首席会员数据库值: -
member_id = 1;
member_no= 001;
status = '2';
第二个会员数据库值
member_id = 2;
member_no= 002;
status = '1';
第三个会员数据库值
member_id = 3;
member_no= 003;
status = '2';
我的Php查询
$select_query_sender = "select * from member where status='1' ORDER BY member_id ASC LIMIT 1";
$run_sender = mysqli_query($con,$select_query_sender);
while ($row_sender = mysqli_fetch_array($run_sender)) {
$member_no_sender = $row_sender['member_no'];
echo $member_no_sender; }
我尝试使用限制查询来回复状态1成员列表,除了输出到 member_no = 002; ,但输出什么都不显示,我错过了请帮助任何一个。
答案 0 :(得分:1)
您的代码没问题,但请检查您的查询是否从db获取任何行。
$select_query_sender = "select * from member where status='1' ORDER BY
member_id ASC LIMIT 1";
$run_sender = mysqli_query($con,$select_query_sender);
if($run_sender->num_rows > 0 ){
while ($row_sender = mysqli_fetch_array($run_sender)) {
echo $row_sender['member_no']; // echo member_no
}
}else{
echo "no row";
}
答案 1 :(得分:0)
...试
$select_query_sender = "select * from member where status='1' ORDER BY member_id ASC LIMIT 1";
$run_sender = mysqli_query($con,$select_query_sender);
$row_sender = mysqli_fetch_object($run_sender);
echo $row_sender->member_no ;