可以显示这样的内容吗
$con=mysqli_connect("localhost","root","","login");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
////////////////////////////////user1
if($session->username=='user1')
{
echo "<table border='1'>
<tr>
<th><b>column1</b></th>
<th><b>column2</b></th>
<th><b>column3</th>
/tr>";
$result = mysqli_query($con,"SELECT * FROM test WHERE user1 == 0 ");
//if($result =='0')
//{
//echo"Nothing to show";
//}
//else
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
if ($row['user1'] == '1'){echo "<td> confirmed </td>";} elseif ($row['user1'] == '2') {echo "<td> not confirmed </td>";} else{echo "<td>waiting</td>";}
if ($row['user2'] == '1'){echo "<td>confirmed </td>";} elseif ($row['user2'] == '2') {echo "<td> not confirmed </td>";} else{echo "<td>waiting</td>";}
if ($row['user3'] == '1'){echo "<td> confirmed </td>";} elseif ($row['user3'] == '2') {echo "<td> not confirmed </td>";} else{echo "<td>waiting</td>";}
echo "</tr>";
}
echo "</table>";
}
帮帮我吧出现了错误警告:mysqli_fetch_array()期望参数1为mysqli_result,布尔值
mysql中的比较运算符是=
,而不是像PHP中的==
。
这么写:
$result = mysqli_query($con,"SELECT * FROM test WHERE user1 = 0 ");
// Only one `=` sign here ----------------------------------^
如果您还遇到这种类型的问题,请在执行查询后先查看$con->error
的值