Question

我在index.php页面中没有看到任何数据。我用这段代码从mysql数据库中获取具有唯一id的数据，但我不能。

include "db/db.php";    
$upload_path = "secure/content/blogpostimg";

<?php
if (isset($_GET['post_id']) && $_GET['post_id'] != '')
{
$p_id = (int) $_GET['post_id'];
}   
$sql = mysql_query("SELECT * FROM blog_post WHERE post_id = '$pid' ORDER BY post_id 
DESC ");

while ($rel = mysql_fetch_assoc($sql))
{
$id = $rel['post_id'];
$sub = $rel['subject'];
$imgname = $rel['img_name'];
$msg = $rel['message'];
$date = $rel['date'];
$poster = $rel['poster'];
$cat_name = $rel['cat_name'];

echo "<h1>". "$sub" ."</h1>". "<br/>";
echo '<img src="' . $upload_path . '/' . $imgname . '" width="200"  />  ';       
include_once("func.php");
echo truncate($rel['message'],"index.php","post_id",$rel['post_id']);      
echo "$date&nbsp;&nbsp;" . "<b>Category:</b>    
&nbsp;$cat_name". "&nbsp;&nbsp;".  "<b>by:</b>&nbsp;&nbsp;" . "$poster&nbsp;&nbsp;" . 
"<b>Comemnts</b>[&nbsp;  ]" ;
}
?>

Answer 1

前两行代码在open标记之前：＆LT; PHP

Answer 2

error_reporting(E_ALL);
ini_set('display_errors',1);

在代码的最顶部添加这两行，并阅读问题的答案这将让你发现你开始的那些愚蠢的错别字。

现在，让我们对SQL做同样的事情

使你的代码像这样

if (isset($_GET['post_id']))
{
  $p_id = (int) $_GET['post_id'];
}
else
{
  die('Required variable is not set');
}  

$sql = "SELECT * FROM blog_post WHERE post_id = $p_id";
$res = mysql_query($sql) or trigger_error(mysql_error()." ".$sql);
if (!mysql_num_rows($res))
{
  die("No records found");
}

Answer 3

您有一个变量 $ p_id

但在查询中，您使用 $ pid 变量名。

我的帖子中没有看到任何数据

3 个答案: