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任何人都可以幫忙嗎?所以我正在用wamp服務器做一個php練習。我的表單只接受記錄,當我只有3列時,但是當我添加更多列時記錄不會被添加。Phpmyadmin數據庫不接受記錄
代碼如下
HTML表單
<body>
<form action = "Form.php" method = "post">
<label> Title: </label> <input type = "text" name = "title"/>
<label> First Name: </label> <input type = "text" name = "fname"/>
<label> Last Name: </label> <input type = "text" name = "lname"/>
<label> EmailAddress: </label> <input type = "text" name = "address"/>
<label> Hobby: </label> <input type = "text" name = "hobby"/>
<label> Sex: </label> <input type = "text" name = "gender"/>
<label> UserName: </label> <input type = "text" name = "uname"/>
<input type = "submit" name = "submitbtn" value = "Submit"/>
</form>
</body>
腓頁
<?php
if (isset($_POST['submitbtn'])){
$title = $_POST['title'];
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$address = $_POST['address'];
$hobby = $_POST['hobby'];
$sex = $_POST['gender'];
$uname = $_POST['uname'];
$connect_db = mysqli_connect("localhost","root","","Humans");
if(!$connect_db){
echo "Connection failed";
} else
echo "Connection successful";
if(!mysqli_select_db($connect_db, 'Humans')){
echo "Database not selected";
} else
echo "Database selected";
$sql = "INSERT INTO People(Title, FirstName, LastName, EmailAddress, Hobby, Sex, UserName)VALUES('$title','$fname','$lname','$address','$hobby','$sex','$uname')";
if(mysqli_query($connect_db, $sql)){
echo "Record was added";
}else
echo "Record was not added";
}
?>
輸出:
Connection successful
Database selected
Record was not added
'phpMyAdmin' **不是數據庫**公司使用PHP編寫的一個工具,它允許您維護** MYSQL **數據庫 – RiggsFolly
打印查詢並在PhpMyadmin中運行它以查看錯誤,如echo $ sql; die;在myadmin中複製並運行此查詢。 –
簡單代碼縮進使您的代碼更易於閱讀,更重要的是**更易於調試** – RiggsFolly