【问题标题】:JQuery JEditable json select issueJQuery JEditable json选择问题
【发布时间】:2019-12-28 16:57:27
【问题描述】:

我正在使用 JEditable,但在 json 选择中我遇到了问题,因为在控制台中显示 json 但在页面中选择输入为空。

这里是捕获:

empty select

console showing the id

json result

代码如下:

require($_SERVER['DOCUMENT_ROOT'] . '/config.php');
$result = $conn->query("SELECT id_user_details, first_name, last_name
    FROM users");
$return_arr = [];
while($row = $result->fetch(PDO::FETCH_ASSOC)) {
    $return_arr[] = array(
        $row['id_user_details'] => $row['first_name'].' '.$row['last_name'],
    );
};
$conn = null;

header('Content-type: application/json');
echo json_encode($return_arr);

这里是javascript:

$(".status").editable("include/users.php", {
            type   : "select",
            loadurl : "include/select.php",
            submit : "OK",
            submitdata : {pk : $(this).attr('id'), _as_csrf_token: token},
            indicator : '<img src="images/spinner.gif" />',
            style  : "inherit"
        });

如果客户端需要更改用户,选择显示的位置

<td>
    <a href="javascrip:void(0);" class="status" id="<?php echo $row['id_ch']; ?>"  style="font-size:11px;">
        <?php echo $row['doctor']; ?>
    </a>
</td>

【问题讨论】:

    标签: json select jeditable


    【解决方案1】:

    好吧,我终于做到了!

    我改了之前的代码:

    require($_SERVER['DOCUMENT_ROOT'] . '/config.php');
    $result = $conn->query("SELECT id_user_details, first_name, last_name
        FROM users");
    $return_arr = [];
    while($row = $result->fetch(PDO::FETCH_ASSOC)) {
        $return_arr[] = array(
            $row['id_user_details'] => $row['first_name'].' '.$row['last_name'],
        );
    };
    $conn = null;
    
    header('Content-type: application/json');
    echo json_encode($return_arr);
    

    到这个:

        require($_SERVER['DOCUMENT_ROOT'] . '/config.php');
    $result = $conn->query("SELECT id_user_details, first_name, last_name
        FROM users");
    while($row = $result->fetch(PDO::FETCH_ASSOC)) {
            $id_user_details = $row['id_user_details'];
            $array[$id_user_details] = $row['first_name'].' '.$row['last_name'];
    };
    $conn = null;
    
    header('Content-type: application/json');
    print json_encode($array);
    

    【讨论】:

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