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我的URL目前擁有:飼養已經設置GET參數值
/add_products_flavour.php?pro_id=2
它
。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;
}
}
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
結果仍然是/add_products_flavour.php?pro_id= – kalvin
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