PHP输出结果到一个页面

Question

我有一个系统，用户搜索电影和评论出现在页面上，每个评论旁边有一个按钮。该按钮可以选择看个人的评论，但我基本上要一个按钮，选择的时候会看一个页面上的所有评论，我使用的个别审查的代码是这样

<?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

可以取电影评论，因为检查表内膜ID
你可以修改上面的代码，并添加另一种形式得到所有电影评论 当上一个按钮，它会将他用户点击/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>"; 
     } 
    } 
}