2016-04-23 71 views
-1

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 
+2

'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

+1

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。 –

+0

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

回答

0

的PHP MySQL數據庫的擴展提供了有用的信息,當事情出錯代碼,這將是一個好主意,學會利用它們。

因此,如果您將mysqli_error($connect_db)添加到您的消息中,您將會發現查詢中實際發生了什麼問題。

$sql = "INSERT INTO People(Title, FirstName, LastName, EmailAddress, 
          Hobby, Sex, UserName) 
     VALUES('$title','$fname','$lname','$address', 
       '$hobby','$sex','$uname')"; 

if(mysqli_query($connect_db, $sql)){ 
    echo "Record was added"; 
}else 
    // add the error message so you see something useful 
    echo "Record was not added " . mysqli_error($connect_db); 
} 

現在運氣好的話,你可以解決你自己的問題,但如果不是在你的問題發佈錯誤消息,我可以在這個答案擴大

+0

thnks我發現問題 – Pr0digy