我使用UDP连接将数据传输到服务器上的程序。数据由Quectel BC66的调制解调器传输。终端的AT命令如下所示:
Warning: mysqli_query() expects parameter 1 to be mysqli, boolean given
in /storage/ssd3/691/11050691/public_html/signin.php on line 255
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result,
null given in /storage/ssd3/691/11050691/public_html/signin.php on line
256
当数据出现在服务器上而不是显示已发送的数据时,它会显示问号:
该程序的代码如下所示:
the connection code:
<?php
$db=mysqli_connect("localhost","id11050691_kimo","025586381") or die
("Couldn't connect to Mysql");
$db=mysqli_select_db($db,"id11050691_daowat_db") or die ("Couldn't select
database");
//time formate
function formatDate($date){
return date('F j, Y, g:i a', strtotime($date));
}
?>
the code wich I have problem with :
//getting daowat post photo
$getposts = mysqli_query($db,"SELECT * FROM daowat WHERE photos != ''
ORDER BY RAND()");
$row = mysqli_fetch_assoc($getposts);
$photos_db = $row['photos'];
$photosrow = "./userdata/daowat_pics/".$photos_db;
您对如何在服务器程序上显示相同的发送数据有何建议?
答案 0 :(得分:0)