我正在尝试在我的SQL服务器中显示导航表中的链接。我选择链接的查询如下,但是当使用时,没有链接显示或只有一个链接(IE不考虑循环)
$query = "SELECT link FROM navigation WHERE permission<='3'";
$result = mysqli_query($link, $query);
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC));
echo '<li>';
echo $row['link'];
echo '</li>';
我也尝试过使用
$query = "SELECT link FROM navigation WHERE permissions<='3'";
$result = mysqli_query($link, $query);
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
echo '<li>';
echo $row['link'];
echo '</li>';
我原来的代码是一个简单的mysql_select,而且是
$result=mysql_query("SELECT * FROM navigation WHERE enabled='1' AND permission<='3'");
while ($row=
mysql_fetch_array($result))
{
echo '<li>';
echo $row['link'];
echo '</li>';
}
?>
答案 0 :(得分:2)
此:
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC));
echo '<li>';
echo $row['link'];
echo '</li>';
应该是这样的:
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
echo '<li>';
echo $row['link'];
echo '</li>';
}
你必须这样定义代码块。