Question

我正在尝试编写PHP代码（req_form.php），它应该插入到SQL数据库“请求”而不刷新页面，通过将插入的数据发送到（insert.php）并且它只是不工作:( < / p>

这是我的“请求”表： Request

req_form.php：

<form action="req_form.php" method="post">
  <textarea id="des" name="des" placeholder="Problem description" type="text" required></textarea>
  <input id="date" name="date" placeholder="Date (YEAR-MONTH-DAY)" type="date"/>
  <input id="time" name="time" placeholder="Time (Hour:Minute)" type="time"/>
  <button type="submit" name="send" id="send" onclick="Send_req()"> Send</button>
</form>

<script type="text/javascript">
   function Send_req() {
     var des= $("#des").value;
     var time= $("#time").value;
     var date= $("#date").value;
     $.ajax({
        type: "POST",
        url: 'insert.php',
        data: {des:des, time:time, date:date;},
        success: function(html) 
        {
            if (html=='1')
            {
                alert ("Congratulations, Request Sent Successfuly! ");
            }

            else
            {
                alert ("Sorry, Error in sending request!");
            }
        }
    }); 
      }
</script>

insert.php：

<?php 
$connect= mysql_connect("localhost","root","") or die("Can't connect to database!");
$db = mysql_select_db("hwp",$connect) or die ("Can't select database!");

if($_POST["des"] !="" && $_POST["date"] != "" && $_POST["time"] !="" )
{
    $des= mysql_real_escape_string($_POST["des"]);
    $date= mysql_real_escape_string($_POST["date"]);
    $time= mysql_real_escape_string($_POST["time"]);
    $sender= '1';
    $receiver= '2';
    $sql = mysql_query("INSERT INTO request (Req_sender_id, Req_receiver_id, Req_date, Req_time, Req_msg) VALUES ( '$sender','$receiver','$date','$time','$des');") or die(mysql_error());
      if  ($sql)  
    {
        echo '1';
    }

    else
    {
        echo '0';
    }
}
?>

有什么问题??！

Answer 1

纯javascript的值。如果你想在jquery中获取值，你必须使用.val（）

var des= $("#des").val();
var time= $("#time").val();
var date= $("#date").val();

使用AJAX插入数据库而不刷新页面

1 个答案: