我试过使用$_POST、$_GET和$_REQUEST,但它们都不起作用。我应该使用什么来保存输入的答案?
<!DOCTYPE html>
<html>
<head> <title>Sign Up</title> </head>
<body>
<form method="POST">
<input type="text" name="first" id='first'>
<input type="text" name='last' id='last'>
<button type='submit' name='submit'>Click</button>
</form> //WHERE THE INFORMATION IS ASKED
<?php //WHERE I ATTEMPT TO SAVE THE INFORMATION BUT I FAIL
include_once 'dbh.inc.php';
$first = $_POST['first'];
$last = $_POST['last'];
/*Notice: Undefined index: first in C:\xampp\htdocs\index.php on line 14 Notice: Undefined index: last in C:\xampp\htdocs\index.php on line 15*/
$sql = "INSERT INTO keep1(Firstname, Lastname) VALUES('$first', '$last');"; mysqli_query($conn, $sql); ?>
</body>
</html>
附加的代码段:
注册单击//在询问信息的位置
发布于 2018-07-09 00:57:53
你应该注意你的代码,使它易于阅读。所以其他人可能会读到你的代码。
尝试此操作,可能会出现错误,因为您尚未发布数据。
<!DOCTYPE html>
<html>
<head>
<title>Sign Up</title>
</head>
<body>
<form method="POST">
<input type="text" name="first" id="first">
<input type="text" name="last" id="last">
<button type="submit">Click</button>
</form>
<?php
include_once 'dbh.inc.php';
if(isset($_POST['first']) && isset($_POST['last'])){
$first = $_POST['first'];
$last = $_POST['last'];
$sql = "INSERT INTO keep1(Firstname, Lastname) VALUES('$first', '$last');";
mysqli_query($conn, $sql);
}
?>
</body>
</html>
https://stackoverflow.com/questions/51233739
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