首页猿问必须按下按钮才能激活php代码？

必须按下按钮才能激活php代码？

PHP

千万里不及你 2022-08-05 16:07:01

因此，该网站的想法是，当您按下按钮时，它会从数据库中删除相应的行。但是，我的代码的问题在于，在第一次按下按钮后，按钮将填充id变量，然后在下次按下时执行php。我该如何避免这种情况，并让他php适合并在第一次按下按钮时执行php？包括嵌入了php的HTML页面：<!doctype html><html><head><meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.4.1/css/bootstrap.min.css"> <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script> <script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.16.0/umd/popper.min.js"></script> <script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.4.1/js/bootstrap.min.js"></script><title>Untitled Document</title> <style>table { font-family: arial, sans-serif; border-collapse: collapse; width: 100%;}td, th { border: 1px solid #dddddd; text-align: left; padding: 8px;}tr:nth-child(even) { background-color: #dddddd;}</style></head><body> <div class="container-fluid"> <h2 style"text-align:center";>Please enter the item you want to add to the list below </h2><?php$servername = "localhost";$username = "root";$password = "";$dbname = "myDB";$sql = "CREATE DATABASE myDB";$conn = new mysqli($servername, $username, $password, $dbname);if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error);} $sql = "CREATE TABLE freezerinventory (id INT AUTO_INCREMENT PRIMARY KEY, item VARCHAR(30) NOT NULL,reg_date TIMESTAMP DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP)";if(isset($_POST['itemx'])) {$itemvar = $_POST["itemx"];$sql = "INSERT INTO freezerinventory (item)VALUES ('$itemvar')"; $add = mysqli_query($conn, $sql);}else { $sql = "";}

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3 回答

烙印99

TA贡献1829条经验获得超13个赞

您可以创建单独的处理脚本，并将表单提交到按钮的 URL 路径“onClick”。不能使用 JS 代码直接调用 PHP 函数。

反对回复 2022-08-05

忽然笑

TA贡献1806条经验获得超5个赞

PHP是一种仅在服务器端运行的语言。它不同于在客户端（浏览器上）运行的javascript。

如果您可能注意到，您将永远不会在其他网站上看到PHP代码，因为（如果网站使用PHP）它是在服务器端处理的，并且返回的只是HTML + CSS。

如今，我们不会创建这样的页面。我们将使用PHP创建我们所谓的API。我们将使用 REST-API 策略。

假设我们将在此地址上托管此 APIwww.mywebsite.com/myapi/myfreezerendoint.php

我们将使用JS请求或发布到这个端点，并使用名为.fetch("www.mywebsite.com/myapi/myfreezerendoint.php", { ... })

我发现了这个关于以MySQL为数据库的PHP API的有趣教程：https://webdamn.com/create-simple-rest-api-with-php-mysql/

反对回复 2022-08-05

largeQ

TA贡献2039条经验获得超7个赞

我已经纠正和美化了所有代码，最小化了它，现在应该是它：

<?php

$servername = "localhost";

$username = "root";

$password = "";

$dbname = "myDB";

$sql = "CREATE DATABASE myDB";

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {

die("Connection failed: " . $conn->connect_error);

}

$sql = "CREATE TABLE freezerinventory (

id INT AUTO_INCREMENT PRIMARY KEY,

item VARCHAR(30) NOT NULL,

reg_date TIMESTAMP DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP)";

if(isset($_POST['itemx'])) {

$itemvar = $_POST["itemx"];

$stmt = $conn->prepare("INSERT INTO freezerinventory (item) VALUES ('?')");

$stmt->bind_param("s", $itemvar);

$stmt->execute();

} else {

//the 'itemx' post parameter isn't set, make an alert or something

}

mysqli_close($conn);

<!doctype html>

<html>

<head>

<title>Untitled Document</title>

<style>

table {

font-family: arial, sans-serif;

border-collapse: collapse;

width: 100%;

}

td, th {

border: 1px solid #dddddd;

text-align: left;

padding: 8px;

}

tr:nth-child(even) {

background-color: #dddddd;

}

</style>

</head>

<body>

<h2 style"text-align:center";>Please enter the item you want to add to the list below </h2>

<form action="<?php echo $_SERVER["PHP_SELF"];?>" class="needs-validation" novalidate method="post">

<div class="valid-feedback">Valid.</div>

<div class="invalid-feedback">Please fill out this field.</div>

</div>

</form>

</div>

<?php

$servername = "localhost";

$username = "root";

$password = "";

$dbname = "myDB";

$aVar = mysqli_connect('localhost','root','','myDB');

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {

die("Connection failed: " . $conn->connect_error);

}

if(isset($_POST['id'])) {

$delete = $con->prepare("DELETE FROM freezerinventory WHERE id= ?");

$delete->bind_param("s", $_POST['id']);

$delete->execute();

}

$sql = "SELECT * FROM freezerinventory";

$result = mysqli_query($aVar, $sql);

echo "<table><tr>

<th>Date added</th>

<th>remove</th>

</tr> ";

while($row = mysqli_fetch_array($result)) {

echo " <tr><td>" . $row['item'] . "</td><td>" . $row['reg_date'] . "</td><td><form action='' method='POST'>

<button class='btn btn-danger' onclick='deleteitem(".$row['id'].")' id='delete'>Remove</button>

</div></div>

</form></td></tr>";

}

echo "</table>";

function deleteitem (id_data) {

$.post('<?php echo $_SERVER["PHP_SELF"];?>', {id: id_data}, function(data) {

console.log(data); //callback data

});

}

function myFunction() {

confirm("I am an alert box!");

}

(function() {

'use strict';

window.addEventListener('load', function() {

// Get the forms we want to add validation styles to

var forms = document.getElementsByClassName('needs-validation');

// Loop over them and prevent submission

var validation = Array.prototype.filter.call(forms, function(form) {

form.addEventListener('submit', function(event) {

if (form.checkValidity() === false) {

event.preventDefault();

event.stopPropagation();

}

form.classList.add('was-validated');

}, false);

});

}, false);

})();

</script>

</body>

</html>

它可能仍然需要一些修复，我在那里留下了一些未使用的html元素，但从理论上讲，这应该有效。它利用将数据发送到。您不必使用，但从理论上讲，这应该可以解决您的问题。如果您遇到任何问题，请在我的答案下发表评论。ajaxphpajax

注意：只是为了澄清，我没有尝试这样做，因此可能存在一些错误。

反对回复 2022-08-05

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必须按下按钮才能激活php代码？

必须按下按钮才能激活php代码？

3 回答

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