搜索篩選繼續

Question

這就是我迄今爲止所擁有的。 search_db.php現在將用戶字段顯示爲超鏈接，這是非常好的，當鏈接被遵循時，我沒有搜索結果。

search_db.php

$term = $_POST['term']; 

$data = mysql_query("select * FROM mordred13 WHERE alliance like '%$term%' ORDER BY alliance, might DESC"); 



    echo "<table border='1' cellpadding='5'>"; 
    echo "<tr> <th>Alliance</th> <th>User</th> <th>Might</th>"; 

    // loop through results of database query, displaying them in the table 

    while($row = mysql_fetch_array($data)) { 

    // echo out the contents of each row into a table 

    echo "<tr>"; 
      echo '<td>' . $row['alliance'] . '</td>'; 
      echo '<td><a href="userDetail.php?userID='.$row['id'].'">' . $row['user'] . '</td>'; 
    echo '<td>' . $row['might'] . '</td>' 
      echo "</tr>"; 
    } 

    // close table> 
    echo "</table>"; 
?> 

我試圖在userDetails.php查詢的不同變化，但只是無法得到它顯示我的篩選結果

userDetails.php

$term = $_POST['term']; 

$data = mysql_query("SELECT * FROM mordred13 WHERE user='user'"); 



    echo "<table border='1' cellpadding='5'>"; 
    echo "<tr> <th>Alliance</th> <th>User</th> <th>Might</th>"; 

    // loop through results of database query, displaying them in the table 

    while($row = mysql_fetch_array($data)) { 

    // echo out the contents of each row into a table 

    echo "<tr>"; 
      echo '<td>' . $row['alliance'] . '</td>'; 
      echo '<td>' . $row['user'] 
    echo '<td>' . $row['might'] . '</td>' 
      echo "</tr>"; 
    } 

    // close table> 
    echo "</table>"; 
?> 

Answer 1

您需要替換如下查詢，

更換

$data = mysql_query("SELECT * FROM mordred13 WHERE user='user'"); 

與

$data = mysql_query("SELECT * FROM mordred13 WHERE id ='".$_REQUEST['userID']."'");