Question

我有一个提交给MySQL的php表单，以及第二个使用第一个存储的值的php表单。

数据库中booth_member的最高值目前为3.在下面的示例中，= $ booth_member应该等于4，但总是返回2.

$membersql = $mysqli->query("SELECT booth_member FROM users WHERE booth_number = '{$user['booth_number']}' ORDER BY booth_member DESC LIMIT 1");
$booth_member = $membersql + 1;
echo $booth_member;

如果我尝试回显$ membersql，它只返回＆＃39;数组。＆＃39;

Answer 1

mysqli::query会返回一个资源对象，这就是它无法正常工作的原因。一些代码来解决它：

$membersql = $mysqli->query("SELECT booth_member FROM users WHERE booth_number = '{$user['booth_number']}' ORDER BY booth_member DESC LIMIT 1");
if($membersql->num_rows < 1)
{ /* error handling goes here */ }
else
{
  $res_array = $membersql->fetch_assoc(); // save result as array
  $booth_member = $res_array["booth_member"] + 1;
  echo $booth_member;
}

请务必先转义$user['booth_number']以防止sql注入。

有关详细信息，请查看mysqli::query文档：）

Answer 2

对于生产系统，您需要一堆try / catch代码，但要测试：

$result = $mysqli->query("SELECT booth_member FROM users WHERE booth_number = '{$user['booth_number']}' ORDER BY booth_member DESC LIMIT 1");

$booth_member = $result->fetch_object('booth_member');
++$booth_member;
echo $booth_member;

Answer 3

那是因为它是一个数组。我相信你是想做的：

$booth_member = $membersql['booth_member'] + 1;

在不解析数组的情况下递增整数

3 个答案: