Question

我有一个global dvars var1 var2 var3 local append "replace" foreach dvar in $dvars { reg `dvar' treatment outreg2 using "file", `append' local append "append" }的div，我希望在ajax调用成功时刷新。我试过这个。但除非我刷新整个页面，否则它不会刷新div。这是代码。

股利

index.php

的Ajax

<div id="cartContainer">
    <div id="cart">
        <li style="color: #515151">
            <img id="cart_img" src="images/cart.png"> Cart <span class='badge' id='comparison-count'>
      <?php
      if(isset($_SESSION['cart'])&& !empty($_SESSION['cart']))
      {
        $cart_count=count($_SESSION['cart']);
        echo $cart_count;
      } else {
        $cart_count=0;
        echo $cart_count;
      }
        ?>
      </span>
        </li>
        <div id="sidebar">
            <?php if(isset($_SESSION[ 'cart'])&& !empty($_SESSION[ 'cart'])){ ?>
            <table id="s_table">
                <?php foreach($_SESSION[ 'cart'] as $id=> $value){ ?>
                <form class="product" method="get" action="index.php?">
                    <tr id="tr_s">

                        <input type="hidden" name="action" value="remove">
                        <input type="hidden" name="id" value="<?php echo $value['id'] ?>">
                        <td class="s_th">
                            <?php echo $value[ 'name']; ?>
                            <input type="hidden" name="name" value="<?php echo $value['name'] ?>">
                        </td>
                        <td class="s_th">
                            <?php echo $value[ 'quantity'] ?>
                            <input type="hidden" name="quantity" value="<?php echo $value['quantity'] ?>">
                        </td>
                        <td class="s_th">
                            <?php echo $value[ 'color']; ?>
                            <input type="hidden" name="color" value="<?php echo $value['color'] ?>">
                        </td>
                        <td class="s_th">
                            <?php echo $value[ 'size'] ?>
                            <input type="hidden" name="size" value="<?php echo $value['size'] ?>">
                        </td>

                        <td>
                            <button type="submit" id="btn_submit">Remove</button>
                        </td>
                    </tr>
                </form>
                <?php } ?>
                <tr>
                    <td class="cart_btn" colspan="2"><a href="index.php?page=cart" style="">GO TO CART</a>
                    </td>
                    <td class="cart_btn" colspan="2"><a href="index.php?page=checkout">CHECK OUT</a>
                    </td>
                </tr>
            </table>
            <?php } else { ?>
            <p id="p_s"><i> "Your Cart is Empty"</i>
            </p>
            <?php } ?>
        </div>
    </div>
</div>

Answer 1

如果你想用AJAX的回复来更新div，你可以这样做：

 $.post('add_cart.php', post_data, function(response) {
     $("#cartContainer").html(response);
 }

当然，这取决于很多事情：您返回的数据类型，您希望更新的内容等等。很难说明您发布的内容是什么意思。

div id刷新ajax成功

1 个答案: