Question

当我从第一个下拉菜单中选择一个选项然后自动显示表而不点击提交时，如何进行onchange事件？

每当我点击编辑数据提交时，如何刷新上表？

        <?php

    $conn = new mysqli('localhost', 'root', 'jared17', 'hbadb') 
    or die ('Cannot connect to db');

    $result = $conn->query("select * from english");

    echo "<html>";
    echo "<body>";
    echo "<form name='form' method = POST>";
    echo "<select name = 'Students'>";
    while ($row = $result->fetch_assoc()) {
        $LRN = $row['LRN'];
        $Last = $row['Last_Name']; 
        $First = $row['First_Name'];
        if ($LRN == $_POST['Students']) $selected = 'selected="selected"';
        echo '<option value="'.$LRN.'">'.$Last.', '.$First.'</option>';

    }
     echo "</select>";
    echo "<input type='submit' name='submit' value='Show'>";


    if (isset($_POST['Students'])) {
        $lrn = $_POST['Students'];
        $stmt = $conn->prepare("SELECT Last_Name, First_Name, Level, Q1, Q2, Q3, Q4, FINAL FROM english WHERE LRN = ?");
        $stmt->bind_param('i', $lrn);
        $stmt->execute();
        $stmt->bind_result($last, $first, $level, $q1, $q2, $q3, $q4, $final);
        $stmt->fetch();
        echo "<table><tr><th>LRN</th><th>Name</th><th>Level</th><th>Q1</th><th>Q2</th><th>Q3</th><th>Q4</th><th>Final</th></tr>";
        echo "<tr><td>$lrn</td><td>$last, $first</td><td>$level</td><td>$q1</td><td>$q2</td><td>$q3</td><td>$q4</td><td>$final</td></tr></table>";
    }
    echo "</form>";

    echo "<form name='form2' method = POST>";
    ///////////EDIT DATA
    echo "Edit Data: ";
    echo "<select name = 'Edit'>";

    echo '<option value=Q1>Q1</option>';
    echo '<option value=Q2>Q2</option>';
    echo '<option value=Q3>Q3</option>';
    echo '<option value=Q4>Q4</option>';
    echo '<option value=FINAL>FINAL</option>';
    echo '<input type="number" name="editdata">';
    echo "</select>";
    echo "<input type='submit' name='submit2' value='Edit Now'>";


    if (isset($_POST['Edit'])) {
        $conn2 = new mysqli('localhost', 'root', 'jared17', 'hbadb') 
        or die ('Cannot connect to db');
        $upd = $_POST['Edit'];
        $txt = $_POST['editdata'];
        $now = "UPDATE english SET $upd='$txt' WHERE LRN='$lrn'";
        $res = $conn2->query($now);
        if (!$conn2->error) {
            echo "Errormessage: $conn->error";
    }
        echo $now;

    }



    echo "</form>";



    echo "</body>";
    echo "</html>";
    ?>

Answer 1

您的所有问题都必须由ajax修复。

$.ajax({
    url: "example.php", //file which has query select to db table
    data: {id:theid},   //describe your value of select option here
    dataType: 'json',   // type of data that will you get (JSON/HTML).
    type: 'POST',       //sending type (POST/GET)
    success: function(data) {
       showTable();
    }
});

提交后的onchange事件和自动刷新表

1 个答案: