Question

display_art.php

print"<td><a href='member.php?username=$username'>$username</td>";

member.php： 代码显示由一位用户完成的所有艺术风格。当试图在“ where子句”中显示未知列“用户名”

$db = mysqli_connect("localhost", "root", "","Artworks" ); 

 $results = mysqli_query($db, "select * from artwork where username = 
{$_GET['username']}")or die(mysqli_error($db));

表格：

create table artwork(
    artwork_id serial primary key,
    username text,
    title text,
    category text,
    description text,
    tags text,
    filename text
    );

Answer 1

请确保输入正确的区分大小写的匹配列名，并使用准备好的语句

$db = mysqli_connect("localhost", "root", "","Artworks" ); 

$stmt = mysqli_prepare($db, "select * from artwork where username =  ?") or die(mysqli_error($db));
mysqli_stmt_bind_param($stmt, 's',$_GET['username']);

mysqli_stmt_execute($stmt);

这样，您应该避免引号和sqlinjection问题

“ where子句”中有一个未知的列“ username”

1 个答案: