我试图将表格数据从我的html页面发布到PHP,但我无法看到我发布的数据。请帮助。这是我的代码:
html文件的代码:
<!doctype html>
<html>
<head>
<title>
Intro Page.
</title>
</head>
<body>
<form action="receive.php" method="POST">
Name:<input type="text" name="username">
Password:<input type="text" name="password">
<input type="submit" value="submit">
</body>
</html>
我的receive.php文件的代码是:
<?php
$name=$pass="";
$name=$_POST["username"];
$pass=$_POST["password"];
if($_SERVER["REQUEST_METHOD"]=="POST")
{
$name= test_input($_POST["username"]);
$pass= test_input($_POST["password"]);
}
function test_input($data)
{
$data=trim($data);
$data=stripslashes($data);
$data=htmlspecialchars($data);
return $data;
}
echo $name;
echo $pass;
?>
答案 0 :(得分:0)
我首先更改您的表单提交按钮以获得名称:
<input type="submit" name="submit" value="submit>
允许您将PHP修改为:
function test_input($data) {
$data=trim($data);
$data=stripslashes($data);
$data=htmlspecialchars($data);
return $data;
}
$name=$pass = '';
if(isset($_POST['submit'])) {
$name = test_input($_POST['name']);
$pass = test_input($_POST['pass']);
}
echo $name;
echo $pass;
您可能会收到您不了解的错误。在调试/编写代码时,最好通过在PHP脚本的顶部添加以下内容来启用错误报告:
<?php
ini_set('display_errors', 1);
error_reporting(-1); // or E_ALL
答案 1 :(得分:0)
1)检查php安装是否正确
2)知道该文件位于apache根目录下
测试:
直接打开receive.php并检查第一个php是否工作。
<强> HTML 强>
<input type="submit" name="form1_submit" value="submit">
根据@kunruh提到
You are missing your closing </form>
<强> PHP 强>
<?php
$name=$pass="";
function test_input($data)
{
$data=trim($data);
$data=stripslashes($data);
$data=htmlspecialchars($data);
return $data;
}
//you have to check the value post like because some times more than one form in your html means it make conflict so you have to use name isset .
if(isset($_POST['form1_submit'])) {
{
$name= test_input($_POST["username"]);
$pass= test_input($_POST["password"]);
echo $name;
echo $pass;
}
?>
答案 2 :(得分:0)
确保将receive.php文件放在html文件附近。例如:
如果html文件位于以下路径/var/www/html/YOUR_HTML_FILE.html
然后receive.php文件也应放在同一路径/var/www/html/receive.php中。否则,请在表单操作<form action="/var/www/html/receive.php" method="POST">
中提及正确的路径。同时将密码字段从文本更改为密码。
Text field - <input type="text" name="password">
<br>
Password field - <input type="password" name="password">
答案 3 :(得分:0)
尝试这样的事情......
Collections.sort(resultList, new Comparator<Item>() {
@Override
public int compare(Item lhs, Item rhs) {
return lhs.getDistance().compareTo(rhs.getDistance());//mungkin valuenya null
}
});
和“welcome_get.php”如下所示:
<html>
<body>
<form action="welcome_get.php" method="get">
Name: <input type="text" name="name"><br>
E-mail: <input type="text" name="email"><br>
<input type="submit">
</form>
</body>
</html>