Question

我正在尝试从页面上的mysqli循环中显示一组PHP的结果，并提供一个链接选项，一次显示10个，直到达到50个结果为止。显示指向完整结果页面的链接。

我目前有一个令人费解的部分解决方案，该解决方案很讨厌，但是可以正常工作。

function get_review_by_id($review_id) {

    $mysqli = new mysqli($GLOBALS['db_host'], $GLOBALS['db_user'], $GLOBALS['db_pass'], $GLOBALS['database']);

    if ($mysqli->connect_errno) {
        printf("Connect failed: %s\n", $mysqli->connect_error);
        exit();
    }

    if ($result = $mysqli->query("SELECT item_id, entry_date, name FROM tbl_item_entry WHERE post_id = $review_id")) {
        while ($row = $result->fetch_assoc()) {
            $review_id  = $row['item_id'];
            $date           = $row['entry_date'];
            $name           = $row['name'];
            if ($user = $mysqli->query("
                SELECT tbl_15.value as role, tbl_19.value as environment, tbl_17.value as title, tbl_18.value as review
                FROM tbl_15, tbl_19, tbl_17, tbl_18
                WHERE tbl_15.id = tbl_19.id AND tbl_17.id = tbl_18.id AND tbl_15.id = tbl_17.id AND tbl_17.id = tbl_19.id
                AND tbl_15.id = $review_id ")) {
                while ($value = $user->fetch_assoc()) {
                    $role                   = $value['role'];
                    $environment        = $value['environment'];
                    $title                  = $value['title'];
                    $review                 = $value['review'];
                    $item_instance  = "
                    <div class='row'>
                        <div class='col-sm-12'>
                            <div class='item-instance'>
                                <div class='row'>
                                    <div class='col-sm-3'>
                                        <span class='item-title'>$name</span>
                                        <p>$date</p>
                                        <p>$environment</p>
                                        <p>$role</p>
                                    </div>
                                    <div class='col-sm-9'>
                                        <span class='item-title'>$title</span>
                                        <p>$review</p>
                                    </div>
                                </div>
                            </div>
                        </div>
                    </div>

                    ";
                    echo $item_instance;
                }
            }
        }

        $result->close();
    }

    $mysqli->close();
}

然后在我的页面上，将其按以下方式加载到隐藏的div中：

<div id="reviews" style="display: none;">
    <?php get_review_by_id($product_id); ?>
  </div>

<script>
  jQuery(function($) {
    $( document ).ready(function() {
      $('#reviews').fadeIn(1200);
    });
  });
</script>

我已经尝试将结果加载到部分PHP文件中，并使用jQuery $.ajax方法，但是到目前为止还没有运气。以上是我要显示的唯一方法。

延迟使用JavaScript加载PHP MySQL结果

0 个答案: