我的代码被卡住了，因为无论项目添加成功与否，我都无法显示通知。如何使用ajax和php显示它？
这个div将显示通知，但我发现很难解决，因为它在另一个php文件中：
<center><div class="notify"></div></center>
ajax代码：
dialog = $( "#dialog-form" ).dialog({
buttons: {
"Add": function() {
var reg=$("#registration-form").valid();
if(reg==false){
$('#registration-form .submit').click();
}
else if(reg==true){
var product_code = $("#product_code_txt").val();
var product = $("#product_name_txt").val();
var unit = $("#unit_txt").val();
var quantity = $("#quantity_txt").val();
var c_price = $("#c_price_txt").val();
var s_price = $("#s_price_txt").val();
var date = $(".date_txt").val();
var cat_id = $("#drpdown").val();
$.ajax({
type: "POST",
url: "getAddMedicine.php",
data: {product_code_txt: product_code, product_name_txt: product, unit_txt: unit, quantity_txt: quantity, c_price_txt: c_price, s_price_txt: s_price, date_txt: date, category_txt: cat_id},
success: function(html){
$(".displayData").load("getMedicine.php");
dialog.dialog("close");
}
});
return false;
}
},
Cancel: function() {
dialog.dialog( "close" );
}
},
插入数据库的代码：
<?php
$product_code=mysqli_real_escape_string($conn, $_POST['product_code_txt']);
$check="Call checkMedicine('$product_code')";
$result=mysqli_query($conn, $check);
$rowcount=mysqli_num_rows($result);
if($rowcount==0){
include("db.php");
$product_code=mysqli_real_escape_string($conn, $_POST['product_code_txt']);
$product=mysqli_real_escape_string($conn, $_POST['product_name_txt']);
$cat_id=mysqli_real_escape_string($conn, $_POST['category_txt']);
$unit=mysqli_real_escape_string($conn, $_POST['unit_txt']);
$date=mysqli_real_escape_string($conn, $_POST['date_txt']);
$quantity=mysqli_real_escape_string($conn, $_POST['quantity_txt']);
$s_price=number_format($_POST['s_price_txt'], 2);
$s_price=mysqli_real_escape_string($conn, $s_price);
$c_price=number_format($_POST['c_price_txt'], 2);
$c_price=mysqli_real_escape_string($conn, $c_price);
$sql = "Call addMedicine('$product_code','$product','$unit','$quantity','$c_price','$s_price','$date','$cat_id')";
if (mysqli_query($conn, $sql)) {
echo "1"; //how to display it into the div
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
}
else
echo "2"; //how to display it into the div
mysqli_close($conn);
?>
在success函数中，html参数是来自php页面的响应。使用它来设置这样的通知：
success: function(html){
$(".displayData").load("getMedicine.php");
$('. notify').html(html); // add this line
dialog.dialog("close");
},
如果你想做更多的事情，比如用不同的颜色显示通知，你需要在发送