從jquery ajax調用中獲取數據回php以在將來的SQL查詢中使用

Question

我希望從下拉框中獲取所選字段，以便在以后的下拉框中使用它，但是我無法弄清楚如何通過ajax將變量回顯到html。

<p>Trainer</p>
<select name = "trainer_has_update_pokemon">
<option>Select Trainer</option>
<?php
$query = "SELECT name FROM Trainer";

if ($stmt = $mysqli->prepare($query)) {
$stmt->execute();
$stmt->bind_result($name);
while ($stmt->fetch()) {
    echo"<option>$name</option>";
}


$stmt->close();
}


?>
</select>

寵物小精靈

<select name = "type_of_update_pokemon">
</select>

<script type="text/javascript" src="http://code.jquery.com/jquery-1.8.3.min.js"></script>
<script type="text/javascript">
$(function(){ 
    $('select[name="trainer_has_update_pokemon"]').change(function(){ // when trainer_has_update_pokemon changes
        $.ajax({
            type:"POST", //send a post method
            url:'pkmn_dropdown.php', // path to ajax page
            data:"trainer_name="+$(this).val(), //set trainer_name to value
            success:function(response){ // retrieve response from php
                $('select[name="type_of_update_pokemon"]').html(response); // update select
            }
        });
    });
});
</script>


<select name="nickname_of_update_pokemon">
</select>

<script type="text/javascript" src="http://code.jquery.com/jquery-1.8.3.min.js"></script>
<script type="text/javascript">
$(function(){ 
    $('select[name="type_of_update_pokemon"]').change(function(){ // when trainer_has_update_pokemon changes
        $.ajax({
            type:"POST", //send a post method
            url:'nickname_dropdown.php', // path to ajax page
            data:"pkmn_name="+$(this).val() & "trainer_name=" +$trainer_name,//set trainer_name to value
            success:function(response){ // retrieve response from php
                $('select[name="nickname_of_update_pokemon"]').html(response); // update select
            }
        });
    });
});
</script>

名稱下拉列表的php：

<?php

//connect to db

    $trainer_name = $_POST['trainer_name']; 
    $query = "SELECT DISTINCT p.name FROM Pokemon p WHERE p.owner_id = (SELECT t.trainer_id FROM Trainer t WHERE t.name = '$trainer_name')";
    if ($stmt = $mysqli->prepare($query)) {
        $stmt->execute();
        $stmt->bind_result($pkmn_name);
        while ($stmt->fetch()) {
            echo"<option>$pkmn_name</option>";
        }
        $stmt->close();
        echo $trainer_name;
    }?>

昵稱下拉列表的php：

<?php

//connect to db

$pkmn_name = $_POST['pkmn_name']; 
$query = "SELECT p.nickname FROM Pokemon p, Trainer t WHERE p.name = '$pkmn_name AND p.owner_id = t.trainer_id AND t.name = $trainer_name";
if ($stmt = $mysqli->prepare($query)) {
    $stmt->execute();
    $stmt->bind_result($nickname);
    while ($stmt->fetch()) {
        echo"<option>$nickname</option>";
    }
    $stmt->close();
}?>

任何想法如何從第一個下拉框中獲取選定的教練名稱，以便我可以在昵稱下拉框中使用它

Answer 1

當您運行上面的代碼時，到底發生了什么？ 選擇昵稱會丟失所有選項嗎？ 您的ajax響應是否包含預期的選項字符串？

修改選擇的innerHTML在IE中不起作用。 您將需要添加實際元素。 嘗試這樣的事情。

$( 'select[name="nickname_of_update_pokemon"]' ).empty().append( $( response ) );

在你的ajax成功函數中。