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I want to refresh two variables named profittext and sumtext which will be refreshed and echoed in the following places every few seconds. I know AJAX is needed to do this but how do i actually make it work ? The only way i found out was to refresh the content of the whole file. is there any way to refresh specific variables? Any answers will be greatly appreciated . Thank you very very much.

<table>
if($profitandloss<$zero) {
    $profitText = "<div style=\"color: red;\">$profitandloss</div>";
} elseif ($profitandloss>$zero) {
    $profitText = "<div style=\"color: green;\">$profitandloss</div>";
}
// for profit and loss counting

$sum+= $profitandloss;
//

echo "<tr><td>" . $row['trade_id'] .         
        "</td><td>" . $row['selection'] . 
        "</td><td>" . $row['date'] .
        "</td><td>" . $row['type'] .
        "</td><td>" . $row['size'] .
        "</td><td>" . $row['bidprice'] .
        "</td><td>" . $row['offerprice'] .
        "</td><td>" . $row['stoploss'] .
        "</td><td>" . $row['takeprofit'] .
        "</td><td>" . $profitText . 
        "</td><td><a href ='delete.php?id=".
        $row['trade_id']."'>X</a>
    </td></tr>";  

$profitandloss=0;

if($sum<$zero) {
    $sumText = "<div style=\"color: red;\">$sum</div>";
} elseif ($sum>$zero) {
    $sumText = "<div style=\"color: green;\">$sum</div>";
}
}
echo "</table><br>";
?>
<!DOCTYPE html>
<html>
<table style="border:1px solid black;">
<tr>
<th style="border:1px solid black;">Profit/Loss</th>
</tr>
<tr>
<td style="border:1px solid black;"><?php echo $sumText ;?></td>
</tr>
</table>
</html>
4

3 回答 3

2

当我刚开始的时候,我也在为如何构建这样的代码的概念而苦苦挣扎。虽然它不是特定于您的特定变量,但这里有一个快速示例,说明如何使用 jQuery/PHP 通过 AJAX 更新 var。

序言:如果这是你经常要做的事情,你会想要学习jQuery,而不是单独使用普通的 javascript。关于如何学习 jQuery,有很多很棒的免费资源。或者,如果您对在线免费教程不满意,是一本很棒的书。我将用 jQuery 编写示例。

设计:好的,它的工作方式是这样的:

  • 在 javascript 中设置一个计时器以每 X 秒执行一次特定功能(您不想每秒执行一次)。

  • 该函数对服务器上的 .PHP 文件进行 AJAX 调用(使用 jQuery),向其发送必要的数据,以便 .PHP 代码知道要发送回什么。

  • .PHP 代码获取所需的数据(例如,使用 MySQL)以 JSON 格式对其进行编码,然后退出。

  • AJAX 调用上的一个 promise 被触发并且从 PHP 发送的数据被接收。随心所欲地处理它。

  • 从第 2 步开始重复。

如果您对代码的作用有任何疑问,请询问。

AJAX.PHP

<?php

$mysqli = new mysqli("localhost", "my_user", "my_password", "world");

$return_obj = array();
$request_obj = NULL;

// our AJAX call used "POST" as it's 'type', so we look in that
// variable.
if ( array_key_exists("func",$_POST) ) {

    if ( $_POST['func'] === "get_update" ) {

        if ( array_key_exists("which_var",$_POST) ) {

            $which_var = $_POST['which_var'];
            $which_var = $mysqli->real_escape_string($which_var); // should use prepared statements

            // we sent 'num_people_logged_in' as our value here, so we'll be looking for a column/field
            // with that value. this assumes that some other code, somewhere else,
            // is regularly updating the table. it also assumes that there will only
            // be a single row returned, which will hold the value.
            $query = "SELECT '$which_var' FROM site_stats ";
            if ( $result = $mysqli->query($query) ) {

                if ( $row = $result->fetch_assoc() ) {
                    $request_obj[$which_var] = $row[$which_var];
                }
            }
        }
    }
}

$return_obj['request'] = $request_obj;
echo json_encode($return_obj);
die();

?>

MYCODE.JS

// this actually sends the AJAX request to the server.
function getUpdate() {

    var jqXHR = $.ajax({        
        url : "ajax.php",
        data : {
            'func' : 'get_update',
            'which_var' : 'num_people_logged_in'
        },
        dataType : 'json',
        type : 'POST',
        timeout : 10000
    });

    // attach 'promises' to the jqXHR object, which represents
    // the AJAX call itself. 'promises' are, in this context,
    // just events.

    jqXHR.done(function(data,textStatus,jqXHR) {        

        // this executes if the AJAX call succeeded.
        // the variable 'data' holds what the server
        // sent us.
        if ( ( data ) && ( data.request ) ) {
            receiveUpdate(data.request);
        }
    });

    jqXHR.fail(function(jqXHR,textStatus,errorThrown) {

        // this executes if it failed
        console.log("Fail: " + textStatus + " (" + errorThrown + ")");
    });

    jqXHR.always(function(a,textStatus,c){                    

        // this executes either way, after .done or .fail
    });
}

// this is called from jqXHR.done, on success
function receiveUpdate(request_obj) {
    if ( request_obj.num_people_logged_in ) {
        updateDOM(request_obj.num_people_logged_in);
    }
}

function updateDOM(num_people_logged_in) {
    if ( num_people_logged_in ) {
        $("#mydiv > p.update").html("The updated value is: " + num_people_logged_in);
    }
}

var timeoutID = null;

// setup our timer, to periodically make an
// AJAX call
function init() {
    timeOutID = setInterval(function(){
        getUpdate();
    },5000);
}

// stop the timer
function cleanup() {
    clearTimeout(timeoutID);
}

索引.HTML

<!DOCTYPE html>
<html>
<head>
    <meta charset='utf-8'>
    <title>AJAX practice</title>

    <!-- <link href="mycss.css" rel='stylesheet'> if needed -->

    <script src="http://code.jquery.com/jquery-1.9.1.js"></script>
    <script src="mycode.js"></script>
    <script>

        $(document).ready(function() {

            init();

            $("#cleanup").on("click",function(){
                cleanup();
            });

        }); // end ready

    </script>
</head>

<body>

    <div id='mydiv'>
        <p>
            How many people are online?
        </p>
        <p class='update'>
        </p>
    </div>

    <button id='cleanup'>Stop updating!</button>

</div>
</body>
</html>
于 2013-08-18T17:06:00.370 回答
0

使用计时器:https ://developer.mozilla.org/en/docs/DOM/window.setInterval

setInterval(function(){
  //update your var here
},1000);
于 2013-08-18T16:34:07.320 回答
0

您将需要两个 PHP 页面: - 一个带有 HTML 和 JS 的页面,它会定期进行 ajax 调用并将结果放入 HTML - 第二个带有动态数据片段的 json(甚至纯文本)输出

不幸的是,在答案中编写完整的代码并不是人们在 stackoverflow 所做的事情,所以只需看看下面的小示例,并尝试找出缺失的部分。

http://jsfiddle.net/AMEqz/

var xhr = new XMLHttpRequest();
xhr.onload = function(r) {
   // your render logic HERE
   setTimeout(send, 1000);
}
function send() {
    xhr.open("GET", "/", true);
    xhr.send();
}
send();

ps:请记住,每个 ajax 请求都意味着与您的服务器的额外连接,因此请确保它可以处理负载;)

于 2013-08-18T16:35:07.927 回答