我在使用这段代码时遇到问题
include '01.php';
include 'header.php';
$sql = "SELECT cat_id, cat_name, cat_description FROM categories WHERE cat_id = " . mysql_real_escape_string($_GET['id']) . "";
$result = mysql_query($sql);
if(!$result)
{
echo 'The category could not be displayed, please try again later.' . mysql_error();
}
khác
{
if(mysql_num_rows($result) == 0)
{
echo 'This category does not exist.';
}
khác
{
//display category data
while($row = mysql_fetch_assoc($result))
{
echo 'Topics in ′' . $row['cat_name'] . '′ category
';
}
//do a query for the topics
$sql = "SELECT
topic_id,
topic_subject,
topic_date,
topic_cat
FROM
topics
WHERE
topic_cat = " . mysql_real_escape_string($_GET['id']);
$result = mysql_query($sql);
if(!$result)
{
echo 'The topics could not be displayed, please try again later.';
}
khác
{
if(mysql_num_rows($result) == 0)
{
echo 'There are no topics in this category yet.';
}
khác
{
//prepare the table
echo '
Topic |
Created at |
';
while($row = mysql_fetch_assoc($result))
{
echo '';
echo ''; echo ' | ';
echo ''; echo date('d-m-Y', strtotime($row['topic_date'])); echo ' | ';
echo '
';
}
}
}
}
}
include 'footer.php';
?>
它给我的错误是这个 The category could not be displayed, please try again later.你的SQL语法有错误;查看与您的 MySQL 服务器版本对应的手册,了解在第 1 行 strong text ** 附近使用的正确语法 $sql = "SELECT cat_id, cat_name, cat_description FROM categories WHERE 由于某种原因一直无法工作strong text**非常感谢您的帮助,我不知道该怎么做,到处搜索并尝试了不同的组合,但无法从 cat_id 获取 ID 请帮助
您似乎没有合适的报价。您应该将 mysql_real_escape_string($_GET['id'])
的结果包含在 '
中像下面的示例
"SELECT cat_id, cat_name, cat_description
FROM categories WHERE cat_id = '" . mysql_real_escape_string($_GET['id']) . "'";
如果您想尝试使用简化的查询,请记住这样的引号:
$sql= mysql_query("SELECT cat_id, cat_name, cat_description
FROM categories WHERE cat_id = '" .$id . "';");
但为了调试尝试
var_dump($id);
var_dump($sql);
并检查 $id 是否包含正确的值以及 $sql 的格式是否正确(尝试复制结果查询并在您的控制台中执行)
然后如果查询在控制台中给你正确的结果,这意味着结果查询是正确的..
PS 网址应该是
xxxxx.com/category.php?id=1
记得给你的id赋值
Tôi là một lập trình viên xuất sắc, rất giỏi!