在條款

Question

中的未知列'chenzhen'我有一個PHP腳本，它使用mysqli擴展連接到MySQL數據庫，以根據用戶名或ID搜索博客帖子。我創建了一個名爲BlogSearch查看使用聯接形成其他表彙總是這樣表示我需要共同的信息：

它被稱爲Profiles具有用戶信息，BlogPosts和BlogCategory拉表

每次我搜索我的錯誤：我用下面 Unknown column 'chenzhen' in 'where clause'

的PHP代碼：

require 'database.php'; 

     $query = "SELECT * FROM BlogSearch"; 

    echo <<<EOF 
<form method='post' action='' style="padding: 30px 0;"> 
     <table cellspacing="0" border="0" style="float: left;"> 
     <tr> 
     <td>Search Blog Posts by Username/ID</td> 
     <td><input type="text" id="search" name="search" style="width: 300px;"/></td> 
     <td><input type="submit" id="submit_button" value="Search" name="submit_button" style="float: right;" /></td> 
     </tr> 
     </table> 
    </form> 
EOF; 

     if(isset($_POST['submit_button'])) 
    { 
     $search_term = $_POST['search']; 
      $query = $query . " WHERE `NickName` LIKE '%$search_term%' OR ID = $search_term "; 


     // run the query and store the results in the $result variable. 
     $result = $mysqli->query($query) or die(mysqli_error($mysqli)); 

    } 

    if ($result) { 

    // create a new form and then put the results 
    // into a table. 



    echo "<form method='post' action='delete.php' style='clear: both;'>"; 
    echo "<table cellspacing='0' cellpadding='15'> 
     <th width='5%'> 
     <input type='checkbox' id='allcb' onclick='checkAll(this)' name='allcb' />Check All 
     </th> 
     <th width='10%'>User</th> 
     <th width='85%'>Blog Post Title</th> 
     "; 


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

     $title = substr($row->PostCaption,0,50); 
     $id = $row->PostID; 
     $user = $row->NickName; 

     //put each record into a new table row with a checkbox 
    echo "<tr> 
      <td><input type='checkbox' name='checkbox[]' id='checkbox[]' value=$id /> 
      <td>$user</td> 
      <td>$title</td> 
     </tr>"; 

    } 

    // when the loop is complete, close off the list. 
    echo "</table><p><input id='delete' type='submit' class='button' name='delete' value='Delete Selected Items'/></p></form>"; 
} 

我不知道爲什麼它甚至將用戶名確定爲列。任何人都可以指出我正確的方向來解決這個問題嗎？

在此先感謝。

Answer 1

SQL查詢中不是SQL關鍵字或文字（用單引號表示）中的任何元素都假定爲對象（例如表，列）名稱。

你的問題是圍繞$search_term缺失報價在WHERE條款：

$query = $query . " WHERE `NickName` LIKE '%$search_term%' OR ID = $search_term "; 

您應該添加它們，這樣的：

$query = $query . " WHERE `NickName` LIKE '%$search_term%' OR ID = '$search_term' "; 

Answer 2

附上您的$ SEARCH_TERM在在單引號像這樣的條款'$search_term'