因此，該網站的想法是，當您按下按鈕時，它會從數據庫中刪除相應的行。但是，我的代碼的問題在于，在第一次按下按鈕后，按鈕將填充id變量，然后在下次按下時執行php。我該如何避免這種情況，并讓他php適合并在第一次按下按鈕時執行php？包括嵌入了php的HTML頁面：<!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 = "";}
3 回答
忽然笑
TA貢獻1806條經驗 獲得超5個贊
PHP是一種僅在服務器端運行的語言。它不同于在客戶端（瀏覽器上）運行的javascript。
如果您可能注意到，您將永遠不會在其他網站上看到PHP代碼，因為（如果網站使用PHP）它是在服務器端處理的，并且返回的只是HTML + CSS。
如今，我們不會創建這樣的頁面。我們將使用PHP創建我們所謂的API。我們將使用 REST-API 策略。
假設我們將在此地址上托管此 APIwww.mywebsite.com/myapi/myfreezerendoint.php
我們將使用JS請求或發布到這個端點，并使用名為.fetch("www.mywebsite.com/myapi/myfreezerendoint.php", { ... })
我發現了這個關于以MySQL為數據庫的PHP API的有趣教程：https://webdamn.com/create-simple-rest-api-with-php-mysql/
largeQ
TA貢獻2039條經驗 獲得超8個贊
我已經糾正和美化了所有代碼，最小化了它，現在應該是它：
<?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>
<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>
<form action="<?php echo $_SERVER["PHP_SELF"];?>" class="needs-validation" novalidate method="post">
<div class="form-group">
<label for="uname"></label>
<input type="text" class="form-control" id="itemx" placeholder="Enter an item for the freezer here" name="itemx" required>
<div class="valid-feedback">Valid.</div>
<div class="invalid-feedback">Please fill out this field.</div>
</div>
<button id = "SubmitButton" name = "SubmitButton" type="button submit" class="btn btn-primary">Add to list</button>
</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, $sql);
echo "<table><tr>
<th>Item name</th>
<th>Date added</th>
<th>remove</th>
</tr> ";
while($row = mysqli_fetch_array($result)) {
echo " <tr><td>" . $row['item'] . "</td><td>" . $row['reg_date'] . "</td><td><form action='' method='POST'>
<div class= 'input-group' > <div class='input-group-append'>
<button class='btn btn-danger' onclick='deleteitem(".$row['id'].")' id='delete'>Remove</button>
<input type='hidden' name='id'/>
</div></div>
</form></td></tr>";
}
echo "</table>";
?>
<script>
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復，我在那里留下了一些未使用的html元素，但從理論上講，這應該有效。它利用 將數據發送到 。您不必使用，但從理論上講，這應該可以解決您的問題。如果您遇到任何問題，請在我的答案下發表評論。ajaxphpajax
注意：只是為了澄清，我沒有嘗試這樣做，因此可能存在一些錯誤。
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