PHP - 调用自己的表单动作，如何在1页中显示所有内容？ [英] PHP - form action calling itself, how to display everything in 1 page?
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问题描述
test1.php
< HTML>
< head>
< title>样本< / title>
< / head>
< body>
< form action =test2.phpmethod =post>
请输入一个数字< input type =numbername =userNumber>< br>
< input type =submit>
< / form>
< / body>
< / html>
test2.php
< HTML>
< head>
< title>示例显示页面< /标题>
< / head>
< body>
<？php
$ user_number = $ _POST [userNumber];
echo您已选择$ user_number;
？>
< / body>
< / html>
我想知道如何才能将它全部显示在单个页面上，即在test1上.php，而不必拥有两个文件。
解决方案
您的 test1.php 将需要看起来像这样
< html>
< head>
< title>样本< / title>
< / head>
< body>
< form action =test1.phpmethod =post>
请输入一个数字< input type =numbername =userNumber>< br>
< input type =submit>
< / form>
<？php
if（isset（$ _ POST [userNumber]））{
$ user_number = $ _POST [userNumber];
echo您已选择$ user_number;
}
？>
< / body>
< / html>
I have two .php files as such:
test1.php
<html>
<head>
<title>Sample</title>
</head>
<body>
<form action="test2.php" method="post">
Please enter a number <input type="number" name="userNumber"><br>
<input type="submit">
</form>
</body>
</html>
test2.php
<html>
<head>
<title>Sample Display Page</title>
</head>
<body>
<?php
$user_number = $_POST["userNumber"];
echo "You have chosen $user_number";
?>
</body>
</html>
I want to know how can I get it all to display on a single page, i.e. on the test1.php, without having to have two files.
解决方案
Your test1.php will need to look like this
<html>
<head>
<title>Sample</title>
</head>
<body>
<form action="test1.php" method="post">
Please enter a number <input type="number" name="userNumber"><br>
<input type="submit">
</form>
<?php
if(isset($_POST["userNumber"])) {
$user_number = $_POST["userNumber"];
echo "You have chosen $user_number";
}
?>
</body>
</html>
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