我有一个下拉菜单。当选择该选项时,它会调用 ajax 脚本并根据选择从另一个 php 页面传递值。我想在单击PayPal按钮时将该值传递给PayPal帐户。
阿贾克斯
function showUser(id) {
//get the selected value
//make the ajax call
$.ajax({
url: 'ajax_cat.php',
type: 'GET',
data: {option : id},
success: function(data) {
document.getElementById('cat_cost').innerHTML =data;
}
});
}
形式
<form action="https://www.sandbox.paypal.com/cgi-bin/webscr" method="post">
<div >
<select name="category" class="choose_list" onchange="showUser(this.value)" >
<?php
$query= "SELECT * FROM requests
WHERE cat_action='Waiting'" ;
$result= mysql_query($query);
while($row = mysql_fetch_assoc($result)){
echo '<option value="'.$row['package_id'].'">'.$row['cat'].'</option>';
}
?>
</select>
</div>
<div id="cat_cost">
</div>
<input type="hidden" name="cmd" value="_xclick" />
<input type="hidden" name="business" value="abc@abc.com" />
<?php /*?><input type="hidden" name="item_name" value="<?php echo $type; ?>" /><?php */?>
<input type="hidden" name="item_number" value="<?php ?>" />
<input type="hidden" name="amount" value="<?php ?>" />
<input type="image" src="https://www.paypalobjects.com/en_US/i/btn/btn_buynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!">
<img alt="" border="0" src="https://www.paypalobjects.com/en_US/i/scr/pixel.gif" width="1" height="1">
</form>
制作此 HTML
<select name="category" class="choose_list" id="select_choose_list">
<?php
$query= "SELECT * FROM requests WHERE cat_action='Waiting'" ;
$result= mysql_query($query);
while($row = mysql_fetch_assoc($result)) {
echo '<option value="'.$row['package_id'].'">'.$row['cat'].'</option>';
}
?>
</select>
现在在 AJAX 请求中
$('#select_choose_list').change(function(){
$.ajax({
type: 'GET',
url: 'ajax_cat.php&id=' + $('#select_choose_list').val(),
success: function(data) {
$('#cat_cost').html(data);
}
});
});
试试这个:)