Question

我正在制作食谱网站，为了处理搜索查询，我正在一个名为“search.php”的单独页面上处理表格：

<form id="recipehome" form method="post" action="search.php">
 <p>I want a 
         <Select name="dropoption">
             <option value="empty">----</option>     
            <?php echo $cuisinetype;?>    
            </Select>

            <Select name='meal'>
                    <option value="B">Breakfast</option>
                    <option value="l">Lunch</option>
                    <option value="D">Dinner</option>
            </Select>
        <input type="submit" id="submit" name="Feed Me" value="submit" /> 
       </form>

我的搜索页面：

if(trim($_POST['submit']) =="submit"){
        }else{              
            if (isset($_POST['dropoption']) && ($_POST['dropoption'] != '')){
                if (isset($_POST['meal']) && ($_POST['meal'] != '')) {

                $dropoption = clean_string($db_server, $_POST['dropoption']);
                $meal = clean_string($db_server, $_POST['meal']);
                $quer = "SELECT recipeid FROM `recipename` WHERE `cuisine_type` ='$dropoption' AND `b_l_d` ='$meal'LIMIT 0,1";
                mysqli_select_db($db_server, $db_database);
                $querya= mysqli_query($db_server, $quer); 
                if (!$querya) die("database access failed: " . mysqli_error($db_server));
                while($row = mysqli_fetch_array($querya)){
                            $searchresult .=$row['recipeid'];
                            }
                $searchresult = clean_string($db_server, $searchresult);            
                header("Location:results.php?recipeid=$searchresult");//
                }//if(meal)//
            }//if(cuisine)//    
        } //if(trim)//
    }

然后在那里执行查询，然后我试图通过传递我得到的变量从那里重定向到另一个结果页面我已经设法将其放入url但我现在遇到问题现在将这个整数插入我的数据库作为查询。此页面名为results.php

if (!$db_server){
    die("unable to Connect to MYSQL: " . mysqli_connect_error($db_server));
    $db_status = "not connected";
    $query = "SELECT * FROM `recipename` WHERE `recipeid`='$searchresult'";
            mysqli_select_db($db_server, $db_database);
            $result=mysqli_query($db_server, $query); 
            if (!$result) die("database access failed: " . mysqli_error($db_server));
                    while($row = mysqli_fetch_array($result)){
                        $recipename .="<h1>". "Why dont you have ".$row['mealname']."</h1>";
                        $ingredients .="<p>".$row['ingredients']."</p>";
                        $recipe .="<p>" .$row['recipe']."</p>";
                        $cookingtime .="<h4>" .$row['hours']." Hours".$row['minutes']." Minutes </h4>";
                        $mealpic .="<img src='http://ml11maj.icsnewmedia.net/Workshops/Week%207/".$row['imagepath']."'/>"; 

                        if(trim($_POST['Submit']) =="submit3"){
                            if ($comment != ''){
                                $userid = trim($_SESSION['userid']);
                                $comment = trim($_POST['comment']);
                                $userid = clean_string($db_server, $_SESSION['userid']);    
                                $comment = clean_string($db_server, $_POST['comment']);

                                        $query = "INSERT INTO Comments (comment,userid,recipeid) VALUES ('$comment','$userid','$receipeid')";
                                        mysqli_query($db_server, $query) or
                                        die("Insert failed: " . mysqli_error($db_server));
                                        $message = "Thanks for your comment!";  
                    }
                }
            }
}

Answer 1

首先你需要得到变量。在results.php页面上，尝试在查询之前使用它：

$searchresult = $_GET['recipeid'];

通过标头传递变量

1 个答案: