question

我需要一点帮助，了解如何在PHP中使用OOP来执行表单提交操作。手头的任务...... 我正在尝试学习如何使用OOP编写PHP代码。到目前为止，我理解了类，函数，调用函数，继承等的一般概念。

我创建了一个简单的练习项目，允许用户在某个位置搜索用餐。到目前为止，我有一个包含2个<input>字段的表单。通常对于表单操作，我会<form action="actionFileName.php">但是现在我有一个具有处理表单的函数的类，我将如何使用该操作值？

我想过创建一个类的实例并调用处理表单的函数但是我找不到一个Object！我提交的表单中包含地址栏中显示的echo else语句中的hungryClass.php值。

我该如何解决这个问题？感谢。

我的代码是什么样的： HTML表格

<?php  require_once 'hungryClass.php';
  $newSearch = new hungryClass();
?>
<form action="<?php $newSearch->searchMeal();?>" method="post" id="searchMealForm">
  <input type="search" size="35" placeholder="What Food Are you looking for?" id="mealName" class="meal"/>
  <input type="search" placeholder="City Area" id="mealLocation" class="meal">
  <input type="submit" value="Satisfy Me" id="findMeal" />
</form>

页面处理表单（hungryClass.php）

<?php 
  require_once('dbConnect.php');
  class hungryClass{

       public function searchMeal(){
         //call connection function.
          $connect = new dbConnect();

         //validate input
         if(isset($_POST['mealName'])){
             $meal = $_POST['mealName'];

           //ensure value is a string.
           $cleanse_meal = filter_var($meal, FILTER_SANITIZE_STRING);
           echo $cleanse_meal;
        }
        else{
       echo "Please supply the meal you crave";
        }

      //validate location
       if(isset($_POST['mealLocation'])){
     $location = $_POST['mealLocation'];

         //validate and sanitize input. ensure value is a string.
    $cleanse_location = filter_var($location, FILTER_SANITIZE_STRING);
        echo $cleanse_location;

       }
       else{
     echo "Please supply a location";
       }

}

数据库类

<?php

class dbConnect{
private $host = "localhost";
private $user = "stacey";
private $pass = "";
private $db_name = "menu_finder";

private $connect;
//private static $dbInstance;

public function __construct(){
    try{
        $this->connect = new mysqli($host, $user, $pass, $db_name);
        if(mysqli_connect_error()){
           die('connection error('.mysqli_connect_errno().')' . mysqli_connect_error());
        }
    }

    catch(Exception $e){
        echo $e->getMessage();
    }
}

＆GT;

Answer 1

表单的action属性适用于您要提交的脚本名称。您希望将表单提交给要处理的饥饿类，但在实例化饥饿类之后才能执行此操作。您需要使用脚本名称作为表单中的操作值。假设你想提交给temp.php，你的表单应该是这样的

<form action="temp.php" method="post" id="searchMealForm">
  <input type="search" size="35" placeholder="What Food Are you looking for?" id="mealName" class="meal"/>
  <input type="search" placeholder="City Area" id="mealLocation" class="meal">
  <input type="submit" value="Satisfy Me" id="findMeal" />
</form>

然后当这个表单提交时，它将被发送到temp.php。要让您的饥饿类处理此表单，您需要在temp.php中创建它的实例，并使用此实例调用searchMeal。 temp.php应该看起来像这样

<?php
  require_once 'hungryClass.php';
  $newSearch = new hungryClass();
  $newSearch->searchMeal();
?>

或将所有内容放在一个文件中

<?php
require_once 'hungryClass.php';

if($_SERVER['REQUEST_METHOD'] == 'POST') {
  $newSearch = new hungryClass();
  $newSearch->searchMeal();
  exit();
}
?>
<form action="<? echo $_SERVER['PHP_SELF']?>" method="post" id="searchMealForm">
  <input type="search" size="35" placeholder="What Food Are you looking for?" id="mealName" class="meal"/>
  <input type="search" placeholder="City Area" id="mealLocation" class="meal">
  <input type="submit" value="Satisfy Me" id="findMeal" />
</form>

Answer 2

您应该将表单提交到将处理和处理表单的php文件。

<form action="<?php $newSearch->searchMeal();?>" method="post" id="searchMealForm">

应该是这样的：

<form action="formaction.php" method="post" id="searchMealForm">

在formaction.php中你可以调用你的方法，当然你需要包含所需的文件：

<?php
$newSearch->searchMeal();

希望这有帮助。

Answer 3

这是php中最简单的OOP代码你可以按照以下方式尝试

<强> form.php的

<?php
  require_once 'DbClass.php';
?>
<form action="formaction.php" method="post" id="searchMealForm">
  <input type="search" size="35" name='search1' placeholder="What Food Are you looking for?" id="mealName" class="meal"/>
  <input type="search" placeholder="City Area" name='search2' id="mealLocation" class="meal">
  <input type="submit" value="Satisfy Me" id="findMeal" name='butsearch' />
</form>

<强> formaction.php

<?php
 require_once 'DbClass.php';
 $obj = new DbClass();
 if(isset($_REQUEST['butsearch']))
 {
    $ser = $_REQUEST['search1'];
    $ser2 = $_REQUEST['search2'];
    $inf0 = array('ser1'=>$ser,'ser2'=>$ser2)
    $obj->search($info);
 }
?>

<强> DbClass.php

<?php
//if any file needs to be included, include here
 class Dboper
 {
   public function __construct() {
     //DB Connection Code here
   }
   function serach($params)
   {
     $ser1 = $params['ser1'];
     $ser2 = $params['ser2'];
     //write query to search here
    // call the corresponding page to display the result
   }

 }

?>

如果您有任何进一步的疑问，请告诉我

Answer 4

OOP！..

1.必须有一个条目，其他主要是控制器。以下方式：

action.php的

<?php
   include 'common.inc.php'; //they are hungryClass,dbConnect etc that you need required;
    $do=$_POST['do'];
   $hungry=new hungryClass();
   if(!empty($do)){
   if(method_exists($hugry,$do)){
        $hugry->$do();
   }else
        echo 'method not exists;'
   }
   }
  ?>

2.form.html

<form action="action.php" method="post" id="searchMealForm">
  <input type="search" size="35" placeholder="What Food Are you looking for?" id="mealName" class="meal"/>
  <input type="search" placeholder="City Area" id="mealLocation" class="meal">
  <input type="submit" value="Satisfy Me" id="findMeal" />
</form>

你不能只学习OOP，你也必须学习MCV ....

使用oop进行PHP表单提交

4 个答案: