我开发了一个预订系统,当我点击该按钮时它有一个按钮,它将更新从待定到完成的约会状态。每次我想完成约会,我都要点击几次。我认为只需点击一次即可在ajax中完成。
这是我的代码
if ($_REQUEST['command'] == "send") { // if the command send pressed then
mysql_query(" UPDATE bookings SET status='sent' WHERE id=$_REQUEST[id]"); // update bookings and set the Statu of the statuts field to sent
}
if ($_REQUEST['command'] == "Done") { // if the command Done pressed then
mysql_query(" UPDATE bookings SET status='Done' WHERE id=$_REQUEST[id]"); //update bookings and set the Statu of the statuts field to Done
}
if ($_REQUEST['command'] == "Delete") {
mysql_query(" DELETE FROM bookings WHERE id=$_REQUEST[id]");
}
if($ a [' status'] ==' new'){//如果状态状态等于" new"那么> echo"确认&#34 ;; //打印出发送动作库存!需要完成!! } else if($ a [' status'] ==' sent'){ 回声"当然? &#34 ;; // echo"删除&#34 ;; }
我很感激有人可以帮助我使用ajax类来完成这个功能,
非常感谢提前。
答案 0 :(得分:0)
您可以使用Jquery Ajax函数,它非常易于实现。示例如下所示
<button onclick="changeStatus('send');"></button>
<button onclick="changeStatus('Done');"></button>
<button onclick="changeStatus('Delete');"></button>
<script type="text/javascript">
function changeStatus(command){
$.ajax({
url: "project_url/functoin_name", //where your want to send this request(the php file location)
type: "POST",
data: {command: command},
success: function(response){
alert(response);
}
});
}
</script>
有关详细信息,请查看此http://api.jquery.com/jquery.ajax/