Question

<?php
$age1 = isset($_GET['age1']) ? $_GET['age1'] : "";

if ($age1 <= "18")
{
    $sql = 'select * from events where type_id='.$id.' and id = "1"';
    $tid = mysqli_query($conid, $sql);
}
else
{
    $sql = 'select * from events where type_id='.$id.'';
    $tid = mysqli_query($conid, $sql);
}
?>
-----------
<tr>
    <td class="style1">Age:</td>
    <td class="style2"><input type="text" name="age1" id="age1" class="form-control22" autocomplete="off"></td>
</tr>
<tr>
    <td class="style1">Event Type:</td>
    <td class="style2"><select name="event_id" class="ddl" id="event_id" onchange="geteventprice()">
                       <option value="0">--Select Event Type--</option>
                       <?php while($evt= mysqli_fetch_object($tid))
                        {
                            echo   '<option value="'.$evt->id.'">'.$evt->name.'</option>'; } ?>
                    </select>

    </td>
</tr>

其他声明无效......

Answer 1

算术运算对字符串值不起作用。更新您的代码，如：

 if ($age1 <= 18)
                    {
                        $sql = 'select * from events where type_id='.$id.' and id = "1"';
                        $tid = mysqli_query($conid, $sql);
                    }
                    else
                    {
                        $sql = 'select * from events where type_id='.$id.'';
                        $tid = mysqli_query($conid, $sql);
                    }

Answer 2

<?php

$sql = 'select * from events where type_id='.$id.'';
if (isset($_GET['age1']) <= 18)
{
    $sql.= 'and id = 1';
}
 $tid = mysqli_query($conid, $sql) or die(mysqli_error($conid));
?>

如何使用这种方式。它更干净，更容易理解。您还可以查看查询是否存在问题

if语句不起作用而不提供输出

2 个答案: