Question

我正在尝试创建一个依赖的下拉列表，但在我第一次选择后它似乎没有填充。每个选择都将从MySQL数据库中获取数据。为了使第二个下拉列表具有任何选项（除了默认的“选择选项”值），用户必须首先在第一个下拉列表中进行选择。经过大量的谷歌搜索后，我很难找到一个简单的解决这个问题。

这是我到目前为止所拥有的，

下拉列表（我在此使用PHP和MySQL生成并输出getter.php和require_once中的下拉列表下拉列出index.php和echo

$accountOptions = "";
$facilityOptions = "";

$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);        
if (!$dbc) {
    die("Connection failed: " . mysqli_connect_error());
}

///ACCOUNTS/////
$accountQuery = "SELECT account_id, account_name FROM account";
$accountData = mysqli_query($dbc, $accountQuery);

//loop through data and display all accounts
while ($aRow = mysqli_fetch_array($accountData)) {
         $accountOptions .="<option value=\"".$aRow['account_id']."\">" . $aRow['account_name'] . "</option>";
}

$accountDropDown=" <label>Accounts: </label><br>
                    <select name='account' id='account' onChange='getFacility(this.value)'>
                        <option selected='selected' disabled='disabled' value=''>Select account</option>
                    " . $accountOptions . "
                    </select>";

////FACILITIES/////
$facilityDropDown=" <label>Facility: </label><br>
                    <select name='facility' id='facility'>
                        <option selected='selected' disabled='disabled' value=''>Select facility</option>
                    </select>";

的JQuery / AJAX

function getFacility(val) {
        $.ajax({
        type: "POST",
        url: "getfacility.php",
        data:'account_id='+val,

        success: function(data){
                $("#facility").html(data);
        }
        });
    }

getfacility.php

//db connection..


if(!empty($_POST["account_id"])) {

$accountID = $_POST['account_id'];

$sql = "SELECT *, account.account_name FROM facility "
     . "INNER JOIN account ON account.account_id = facility.account_id "
     . "WHERE facility.account_id = '". $accountID ."'";

$data = mysqli_query($dbc, $sql);

echo "<option selected='selected' disabled='disabled' value=''>Select facility</option>";

while ($fRow = mysqli_fetch_array($data)) {
     $facilityOptions .="<option value=\"".$fRow['facility_id']."\">" . $fRow['facility_name'] . "</option>";
}

    $facilityDropDown=" <label>Facility: </label><br>
                <select name='facility' id='facility'>
                    <option selected='selected' disabled='disabled' value=''>Select facility</option>
                " . $facilityOptions . "
                </select>";
}

现在，当我在第一次下拉时做出选择时，第二次没有填充任何东西，我哪里出错？

Answer 1

在 getfacility.php ，

中进行更改

如果你的ajax在网络（控制台）中显示200 ok状态和预期响应

//db connection..


if(!empty($_POST["account_id"])) {

$accountID = $_POST['account_id'];

$sql = "SELECT *, account.account_name FROM facility "
     . "INNER JOIN account ON account.account_id = facility.account_id "
     . "WHERE facility.account_id = '". $accountID ."'";

$data = mysqli_query($dbc, $sql);

echo "<option selected='selected' disabled='disabled' value=''>Select facility</option>";

while ($fRow = mysqli_fetch_array($data)) {
     echo "<option value=\"".$fRow['account_id']."\">" . $fRow['account_name'] . "</option>";
}

依赖下拉列表Jquery，AJAX，PHP，MYSQL

1 个答案: