这就是我基本上想做的事情:
- 检索列的值
- 将这些值存储到 PHP 中的数组中
- 回显每个值,并在每个值之间换行
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;
};
?>
这就是我想要回显值的地方:
<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:我终于开始编写最新的代码,所以我希望这里没有人评论说我在这段代码中的某个地方有过时的代码(͡°͜ʖ͡°)
最佳答案
$members_array = array();
while ($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>
关于php - 如何检索表的列数据,将其存储到数组中,最后回显该数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36324046/