如果这是重复的，我提前道歉，但我搜索了stackoverflow，没有发现任何关于这个主题的内容。在我的base.html中，我有一个这样的表单：
<form method="post" action="../php/base.php">
<input type="hidden" name="addToDB" value="addToDB" />
<input type="text" id="name" name="name" placeholder="Enter name">
<input type="text" id="address" name="address" placeholder="Enter address">
<button type="submit">Submit</button>
</form>
注意一个伪input元素；它将在PHP中用于针对此表单（这是一个好的做法吗？）base.php具有以下功能：
if (isset($_POST['addToDB'])) {
create_tables(); // creates tables in DB
addToDB(); // inserts content to DB
}
function addToDB() {
try {
// open DB connection
$db = new PDO("sqlite:../db/mydb.sqlite");
$name= $_POST['name'];
$address= $_POST['address'];
$db->exec("INSERT INTO Person (P_Name, P_Address) VALUES ('$name', '$address');");
// The following needs to be redirected back to base.html:
print "<table border=1>";
print "<tr><td>Name</td><td>Address</td></tr>";
$result = $db->query('SELECT * FROM Person');
foreach($result as $row)
{
print "<tr><td>".$row['P_Name']."</td>";
print "<td>".$row['P_Address']."</td></tr>";
}
print "</table>";
}catch(PDOException $e)
{
echo $e->getMessage();
}
}
单击提交按钮后，文件base.php将打开并显示一个表。相反，我希望执行base.php中的脚本，并将输出发送回base.html，它将在那里显示（内容div内的某个位置）。
- 我知道我可能需要使用AJAX，但我不知道如何调用外部PHP文件中的特定函数，并将其返回值（可能是一个带有HTML标记的字符串，用于构造
table）重定向回HTML文件（base.html） - 我使用SQLite进行数据存储；使用
create_tables();创建数据库 - 我是PHP的新手，任何意见/建议都将不胜感激
谢谢
您可以使用HTML、jQuery AJAX和PHP尝试以下操作：
HTML:
<form method="post" id="formid">
<input type="hidden" name="addToDB" value="addToDB" />
<input type="text" id="name" name="name" placeholder="Enter name">
<input type="text" id="address" name="address" placeholder="Enter address">
<!-- take simple button -->
<button type="button" id="submit">Submit</button>
</form>
<div id="response"></div>
JavaScript:
<script type="text/javascript">
$(document).ready(function(){
//on click of submit button
$("#submit").click(function(){
$.ajax({
url:"../php/base.php",
data:{
addToDB:"addToDB",
name:$("#name").val(),
address:$("#address").val()
},
type:"POST",
success:function(res){
$("#response").html(res);
}
});
});
});
</script>
不会对您的PHP进行任何更改。祝你好运。
I want the ... output sent back to the base.html
那你为什么不在里面处理呢。
form.php
<?php
// you can save this two functions in a file 'DBhelper.php';
// then you can 'require(DBhelper.php') here to make this file a little cleaner
// function create_tables(){
// put your codes here...
// };
function addToDBandGetAllUsers($name,$address) {
try {
// open DB connection
$db = new PDO("sqlite:../db/mydb.sqlite");
$db->exec("INSERT INTO Person (P_Name, P_Address) VALUES ('$name', '$address');");
//The following needs to be redirected back to base.html:
$result = $db->query('SELECT * FROM Person');
// this data is for test, even the 1 + 1 solution, I will test it, I don't want to post codes in community with errors....
// $result = array(
// array('P_Name' => 'John','P_Address' => 'Foobar'),
// array('P_Name' => 'Alex','P_Address' => 'Qux')
// );
// array_push($result, array('P_Name' => $name,'P_Address' => $address));
return $result;
}catch(PDOException $e)
{
echo $e->getMessage();
}
}
// if the $_POST['name'] is not empty, we will update db
// this is just a basic validation, if we use isset() instead of empty(),
// the form can be submitted with a empty value;
if (! empty($_POST['name']) && ! empty($_POST['address']) ){
create_tables(); // creates tables in DB
// I change the the function name , since this guy do a lot of things....
$result = addToDBandGetAllUsers($_POST['name'],$_POST['address']);
} else{
echo 'Please fill the form.';
}
?>
<!DOCTYPE html>
<html>
<head>
<title>form</title>
</head>
<body>
// if we got $result, that means the form has been submitted, so we can display the table;
<?php if (isset($result)) :
echo '<h1>All users:</h1>';
print "<table border=1>";
print "<tr><td>Name</td><td>Address</td></tr>";
foreach($result as $row)
{
print "<tr><td>".$row['P_Name']."</td>";
print "<td>".$row['P_Address']."</td></tr>";
}
print "</table>";
echo '<hr>';
endif;
?>
<form method="post" action="form.php">
<input type="hidden" name="addToDB" value="addToDB" />
<input type="text" id="name" name="name" placeholder="Enter name">
<input type="text" id="address" name="address" placeholder="Enter address">
<button type="submit">Submit</button>
</form>
</body>
</html>
form元素的"action"是其自身。
您可以更改base.html文件，如下所示：
<form method="post" action="../php/base.php" id="formid" onsubmit='return false;'>
<input type="hidden" name="addToDB" value="addToDB" />
<input type="text" id="name" name="name" placeholder="Enter name">
<input type="text" id="address" name="address" placeholder="Enter address">
<button type="submit" id="submit">Submit</button>
</form>
<div id="display"></div>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#submit").on("click",function(e){
e.preventDefault();
var formdata=$("#formid").serialize();
var url=$("#formid").attr("action");
$.post(url,formdata,function(data){
$("#display").html(data);
},'json');
});
});
</script>
这可能对你有帮助。