如何使用AJAX我发送数据到我的PHP脚本？我的、如何使用、脚本、数据

由网友(认定了就是永远 ▍)分享简介：我是新来的PHP和JavaScript / AJAX，所以请多多包涵。I'm new to PHP and Javascript/Ajax so please bear with me.所有我需要做的是从阿贾克斯得到一个变量，并将其设置为在PHP变量。我试图用超全局 GET 做到这一点，但东西是不正确的。我不希望这...

我是新来的PHP和JavaScript / AJAX，所以请多多包涵。

I'm new to PHP and Javascript/Ajax so please bear with me.

所有我需要做的是从阿贾克斯得到一个变量，并将其设置为在PHP变量。我试图用超全局 GET 做到这一点，但东西是不正确的。我不希望这种通过提交表单。

All I need to do is get a variable from Ajax and set it as a variable in php. I'm trying to do this with a super global GET but something is not right. I don't want to this by submitting the form.

下面是我的JS：

function myFunction(){
var hora= document.getElementById("hora").value;
$.ajax({ 
        type : 'GET',
         url : 'reservation.php',
         data : {hora: hora},
         success : function(data) {
                console.log(hora);//This is because I was curious as to
                             // what the console would say. I found
                             // that this sets the super global if I
                             // change the url to something else that
                             // doesn't exist. Console would say
                             // -GET http://localhost/bus/(somepage).php?hora=4
                             // 404 (Not Found)-  
                alert(hora);
               }
})
}

下面是我的PHP：

Hora:
 <select name="hora" id="hora" onchange="myFunction()">
     <?php
         $query = "SELECT * FROM vans";
         $horas_result = mysql_query($query);
         while ($horas = mysql_fetch_array($horas_result)) {
        echo "<option value="{$horas["van_id"]}">{$horas["time"]}</option>";
         }
    ?>
 </select>&nbsp;
 Asientos Disponibles:
    <?php echo $_GET["hora"]; ?>
        //Right now I only want to echo this variable..

正如你所看到的，现在我只想呼应这个变量，以后我会用这个来编写一个查询。

As you can see, right now I only want to echo this variable, later on I'll be using this to write a query.

推荐答案

看codeI后，AJAX是用于发布/获取，而不需要刷新页面的数据，但如果你只是想发布的数据，并给出的结果在其他页面中使用的一种形式来代替。

Look at the code i post, ajax is used to post/get data without need to refresh the page but if you just want to post the data and give the result in other page use a form instead.

<?php

    if (isset($_GET["hora"]))
    {
        echo $_GET["hora"];

        exit;
    }
?>

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>Page title</title>

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>

</head>

<body>

<script type="text/javascript">

        $(document).ready(function()
            {
                $("#hora").change(function ()
                    {
                        $.ajax(
                            { 
                                type : 'GET',
                                url : '',
                                data : $('select[name='hora']'),
                                success : function(data)
                                    {
                                        $('#ajax_result').html('Asientos Disponibles:  ' + data);
                                    },
                                error: function(xhr, ajaxOptions, thrownError)
                                    {
                                        alert(thrownError + "rn" + xhr.statusText + "rn" + xhr.responseText);
                                    }
                            }
                        )
                    }
                )

            }
        )

</script>

<select name="hora" id="hora">
     <?php
         $query = "SELECT * FROM vans";
         $horas_result = mysql_query($query);
         while ($horas = mysql_fetch_array($horas_result)) {
        echo "<option value="{$horas["van_id"]}">{$horas["time"]}</option>";
         }
    ?>
 </select>&nbsp;
    <div id="ajax_result">
    </div>
</body>
</html>

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相关专题：我的；如何使用；脚本；数据；Ajax ；PHP ；发布时间：2023-09-10 14:36:01