php - 如何在sql查询中插入php变量？

Question

现在我有两个下拉列表，一个是名称，另一个是时间。我希望当用户在第一个下拉框中单击所需的名称时，第二个下拉框中显示的时间应该属于第一个下拉框中选择的名称。

$con=mysqli_connect("localhost","root","123","fyp");
$query1 = mysqli_query($con,"SELECT fname FROM lecturer");

echo "Select lecturer:<select name= 'fname'>";
$name = 'fname';
$name = mysql_real_escape_string($name);
while($row=mysqli_fetch_array($query1))
{
   echo "<option value='". $row['fname']."'>".$row['fname']. '</option>';
}
   echo '</select>';

$con=mysqli_connect("localhost","root","123","fyp");
$query3 = mysqli_query($con,"SELECT stime FROM studbooking WHERE lecname is '$name'");

echo "Select Booking time:<select name= 'stime'>";
while($row=mysqli_fetch_array($query3))
{
   echo "<option value='". $row['stime']."'>".$row['stime']. '</option>';
}
   echo '</select>';

如何在sql查询中使用php变量？

Answer 1

您可以使用 mysqli->bind_param 来实现此目的。

mysqli::bind_param

示例

$statement = $mysqli->prepare("SELECT stime FROM studbooking WHERE lecname=?");
$statement->bind_param('s', $name);
$statement->execute();

$result = $statement->get_result();
while ($row = $result->fetch_assoc()) {
    print_r($row);
}