下面的代码运行良好。它给出预订ID RRGPGM68
的hotel url
<?PHP
$con=mysqli_connect("localhost","username","pasword","dbName");
// Check connection
if (mysqli_connect_errno()) {
die ("Failed to connect to MySQL: " . mysqli_connect_error());
}
$booking_id_serch_text = RRGPGM68;
$searchroute = "select * from booking_tbl where booking_nmbr = '$booking_id_serch_text'";
$result = $con->query($searchroute);
$row = $result->fetch_assoc();
echo 'Hotel URL: '.$row['hotel_url'];
mysqli_close($con);
?>
但是我想从文本输入字段动态地分配booking_id_serch_text
。所以我修改了代码如下:
<form action="<?php echo get_permalink();?>" method="post">
<input type="text" placeholder="Search..." required name="search_text">
<input type="button" value="search" name="search">
</form>
<?PHP
if($_POST["search"]){
$serch_text = $_POST["search_text"];
$con=mysqli_connect("localhost","username","pasword","dbName");
// Check connection
if (mysqli_connect_errno()) {
die ("Failed to connect to MySQL: " . mysqli_connect_error());
}
$booking_id_serch_text = $serch_text;
$searchroute = "select * from booking_tbl where booking_nmbr = '$booking_id_serch_text'";
$result = $con->query($searchroute);
$row = $result->fetch_assoc();
echo 'Hotel URL: '.$row['hotel_url'];
mysqli_close($con);
}
?>
但是上面的代码什么也没做。我怎样才能让它工作?我将在存档此目的后添加验证。
<input type="button" /> buttons
不会提交表单-默认情况下他们不会做任何事情。它们通常与JavaScript一起使用,作为AJAX应用程序的一部分。
<input type="submit"> buttons
将在用户单击它们时提交它们所在的表单,除非您使用JavaScript另行指定。
所以在这里使用
在按下search
时,您必须使用提交按钮来发布表单
<input type="submit" value="search" name="search">
您没有提交表单
<input type="button" value="search" name="search">
应该是
<input type="submit" value="search" name="search">