我没有得到任何错误，但也没有得到任何结果 - mysql php db

Question

我正在尝试从具有两列的数据库中获取用户提交。一个用于艺术家，一个用于标题。我想从简单的表格中获取他们的输入，并将所有类似的结果输出到下一页的表格中。到目前为止，我已经包含了我编写的整个脚本。我在页面上没有收到任何错误，但我也没有得到任何结果。我花了几天时间在网上寻找我是否可以自己解决这个问题，但我没有这样的运气。很抱歉这么罗嗦，但我是这个网站的新手，想提供尽可能多的细节。

<?php 
include("db_connect.php"); 
// - get form data from "searchform.php"
$song_query = $_GET['song_query'];
// - column 1 and column 2 are all we're looking for in the db 
// - title and artist are currently the two cols. Table is named 
"song_list"
$col1 = "title";
$col2 = "artist";
$tablename = "song_list";
echo "<H1>Search results</H1>";
if ($song_query == "") {
echo "What are you going to sing? You didn't enter a search term! Please 
try again.";
exit;
}
// - pick the db connection out of the included file and give error if 
failed.
mysqli_select_db($conn, $db_name) or die(mysqli_error());
// - cleans up string inputted by user to avoid harmful code insertion 
into form
$song_query = strtoupper($song_query);
$song_query = strip_tags($song_query);
$song_query = trim($song_query);
// - set up parameters for accessing the query of the db
$sql = "SELECT $col1, $col2 FROM $tablename WHERE $col1, $col2 LIKE 
'%$song_query%'";
$result = mysqli_query($conn, $sql);
if (isset($_GET['$result'])){
if (mysqli_num_rows($result) > 0){
    echo "<table><tr>";
    echo "<th>Artist</th>";
    echo "</tr>";

while($row = mysqli_fetch_array($result)){
    echo "<tr>";
    echo "<td>" . $row['$result'] . "</td>";
    echo "</tr>";
    echo "</table>";
    }
    }
}
?>

Answer 1

你错了SQL，它是在运行时构造的

$sql = "SELECT $col1, $col2 FROM $tablename WHERE $col1, $col2 LIKE 
'%$song_query%'";

变成了

$sql = "SELECT title, artist FROM $tablename WHERE title, artist LIKE 
'%$song_query%'";

看WHERE title, artist LIKE这里

$song_query 从$_GET['song_query'] 获取值，该值在运行时发生变化。

Answer 2

这个WHERE $col1, $col2 LIKE '%$song_query%'是你需要说的无效语法

WHERE col1 LIKE '%something%' AND col2 LIKE '%something%'

所以这应该可以解决问题

$sql = "SELECT $col1, $col2 
        FROM $tablename 
        WHERE $col1 LIKE '%$song_query%'
        AND $col2 LIKE '%$song_query%'";

尽管这对SQL Injection Attack 开放 甚至if you are escaping inputs, its not safe! 使用prepared parameterized statements

$sql = "SELECT title, artist 
        FROM songlist 
        WHERE title LIKE ? AND artist LIKE ?";

$stmt = $conn->prepare($sql);
$val = sprintf('%%%s%%', $song_query);
$stmt->bind_param('ss',$val, $val);

$stmt->execute();

$stmt->bind_result($title, $artist);

echo "<table><thead><tr>";
echo "<th>Artist</th><td>Title</th>";
echo "</tr></thead><tbody>";

while ($stmt->fetch()) {
    echo "<tr>";
        echo "<td>$artist</td>";
        echo "<td>$title</td>";
    echo "</tr>";
}
echo "</tbody></table>";

另请注意，您在构建表格时犯了几个错误，我认为我已经修复了。