PHP輸出結果到一個頁面

Question

我有一個系統，用戶搜索電影和評論出現在頁面上，每個評論旁邊有一個按鈕。該按鈕可以選擇看個人的評論，但我基本上要一個按鈕，選擇的時候會看一個頁面上的所有評論，我使用的個別審查的代碼是這樣

<?php 

ini_set ('display_errors', 1); 
error_reporting (E_ALL & ~E_NOTICE); 


$searchfilm=$_POST['searchfilm']; 


//Connect to database 



//Filter search 
$searchfilm = strtoupper($searchfilm); 
$searchfilm = strip_tags($searchfilm); 
$searchfilm = trim ($searchfilm); 

$query = mysql_fetch_assoc(mysql_query("SELECT filmreview FROM review WHERE id = '$id'")); 
$data = mysql_query("SELECT film.filmname, review.filmreview, review.reviewtitle, review.id FROM film, review WHERE film.filmid = review.filmid AND filmname = '$searchfilm'"); 

while($row = mysql_fetch_assoc($data)) 
{ 
    // echo $row['filmname']; 
    // echo "<b>Film Name:</b> " .$searchfilm; 
    echo "<table border=\"2\" align=\"left\">"; 
    echo "<tr><td>"; 
    echo "<b>Review Title:</b> " .$row['reviewtitle']; 
    echo "<tr><td>"; 
    echo $row['filmreview']; 
    echo "<p>"; 
    echo "<form method='post' action='analyse1.php'>"; 
    echo "<input type='hidden' name='reviewid' value='".$row['id']."'>"; 
    echo "<input type='submit' name='submit' value='Analyse'>"; 
    echo "</form>";  
    echo "</table>"; 

} 

?> 

Answer 1

可以取電影評論，因爲檢查表內膜ID
你可以修改上面的代碼，並添加另一種形式得到所有電影評論 當上一個按鈕，它會將他用戶點擊/她film_reviews.php

<?php 
if(isset($_POST['submit']) && $_POST['submit'] == "getAllReviews"){ 

    $filmID = mysql_real_escape_string($_POST['filmid']); 

    $sql = "SELECT * FROM review WHERE filmid = '$filmID'"; 
    $res = mysql_query($sql); 
    if(is_resource($res)){ 
     while($row = mysql_fetch_array($res)){ 
      echo "<p>".$row['reviewtitle']."</p>"; 
      echo "<p>".$row['filmreview']."</p>"; 
     } 
    } 
}