2017-10-11 30 views
2
<!DOCTYPE html> 
<html> 
<body> 
<form action="hw4.php" method="POST"> 
    Name: 
    <input type="text" name="name"> 
    <br> 
    ID: 
    <input type="text" name="ID"> 
    <br> 
    Password: 
    <input type="password" name="pass"> 
    <br> 
    Gender: 
    <input type="radio" name="gender" value="male">Male 
    <input type="radio" name="gender" value="female">Female 
    <br> 
    Major: 
    <input type="checkbox" name="major" value="Computer Science">Computer Science 
    <input type="checkbox" name="major" value="Electronic Engineering">Electronic Engineering 
    <input type="checkbox" name="major" value="Global Edison">Global Edison 
    <input type="checkbox" name="major" value=ICT/>ICT 
    <input type="checkbox" name="major" value="Economics">Economics 
    <input type="checkbox" name="major" value="Management">Management 
    <br> 
    Year: 
    <select name = "year"> 
     <option value="Freshman">Freshman</option> 
     <option value="Sophomore">Sophomore</option> 
     <option value="Junior">Junior</option> 
     <option value="Senior">Senior</option> 
    </select> 
    <br> 
    <br> 
    Message 
    <br> 
    <textarea name="message" rows="10" cols="50"> 
    </textarea> 
    <br> 
    <input type="submit" value="submit"> 

</form> 
</body> 
</html> 


<?php 

    $db_host = "localhost" ; 
    $db_user = "root" ; 
    $db_password = "ha9056"; 
    $db_name = "login" ; 
    $conn = mysqli_connect($db_host, $db_user, $db_password, $db_name) ; 

    if(mysqli_connect_errno($conn)) 
     echo "Database connection failed : ".mysqli_connect_error() ; 
    else 
     echo "Database connected<br><br>" ; 

    $name=$_POST['name']; 
    $id=$_POST['id']; 
    $pass=$_POST['pass']; 
    $gender=$_POST['gender']; 
    $major=$_POST['major']; 
    $year=$_POST['year']; 
    $message=$_POST['message']; 

    $qry = "SELECT * FROM login WHERE NAME = '$name'" ; 

    $result = mysqli_query($conn, $qry) ; 
    $row = mysqli_fetch_array($result) ; 

    $qry = "INSERT INTO login VALUES('$name','$id','$pass','$gender','$major','$year','$message')"; 

    mysqli_query($conn, $qry); 

?> 

大家好。我在這裏有一個註冊表單,並希望添加到MySQL數據庫。我寫下了一個php代碼,但沒有正常工作,任何人都有我的代碼推薦?我會非常感激。 性別必須以註冊形式顯示爲男性或女性,並且應該在mysql中顯示爲M或F.年份必須顯示爲新生,大二等,並應保存爲一個數字,如1,2,3,4在mysql 非常感謝。 下面是現在的mysql的圖像。 enter image description here到mysql的HTML註冊表格

+0

您可以將您的性別輸入的值更改爲'M'和'F'代替,而不是'male'和'female'。與 – Swellar

+0

相同哦,它必須在註冊表格中顯示爲男性和女性,並且應該在mysql中顯示爲F或M – David

+0

我的意思是輸入的值屬性,它仍然呈現爲男性或女性,但你會得到的價值是'M'或'F',這幾年一樣 – Swellar

回答

-1

只要改變值的輸入

<input type="radio" name="gender" value="M">Male 
<input type="radio" name="gender" value="F">Female 
<select name = "year"> 
    <option value="1">Freshman</option> 
    <option value="2">Sophomore</option> 
    <option value="3">Junior</option> 
    <option value="4">Senior</option> 
</select> 
+0

@Downvoter請解釋一下,以便我可以相應地調整 – Swellar