Question

我被困在这里大约4个小时做了很多研究而无法找到答案。我想在使用jquery ajax和php更新内容后重新填充表，所以这是我的代码。

PHP脚本：

    if(isset($_POST['myinputval']))
    {

        $uinput = $_POST['myinputval'];
        $uinput = $mysqli->real_escape_string($uinput);


        $query = ('INSERT INTO tbltest values (" ","'.$uinput.'") ');

        if($mysqli->query($query))
        {

        } else 
            echo 'fail'. $mysqli->error;
        }

        $result= $mysqli->query("Select *from tbltest");

        if($result->num_rows>0)
        {
            while($row=$result->fetch_assoc())
            {
                $sendtojq[]=$row;
            }

            echo json_encode($sendtojq);
        }

        mysqli_close($mysqli);

这是我的剧本

        <script type="text/javascript">
        function ajaxsubmit()
        {
            var myinputval = $("#input1").val();

            $.ajax({url:'myphp.php', type:'post', data:{
                myinputval:myinputval
            },
            success: function(data)
            {
                alert(data);
            }
        });
        }


        </script>

这是提交时的输出

[{"ID":"1","NAME":".$newinput."},{"ID":"2","NAME":".$newinput."},{"ID":"3","NAME":".$newinput."},{"ID":"4","NAME":".$newinput."},{"ID":"5","NAME":".$newinput."},{"ID":"6","NAME":".$newinput."},{"ID":"7","NAME":"$newinput"},{"ID":"8","NAME":"twst"},{"ID":"9","NAME":"twst"},{"ID":"10","NAME":"testtt"},{"ID":"11","NAME":"testtt"},{"ID":"12","NAME":"thisssss"},{"ID":"13","NAME":"thisssss"},{"ID":"14","NAME":"tstet"},{"ID":"15","NAME":"Last"},{"ID":"16","NAME":""},{"ID":"17","NAME":"dddddddd"},{"ID":"18","NAME":"aaaaaaaaaaa"},{"ID":"19","NAME":"gggggggggggggggggggggg"},

现在我要做的是获取存储在变量＆＃39;数据中的数组的每个元素。并使用at作为我的表的值。有什么建议？提前谢谢。

使用jquery ajax和php填充html表

0 个答案: