Php 能给桌子展示一下这样的东西吗?
好的,可以展示这样的东西吗Php 能给桌子展示一下这样的东西吗?,php,mysql,Php,Mysql,好的,可以展示这样的东西吗 $con=mysqli_connect("localhost","root","","login"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } ////////////////////////////////user1 if($session->userna
$con=mysqli_connect("localhost","root","","login");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
////////////////////////////////user1
if($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>";
}
$con=mysqli_connect(“localhost”、“root”、“login”);
//检查连接
if(mysqli\u connect\u errno())
{
echo“未能连接到MySQL:”.mysqli_connect_error();
}
////////////////////////////////用户1
如果($session->username=='user1')
{
回声“
专栏1
专栏2
第3栏
/tr>“;
$result=mysqli_查询($con,“从测试中选择*,其中user1==0”);
//如果($result='0')
//{
//呼应“没什么可展示的”;
//}
//否则
while($row=mysqli\u fetch\u数组($result))
{
回声“;
如果($row['user1']='1'){echo“已确认”;}elseif($row['user1']='2'){echo“未确认”;}其他{echo“正在等待”;}
如果($row['user2']=='1'){echo“已确认”;}elseif($row['user2']=='2'){echo“未确认”;}其他{echo“正在等待”;}
if($row['user3']=='1'){echo“已确认”}elseif($row['user3']=='2'){echo“未确认”;}else{echo“正在等待”}
回声“;
}
回声“;
}
帮我解决这个问题
我出错了
警告:mysqli_fetch_array()希望参数1是mysqli_result,在mysql中的中给出的布尔值比较运算符是
=
,而不像在PHP中那样,=
所以写下:
$result = mysqli_query($con,"SELECT * FROM test WHERE user1 = 0 ");
// Only one `=` sign here ----------------------------------^
如果您遇到这种类型的问题,请在执行查询后先查看<代码> $CON->错误< /代码>。✔) 在答案的分数下。(只有在提问15分钟后才有可能)是的,我在等15分钟嘿@bwoebi我可以做这样的$result=mysqli_查询($con,“SELECT*FROM test,其中user3=0和user2 1”);似乎它不起作用。建议?如果用户3登录,它将仅显示user3=0和user2=not的表格1@Mooho058
从测试中选择*,其中user3=0和user2!=1
应该有效?(如果user2和user3列存在)改变你的问题结构。代替提问,是否可以像这样显示表格?试着解决真正的问题或bug。