下面的代码是给别人打工，但不适合我

Question

当我尝试搜索形式上的东西它让我看到了所有的变量错误...

<form action="search.php" method="get"> 
    <input type="text" size="25" name="search" placeholder="search this site"> 
    <input type="submit" name="submit" value="search"></form> 
    <?php 
    include("includes/connect.php"); 
     if(isset($_GET['submit'])){ 
     $search_id = $_GET['search']; 
     $query = "select * from posts where post_title like '%search_id%'"; 
     $run = mysql_query($query); 
     while ($row=mysql_fetch_array($run)){ 
     $post_id = $row['post_id']; 
     $post_title = $row['post_title']; 
     $post_image = $row['post_image']; 
     $post_content = $row['post_content']; 
    } 
    ?> 
     <h2> 
     <a href="pages.php?id=<?php echo $post_id; ?>"> 
     <?php echo $post_title; ?></a> 
     </h2> 
    <center><img src="images/<?php echo $post_image; ?>" width="100" height="100"></center> 
     <p align="justify"><?php echo $post_content; ?></p> 
    <?php } ?> 

我认为这个问题是有循环的功能。但不知道如何解决这个...

Answer 1

你在发言中忘了$ -sign：

$query = "SELECT * FROM `posts` WHERE `post_title` like '%$search_id%'"; 

顺便说一句：请不要使用的mysql_query了，其弃用。改用mysqli或PDO。 :)

你也很容易mysql注入。尝试跳过$_GET值：mysql_real_escape_string（$ _ GET ['搜索']）

此外，你正在覆盖你的while循环变量，因为变量的调用是在while循环之外。将这些变量放在while循环中，如下所示：

while ($row=mysql_fetch_array($run)){ 
    $post_id = $row['post_id']; 
    $post_title = $row['post_title']; 
    $post_image = $row['post_image']; 
    $post_content = $row['post_content']; 
?> 
<h2> 
    <a href="pages.php?id=<?php echo $post_id; ?>"><?php echo $post_title; ?></a> 
</h2> 
<center> 
    <img src="images/<?php echo $post_image; ?>" width="100" height="100"> 
</center> 
<p align="justify"> 
    <?php echo $post_content; ?> 
</p> 
<?php 
    }//<--- this one closes the while loop 
    }//<--- this one closes the if statement 
?>