您无法将变量值从当前页面的Javascript代码传递到当前页面的PHP代码… PHP代码在服务器端运行，并且对客户端的运行情况一无所知。
您需要使用另一种机制将变量从HTML表单传递给PHP代码，例如，使用GET或POST方法提交表单。
<DOCTYPE html><html> <head> <title>My Test Form</title> </head> <body> <form method="POST"> <p>Please, choose the salary id to proceed result:</p> <p> <label for="salarieids">SalarieID:</label> <?php $query = "SELECt * FROM salarie"; $result = mysql_query($query); if ($result) : ?> <select id="salarieids" name="salarieid"> <?php while ($row = mysql_fetch_assoc($result)) { echo '<option value="', $row['salaried'], '">', $row['salaried'], '</option>'; //between <option></option> tags you can output something more human-friendly (like $row['name'], if table "salaried" have one) } ?> </select> <?php endif ?> </p> <p> <input type="submit" value="Sumbit my choice"/> </p> </form> <?php if isset($_POST['salaried']) : ?> <?php $query = "SELECT * FROM salarie WHERe salarieid = " . $_POST['salarieid']; $result = mysql_query($query); if ($result) : ?> <table> <?php while ($row = mysql_fetch_assoc($result)) { echo '<tr>'; echo '<td>', $row['salaried'], '</td><td>', $row['bla-bla-bla'], '</td>' ...; // and others echo '</tr>'; } ?> </table> <?php endif?> <?php endif ?> </body></html>
