我正在尝试使用我的数据库中的员工动态创建一个复选框列表。该清单是为了参加会议。我尝试使用我在此站点上找到的解决方案来创建列表,但我收到的错误是我的代码中出现意外结束。我找不到了。
<?php
$servername = "localhost";
$username = "root";
$password = "password";
$dbname = "purpletrainer";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
//echo "Connected successfully";
$MeetingID = $_POST["meetingid"];
$sql = "SELECT employee.emplname
FROM purpletrainer.employee, purpletrainer.meeting
WHERE employee.empmngrid=meeting.meetingleader
AND meetingid='$MeetingID'";
$result = $conn->query($sql);
//Iterate over the results that you've gotten from the database
if ($result->num_rows > 0){
while($employee = $result->fetch_assoc())
{
?>
<div>
<form action="processlist.php" method="post">
<span><?php echo $employee['emplname']; ?></span>
<input type="checkbox" name="employees[]" value= '<?php echo $employee[0]; ?>' /><br />
</form>
</div>
}
}
答案 0 :(得分:0)
您需要正确终止PHP代码:
....
</form>
</div>
<?
}
}
?>
答案 1 :(得分:0)
在代码的末尾,您需要一个额外的PHP标记用于结束}
,它当前打印为HTML
答案 2 :(得分:0)
实际上还没有完成你的while循环和括号语法错误。您的if条件代码替换为以下代码。
if ($result->num_rows > 0){
while($employee = $result->fetch_assoc())
{
?>
<div>
<form action="processlist.php" method="post">
<span><?php echo $employee['emplname']; ?></span>
<input type="checkbox" name="employees[]" value= '<?php echo $employee[0]; ?>' /><br />
</form>
</div>
<?php
}
}
?>