DOB mysql插入

Question

我試圖從下拉列表中插入數據到我的數據庫，但是我遇到了一個錯誤，請幫忙。 DOB在我的數據庫中設置爲日期格式。

PS。然而，它插入到我的年月天罰款沒有任何問題。

在此先感謝，併爲貧困英語技能對不起=（

下面是我的PHP代碼

<?php 

require '../ppuyakul/php/db_conn.php'; 

error_reporting(0); 
$message = ''; 
$year = $_POST['year']; 
$month = $_POST['month']; 
$date = $_POST['date']; 
$DOB = date("Y-m-d", mktime(0,0,0,$month, $day, $year)); 



if(!empty($_POST['email']) && !empty($_POST['password']) && !empty($_POST['fullname']) && !empty($_POST['username']) && !empty($_POST['password_confirmation']) && !empty($_POST['gender']) && !empty($_POST['country']) && !empty($_POST['state']) && !empty($_POST['city']) && !empty($_POST['day']) && !empty($_POST['month']) && !empty($_POST['year'])): 

    // Enter the new user in the database 
    $sql = "INSERT INTO assignment2 (fullname,username, email, password, passwordcon, gender, country, state, city, day, month, year, DOB) VALUES (:fullname, :username, :email, :password, :password_confirmation, :gender, :country, :state, :city, :day, :month, :year, DOB)"; 
    $stmt = $conn->prepare($sql); 

    $stmt->bindParam(':fullname', $_POST['fullname']); 
    $stmt->bindParam(':username', $_POST['username']); 
    $stmt->bindParam(':email', $_POST['email']); 
    $stmt->bindParam(':password', password_hash($_POST['password'], PASSWORD_BCRYPT)); 
    $stmt->bindParam(':password_confirmation', password_hash($_POST['password_confirmation'], PASSWORD_BCRYPT)); 
    $stmt->bindParam(':gender', $_POST['gender']); 
    $stmt->bindParam(':country', $_POST['country']); 
    $stmt->bindParam(':state', $_POST['state']); 
    $stmt->bindParam(':city', $_POST['city']); 
    $stmt->bindParam(':day', $_POST['day']); 
    $stmt->bindParam(':month', $_POST['month']); 
    $stmt->bindParam(':year', $_POST['year']); 
    $stmt->bindParam(':DOB', $_POST['year'], $_POST['month'], $_POST['day']); 


    if($stmt->execute()): 
    $message = 'Successfully created new user'; 
    else: 
    $message = 'Sorry there must have been an issue creating your account'; 
    endif; 
endif; 

?> 

這裏是我的HTML

<div> 
      <p style="display: inline; margin-right: 1%"><span style="font-weight: bold;">DATE OF BIRTH :</span></p> 
      <select class="formInputDate" name="day" value="" id="day"></select> 
      <select class="formInputDate" name="month" id="month"> 
           <option value="1">Jan</option> 
           <option value="2">Feb</option> 
           <option value="3">Mar</option> 
           <option value="4">Apr</option> 
           <option value="5">May</option> 
           <option value="6">Jun</option> 
           <option value="7">Jul</option> 
           <option value="8">Aug</option> 
           <option value="9">Sep</option> 
           <option value="10">oct</option> 
           <option value="11">Nov</option> 
           <option value="12">Dec</option> 
      </select> 
      <select class="formInputDate" name="year" value="" id="year"> 
      </select> 
      </div> 

Answer 1

改變這一行下拉代碼：

$stmt->bindParam(':DOB', $_POST['year'], $_POST['month'], $_POST['day']); 

到：

$stmt->bindParam(':DOB', $_POST['year'].'-'. $_POST['month'].'-'. $_POST['day']); 

的原因是因爲日期格式要求正確格式的日期一樣YYYY-MM-DD。