Question

function getfield($get){
    global $connection;

$query = "SELECT id, username, firstname, lastname FROM users WHERE username='".$_SESSION['user_id']."'";
        if ($query_r = mysqli_query($connection, $query)) {
            $num_rows = ($query_r -> num_rows);
            if ($mysqli_result = mysqli_result($query_r, 0, $get)) {
                return $mysqli_result;
            }

我做了一个登录表单，现在一切正常但这个功能没有显示数据。我认为mysql_result在7.1中无效。

Answer 1

函数<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div class="dd" id="nestable" data-id="0"> <ol class="dd-list"> <li class="dd-item" data-id="1"> <div class="dd-handle">Sub-indicator 1</div> <ol class="dd-list"> <li class="dd-item" data-id="2"> <div class="dd-handle">Sub-indicator 2</div> <ol class="dd-list"> <li class="dd-item" data-id="3"> <div class="dd-handle">Sub-indicator 3</div> <ol class="dd-list"> <li class="dd-item" data-id="4"> <div class="dd-handle">Sub-indicator 4</div> </li> </ol> </li> </ol> </li> <li class="dd-item" data-id="3"> <div class="dd-handle">Sub-indicator 3</div> <ol class="dd-list"> <li class="dd-item" data-id="4"> <div class="dd-handle">Sub-indicator 4</div> </li> </ol> </li> </ol> </li> </ol> </div>不存在，因此您需要使用mysqli_result之类的内容。用这个替换你最里面的mysqli_fetch_array() - 子句：

if

Mysql_result ..没有显示数据

2 个答案: