我已經創建了HTML表單,點擊提交按鈕後所有值都存儲在MySQL數據庫中。HTML表單定義插入數據庫
我想創建編輯表單的功能。點擊Edit
後,我正在考慮按鈕表單應該與新表單完全相同,只需將所有填充值插入到數據庫即可。
我該如何做到這一點?我以爲喜歡存儲HTML代碼的值與數據庫(從<form>
到</form>
),但我不知道如果這是好主意,它是可能的。
我真正形成有超過100場,但簡化它看起來像:
HTML表單:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Add Record Form</title>
</head>
<body>
<form action="insert.php" method="post">
<p>
<label for="firstName">First Name:</label>
<input type="text" name="firstname" id="firstName">
</p>
<p>
<label for="lastName">Last Name:</label>
<input type="text" name="lastname" id="lastName">
</p>
<p>
<label for="emailAddress">Email Address:</label>
<input type="text" name="email" id="emailAddress">
</p>
<input type="submit" value="Submit">
</form>
</body>
</html>
PHP代碼插入到數據庫:
<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "root", "", "demo");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Escape user inputs for security
$first_name = mysqli_real_escape_string($link, $_POST['firstname']);
$last_name = mysqli_real_escape_string($link, $_POST['lastname']);
$email_address = mysqli_real_escape_string($link, $_POST['email']);
// attempt insert query execution
$sql = "INSERT INTO persons (first_name, last_name, email_address) VALUES ('$first_name', '$last_name', '$email_address')";
if(mysqli_query($link, $sql)){
echo "Records added successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
// close connection
mysqli_close($link);
?>
是的,它是可能的。你必須點擊編輯獲取所有數據,然後填寫表格。看看這個教程 - http://www.startutorial.com/articles/view/php-crud-tutorial-part-3/ –
另外,閱讀使用PHP的現代連接方法時準備好的語句。 – Strawberry