在php中列出1行而不是所有行

Question

出於某種原因，當我想要的所有內容都從member_details表中返回時，所有行都返回等於下拉列表中所選項目的單個行。

下面是它的PHP：我認爲它會返回等於下拉值的行，但它只是列出它的一切。

<?php 
     if (isset($_POST['members'])) { 
     $ResultSet = getTableResults("member_details"); 
     echo "<h1> Member Details </h1>"; 
     echo "<table border='1' cellpadding='6'>"; 
     echo "<tr> <th>Id</th> <th>Name</th> <th>Job</th> <th>Wage</th> <th>Hobby</th> "; 

     foreach ($ResultSet as $row) { 
      echo "<tr>"; 
      echo "<td>" . $row ['member_id'] . "</td>"; 
      echo "<td>" . $row['first_name'] . " " . $row ['second_name'] . "</td>"; 
      echo "<td>" . $row['job'] . "</td>"; 
      echo "<td>" . $row['wage'] . "</td>"; 
      echo "<td>" . $row['hobby'] . "</td>"; 
      echo "</tr>"; 
     } 

     echo "<table>"; 
    } 
    ?> 

如果你想看到更多的代碼，更有意義的是，請要求和IL編輯和更新的問題。

function getTableResults() { 
    $sql = "SELECT DISTINCT member_details.member_id, members.first_name,  members.second_name, member_details.wage, member_details.job, member_details.hobby 
     FROM members 
     INNER JOIN member_details 
     ON members.member_id=member_details.member_id"; 
    $mysqlConnection = getSQLConnection(); 

    $ResultSet = $mysqlConnection->query($sql); 
    return $ResultSet; 
    } 

Answer 1

你沒有任何條件，同時請求您的信息，getTableResults()獲取存儲在您的表中的所有數據。您需要更改該功能並添加WHERE xy條件。可能是這樣的：

function getTableResults($id) { 
    $sql = "SELECT DISTINCT member_details.member_id, members.first_name,members.second_name, member_details.wage, member_details.job, member_details.hobby 
     FROM members 
     INNER JOIN member_details 
     ON members.member_id=member_details.member_id WHERE member_details.member_id = '$id'"; 
    $mysqlConnection = getSQLConnection(); 

    $ResultSet = $mysqlConnection->query($sql); 
    return $ResultSet; 
    } 

然後以requred member-id作爲參數調用該函數。 無論如何，我建議你通知你自己準備好的語句，以防止SQL注入在這裏！