所以我试图构建一个php页面来编辑记录。现在我正试图让它获取url中的ID,并基于此返回数据。我可以从url中回显id,这很好,但sql查询没有任何作用。我把2个回声放在底部,试图让它们显示值。ID显示得很好,来自URL,但它应该显示与gigigid相关的名称,而不是。
有什么想法吗?
<?php
// query db
$gigid = $_GET['gigid'];
$result = ORM::for_table('gigs')->where('gigid', $gigid)
or die(mysql_error());
$row = mysqli_fetch_array($result);
// check that the 'id' matches up with a row in the databse
if($row)
{
// get data from db
$gig_name = $row['gig_name'];
$gig_type = $row['gig_type'];
$gig_date = $row['gig_date'];
$gig_customer = $row['gig_customer'];
$gig_venue = $row['venue_name'];
$gig_fee = $row['gig_fee'];
$gig_status = $row['gig_status'];
}
mysqli_close($con);
?>
<?php echo $gigid; ?>
<?php echo $gig_name; ?>
我只会使用mysqli_query进行任何数据库操作。
. . .
$con = mysqli_connect("i.p.addr","username","password","database");
$result = mysqli_query($con, "SELECT * FROM gigs WHERE gigid=$gigid") or die(mysqli_error($con));
$row = mysqli_fetch_array($result);
mysqli_close($con);
更新::已更改为mysqli_error()