php - ORDER BY Year DESC 不工作

Question

我的页面上显示了一些信息。我让它完美地显示了我的表格的内容。然而，当我尝试对其进行排序时，一切都崩溃了。任何建议将不胜感激

$sql = "

SELECT * 

 FROM 

 agreements ORDER BY Year DESC 

 WHERE

  1 = 1

  ".($_POST['agreeno'] != '' ? "AND Resolution = '".$_POST['agreeno']."'" : Null)."

  ".($_POST['keyword'] != '' ? "AND Resolution_Name LIKE '%".$_POST['keyword']."%'" : Null)."

  ".($_POST['Year'] != '' ? "AND Year LIKE '%".$_POST['Year']."%'" : Null). "

"

;

这是我的完整代码。

// Create connection

$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection

if ($conn->connect_error) {

    die("Connection failed: " . $conn->connect_error);

} 



echo "<table class='table table-striped publications' border='0' cellspacing='0' cellpadding='0'>

<tbody>

<tr>

<td style='text-align: left;' align='center'><strong>Resolution Name</strong></td>

<td ><strong>Resolution</strong></td>

<td><strong>Year</strong></td>

<td><strong>Department</strong></td>

<td><strong>Document</strong></td>

</tr>"; 













$sql = "

 SELECT * 

 FROM 

 agreements ORDER BY Year DESC 

 WHERE

  1 = 1

  ".($_POST['agreeno'] != '' ? "AND Resolution = '".$_POST['agreeno']."'" : Null)."

  ".($_POST['keyword'] != '' ? "AND Resolution_Name LIKE '%".$_POST['keyword']."%'" : Null)."

  ".($_POST['Year'] != '' ? "AND Year LIKE '%".$_POST['Year']."%'" : Null). "

"

;








$result = $conn->query($sql);



if ($result->num_rows > 0) {



    while($row = $result->fetch_assoc()) {





 echo "<tr><td style='text-align: left' align='center'>" . $row["Resolution_Name"]. "</td><td style='text-align: left;' align='center'> " . $row["Resolution"]. " </td><td style='text-align: left;' align='center'> " . $row["Year"]. " </td><td style='text-align: left;' align='center'> " . $row["Department"]. "</td><td style='text-align: left;' align='center'> <a href='". $row["Document"]. "' target='_empty'>View Document [PDF]</a> </td></tr>";

    }

} else {

    echo "0 results";

}

$conn->close();



echo "</tr></table>";

?>

Answer 1

您的 ORDER BY 位于 WHERE 条件之后。

还可以将 empty() 或 isset() 与发布值一起使用，以防止未定义索引错误。

$sql = "

SELECT * 

 FROM 

 agreements 

 WHERE

  1 = 1

  ".((!empty($_POST['agreeno'])) ? "AND Resolution = '".$_POST['agreeno']."'" : Null)."

  ".((!empty$_POST['keyword'])) != '' ? "AND Resolution_Name LIKE '%".$_POST['keyword']."%'" : Null)."

  ".((!empty($_POST['Year'])) != '' ? "AND Year LIKE '%".$_POST['Year']."%'" : Null). "

ORDER BY Year DESC 
"
;