以下是我基本上要做的事情：
- 检索列的值
- 将这些值存储到PHP中的数组中
- 用每个值之间的换行符回显每个值
这是我尝试的代码：
<?php
$connection = mysqli_connect("localhost", "root", "bruhfrogzombie098", "growtapi_social");
if (!$connection) {
die("Failed to connect to MYSQL: " . mysqli_connect_errno());
};
$members = mysqli_query($connection, "SELECT Username FROM s_users");
$members_status = mysqli_query($connection, "SELECT Status_Content FROM s_users");
$members_array = array();
while ($member = mysqli_fetch_assoc($members)) {
$members_array[] = $member;
};
$members_status_array = array();
while ($status = mysqli_fetch_assoc($members_status)) {
$members_status_array[] = $status;
};
?>
这就是我想要呼应的价值观：
<center>
<h1>Members Directory</h1>
<div style="width: 20%; height; 75%; border: 3px solid black; margin: auto; overflow: hidden; overflow-y: scroll;">
<?php echo $members_array['$member'];
echo "<br />";
?>
</div>
</center>
我没有收到任何错误，但问题是div中没有显示任何内容，这意味着我要么没有正确检索数据，要么没有正确使用数据。
注意：我终于开始写最新的代码了，所以我希望这里没有人评论我在这个代码中的某个地方有过时的代码（͡°͜ʖ''865°）
$members_array = array();
while ($member = mysqli_fetch_assoc($members)) {
$members_array[] = $member;
//$members_array is array and $member is array so $members_array like $members_array[][];
};
<center>
<h1>Members Directory</h1>
<div style="width: 20%; height; 75%; border: 3px solid black; margin: auto; overflow: hidden; overflow-y: scroll;">
<?
$member_count = count($members_array);
for( $i = 0 ; $i < $member_count ; $i++ ){
echo $member_array[$i]['Username'];
echo "<br />";
}
?>
</div>
</center>