2017-03-16 63 views
0

我的URL目前擁有:飼養已經設置GET參數值

/add_products_flavour.php?pro_id=2 

。但是,提交表單後,pro_id值會丟失。我想繼續返回到同一頁面,並保持pro_id的值能夠多次提交表單。

形式:

<form name="flavour_form" method="post" action="add_products_flavour.php"> 
      <label class="add_products_field_label" for="flavour">Aroma:</label> 
      <input class="w-input" data-name="flavour" id="flavour" maxlength="256" name="flavour" required="required" type="text"> 
      <input class="add_products_form_button w-button" type="submit" name="insert_post" value="Adauga"> 
      <?php insertFlavour()?> 
     </form> 

function insertFlavour(){ 

    global $connection; 
    $pro_id=getProductID(); 

     if (isset($_POST['insert_post'])) { 
      $flavour = $_POST['flavour']; 

      $insert_flavour_query = "insert into flavour (name) values ('$flavour')"; 
      $run_insert_flavour = mysqli_query($connection, $insert_flavour_query); 


      if($run_insert_flavour){ 
       echo"<script>alert('flavour added!')</script>"; 
      echo "<script>window.open('add_products_flavour.php?pro_id=$pro_id','_self')</script>"; 
     } 

    } 

} 

function getProductID(){ 
    if(isset($_GET['pro_id'])) { 
     $pro_id = $_GET['pro_id']; 
     return $pro_id; 
    } 
} 
+0

你能不能用'頭解決了這個問題:;'後( '?位置add_products_flavour.php pro_id =' $ pro_id「。)提交已完成。在PHP而不是jQuery – Toxide82

+0

結果仍然是/add_products_flavour.php?pro_id= – kalvin

+0

它是否從你的函數正確返回ID?你有沒有試圖迴應這個確定。 – Toxide82

回答

0

<?php $pro_id = $_GET['pro_id']; 
     echo "<form name='flavour_form' method='post' action='add_products_flavour.php?pro_id=$pro_id'>"; 
      ?>