Question

我有一个无法解决的问题。我有一个简单的页面，我查询所有用户并将其列在表中。当用户单击其中一个表行时，应将其带到另一个页面，用户可以在该页面中编辑他们选择的信息。问题是在我的脚本中，$ _POST值始终是最后一个

的值

CODE

    <?php
    include "conn.php";
    $pquery = "SELECT * FROM Patient NATURAL JOIN User ORDER BY LastName;";
    $patientQuery = $conn->query($pquery);

    if (mysqli_num_rows($patientQuery) == 0)
        echo "<p>No patients found.</p>";

    else{
        while($assoc = $patientQuery->fetch_assoc()){

            echo "<tr onclick = 'sub();'>";
            echo "<td>";
            echo $assoc['UserID'];
            echo "<input type = 'hidden' name = 'UserID' value = '". $assoc['UserID'] ."' />";
            echo "</td>";

            echo "<td>";
            echo $assoc['FirstName'];
            echo "</td>";

            echo "<td>";
            echo $assoc['LastName'];
            echo "</td>";
            echo "</tr>";

        }
    }
?>
<script>
    function sub(){
        document.getElementById("edit").submit();
        return false;
    }
</script>

Answer 1

我稍微修改了你的代码 - 这应该可行：

<?php
    include "conn.php";
    $pquery = "SELECT * FROM Patient NATURAL JOIN User ORDER BY LastName;";
    $patientQuery = $conn->query($pquery);

    if (mysqli_num_rows($patientQuery) == 0)
        echo "<p>No patients found.</p>";

    else{
        while($assoc = $patientQuery->fetch_assoc()){

            echo "<tr onclick = 'sub(". $assoc['UserID'] .");'>";
            echo "<td>";
            echo $assoc['UserID'];
            echo "</td>";

            echo "<td>";
            echo $assoc['FirstName'];
            echo "</td>";

            echo "<td>";
            echo $assoc['LastName'];
            echo "</td>";
            echo "</tr>";

        }
    }
?>

<script>
    function sub(UserID){
        document.location.href = 'http://www.yourdomain.com/something.php?UserID='+UserID;
        return false;
    }
</script>

$ _POST的值始终是我

1 个答案: