Question

类别数据库表：

id  | name  
1   | electronics
2   | Automotive

分类数据库表：

id  user_id  category_id    cover          title       price    description 
102  1       2              iamges/1.jpg    blabla     10      blablabla

我收到错误消息：通知：未定义变量：第49行** \ post.php中的name1

$id = $_GET['id'];
$res2 = mysql_query("SELECT * FROM `classifieds` WHERE `id` = '" . $id . "' LIMIT 1");

if($res2 && mysql_num_rows($res2) > 0){
    while($row = mysql_fetch_assoc($res2)){
        $id = $row['id'];
        $user_id = $row['user_id'];
        $category_id = $row['category_id'];
        $price = $row['price'];
        $cover = $row['cover'];
        $title = $row['title'];
        $description = $row['description'];
        $profile_data = user_data($user_id, 'username');

        $res3 = mysql_query("SELECT * FROM `categories` WHERE `id` = '" . $category_id . "' LIMIT 1");
        if($res3 && mysql_num_rows($res3) > 0){
            while($row1 = mysql_fetch_assoc($res3)){
                $id1 = $row1['id'];
                $name = $row1['name'];
            }
        }  
    }
}else{
    echo 'error';
}echo $name;

为什么我会收到echo $ name1的错误; ？

Answer 1

您在sql查询用户$row和$row2中使用$row3，如下所示

试试这个

$id = $_GET['id'];
$id = mysql_real_escape_string(stripslashes($id));
$res2 = mysql_query("SELECT * FROM `classifieds` WHERE `id` = '" . $id . "' LIMIT 1");

if(mysql_num_rows($res2) > 0){
    while($row2 = mysql_fetch_assoc($res2)){
        $id = $row2['id'];
        $user_id = $row2['user_id'];
        $category_id = $row2['category_id'];
        $price = $row2['price'];
        $cover = $row2['cover'];
        $title = $row2['title'];
        $description = $row2['description'];
        $profile_data = user_data($user_id, 'username');

        $res3 = mysql_query("SELECT * FROM `categories` WHERE `id` = '" . $category_id . "' LIMIT 1");
        while($row3 = mysql_fetch_assoc($res3)){
            $id = $row3['id'];
            $name = $row3['name'];
        }

    }   
}

UDPATE 2：

正如您所说，您正在使用函数user_data

中的查询

$data = mysql_fetch_assoc(mysql_query("SELECT $fields FROM users WHERE user_id = $user_id"));

将其更改为

$data = "";
$res_data = mysql_query("SELECT $fields FROM users WHERE user_id ='$user_id' LIMIT 1 ") or die(mysql_error());
if($row_data = mysql_fetch_assoc($res_data))
{
   $data = $row_data['your_field_name'];
}

Answer 2

错误

"Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in *\post.php on line 20 line 20"

表示mysql_query();函数具有not executed successfully。

可以检查＆＃34; id＆＃34;收到与否的价值。

Answer 3

试试这个......

替换为此..

$res2 = mysql_query("SELECT * FROM `classifieds` WHERE `id` = $id  LIMIT 1");

也

$res3 = mysql_query("SELECT * FROM `categories` WHERE `id` = $category_id  LIMIT 1");

Answer 4

尝试以下代码

    $id = $_GET['id'];
    $res2 = mysql_query("SELECT * FROM `classifieds` WHERE `id` = '" . $id . "' LIMIT 1");

    if($res2 && mysql_num_rows($res2) > 0){
        while($row = mysql_fetch_assoc($res2)){
            $id = $row['id'];
            $user_id = $row['user_id'];
            $category_id = $row['category_id'];
            $price = $row['price'];
            $cover = $row['cover'];
            $title = $row['title'];
            $description = $row['description'];
            $profile_data = user_data($user_id, 'username');

            $res3 = mysql_query("SELECT * FROM `categories` WHERE `id` = '" . $category_id . "' LIMIT 1");
            if($res3 && mysql_num_rows($res3) > 0)
                while($row1 = mysql_fetch_assoc($res3)){
                    $id = $row1['id'];
                    $name = $row1['name'];
                }
            }   

        }   
    }

Answer 5

你应该打印查询并在php myadmin driectly中运行，看看它是否返回正确的结果。

你们三个循环包含相同的$row。请将下面的一个更改为其他人

 while($row = mysql_fetch_assoc($res2)){

and 

  while($row = mysql_fetch_assoc($res3)){
  ------^

更改为$row3

等其他内容

  while($row3 = mysql_fetch_assoc($res3)){
        $id = $row3['id'];
        $name = $row3['name'];
    }

从数据库中检索数据不适用于此代码

5 个答案: