我正在尝试编写 PHP 代码 (req_form.php),该代码应该通过将插入的数据发送到 (insert.php) 来插入 SQL 数据库“请求”而不刷新页面,但它不起作用:(
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>
插入.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';
}
}
?>
问题是什么?!
最佳答案
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();
关于php - 使用AJAX插入数据库,无需刷新页面,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36815877/