我刚刚编写了这段代码,其中有一个表单。您必须写下您的名字、姓氏和国家。您还必须选择自己喜欢的颜色。之后,您按下提交按钮,以便之后可以查看数据。我在 1 页上使用 GET 方法,但我必须在 POST 方法中使用第二个方法,以便每个回显都在第二页上。我怎么能那样做?我的代码是:
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<title>Form</title>
<link rel="stylesheet" type="text/css" href="form.css">
</head>
<body>
<section>
<?php
if (isset($_GET["name"])){
$name = $_GET["name"];
if ($name != ""){
$surname = $_GET["surname"];
$country = $_GET["country"];
$colour = $_GET["colour"];
echo "<p>";
echo "<h2>Data</h2>";
echo $name . " " . $surname . "</br />";
echo $country . "<br />";
echo $colour;
echo "</p>";
}else
echo "<strong>Complete the blank spaces</strong>";
}else{
?>
<h1>Form</h1>
<form class="elegant" method="GET" action="?">
<fieldset>
<legend>Favourite colour</legend>
<div>
<label for="nombre">Name</label>
<input type="text" placeholder="Name" name="name"
id="name" />
</div>
<div>
<label for="surname">Surname</label>
<input type="text" placeholder="Surname" name="surname"
id="surname" size="50" />
</div>
<div>
<label for="country">Country</label>
<input type="text" placeholder="Country" name="country" id="country"
size="10" maxlength="9" />
</div>
<div>
<select name="colour" id="colour">
<option value="yellow" <?php if ($colour == "yellow" ) echo "selected" ?> >yellow</option>
<option value="red" <?php if ($colour == "red" ) echo "selected" ?> >red</option>
</div>
<input class="btn" type="submit" value="Save" />
</fieldset>
</form>
<?php
}
?>
</section>
</body>
</html>
我知道我必须使用指向第二页的链接,但我只知道这些。提前致谢!
jeck猫
九州编程
30秒到达战场