Question

我试图将javascript循环的每次迭代后正在更新的值传递给返回枯萎的php脚本，否则循环将继续。我在循环的第一次迭代后得到了这个错误

ERR: 8) (Undefined index: player2name) (loaction of php file on host) (Line:� 26)

这是有问题的javascript函数

    function startgame()
{
    document.JForm.delay.disabled = true;
    document.JForm.p1name.disabled = true;
    document.JForm.p2name.disabled = true;
    document.JForm.numgames.disabled = true;
    document.JForm.totalpoints.disabled = true; 
//Create the XMLHttpRequest Object

  var xmlhttp;
  // error handling
  if (!xmlhttp && typeof XMLHttpRequest != 'undefined') {
    try {
      xmlhttp = new XMLHttpRequest();
    } catch (e) {
      xmlhttp = false;
    }
  }
  http = xmlhttp;

    var rolling = 1;
    var url = "program.php?player1name=";
    var player1name = document.JForm.p1name.innerHTML; 
    var player2name = document.JForm.p2name.innerHTML;
    var playtopoints = parseInt(document.JForm.totalpoints.value);
    var delay = parseInt(document.JForm.delay.value);
    var numgames = parseInt(document.JForm.numgames.value);
    var gamesplayed = parseInt(document.JForm.gamesplayed.value);
    var p1turn = parseInt(document.JForm.p1turn.value);
    var p2turn = parseInt(document.JForm.p2turn.value);
    var p1total = parseInt(document.JForm.p1total.value);
    var p2total = parseInt(document.JForm.p2total.value);   
    var sdata = url + player1name +"&?player2name=" + player2name + "&?playtopoints=" + playtopoints + "&?delay=" + delay + "&?numgames=" + numgames + "&?gamesplayed=" + gamesplayed + "&?p1turn=" + p1turn + "&?p2turn=" + p2turn + "&?p1total=" + p1total + "&?p2total=" + p2total;

    for (var i = 0; i<numgames; i++){
            http.open("POST", url + sdata, true);

            rolling = "<?php echo $roll; ?>";
            document.JForm.p1name.innerHTML = rolling;
        if (rolling){
            roll(); 
        }
        else{
            //update player stats there was a winner
            alert("not rolling");
            stop();
            if (currentplayer == "P1"){
                currentplayer = "P2";
            }
            else{
                currentplayer = "P1";
            }
        }
    }
    // register with server what function to return data
    http.onreadystatechange = handleHttpResponse;
    // send request to server
    http.send(null);

}

这里是php代码

    <?php

error_reporting(E_ALL);   // Enable all error checking
$return = set_error_handler("MyError");
// Error handler function, called when PHP runtime error detected
function MyError($errno,$errstr,$errfile,$errline) {
    print "(ERR: $errno) ($errstr) ($errfile) (Line:Ê $errline) <br>";
    return true;
}
// End of PHP error handling code

    $p1name=$_GET['player1name'];  // algorithm player 1 is using
    $p2name = $_GET['player2name'];
    $playtopoints = $_GET['playtopoints'];
    $delay = $_GET['delay'];
    $numgames = $_GET['numgames'];
    $gamesplayed=$_GET['gamesplayed'];  // number of games played from JavaScript
    $p1turn = $_GET['p1turn'];
    $p2turn = $_GET['p2turn'];
    $p1total = $_GET['p1total'];
    $p2total = $_GET['p2total'];

    //if statement to find who the current player is 
    if ($numgames % 2 == 0){
        //number is even
        $currentpoints = $p1turn;
        $currenttotal = $p1total;
        $optotal = $p2total;
    }
    else{
        //number is odd
        $currentpoints = $p2turn;
        $currenttotal = $p2total;
        $optotal = $p1total;
    }
    sleep($delay);
    $rolling = algorithm1($playtopoints, $p1turn, $p2turn, $p1total, $p2total);

    print $rolling;

    exit();                 // return to browser
    function algorithm1($wintotal, $p1t, $p2t, $p1total, $p2total)  {
        if (($p1t >= ($wintotal / 2) )||($p2t >= ($wintotal / 2))){
            $continue = 0;
        }
        else if(($p1t <= 20)||($p2t <= 20)){
            $continue = 1;
        }
        return($continue);
    }
?>

Answer 1

JavaScript中存在多个错误。

问题1：以下行?之后的&

var sdata = url + player1name +"&?player2name=" + player2name + "&?playtopoints=" + playtopoints + "&?delay=" + delay + "&?numgames=" + numgames + "&?gamesplayed=" + gamesplayed + "&?p1turn=" + p1turn + "&?p2turn=" + p2turn + "&?p1total=" + p1total + "&?p2total=" + p2total;

问题2：两次发送url

http.open("POST", url + sdata, true);

sdata已包含url个变量。删除重复的url变量。

问题3：混合POST和GET

http.open("POST", url + sdata, true);

您通过POST方法发送了HTTP请求，但在PHP方面，您通过GET收到了数据。更改任意一方，但我建议使用POST，即在PHP中将$_GET更改为$_POST。

旁注：建议使用类似jQuery的库来简化HTTP请求代码。

Answer 2

我可以看到一些额外的'？'在

var sdata = url + player1name +"&?player2name=" + player2name + "&?playtopoints=" + playtopoints + "&?delay=" + delay + "&?numgames=" + numgames + "&?gamesplayed=" + gamesplayed + "&?p1turn=" + p1turn + "&?p2turn=" + p2turn + "&?p1total=" + p1total + "&?p2total=" + p2total;

也许你可以使用

var sdata = url + "?player1name=" + player1name +"&player2name=" + player2name + "&playtopoints=" + playtopoints + "&delay=" + delay + "&numgames=" + numgames + "&gamesplayed=" + gamesplayed + "&p1turn=" + p1turn + "&p2turn=" + p2turn + "&p1total=" + p1total + "&p2total=" + p2total;

我在你的网址中添加了一些额外的'player1name'以及剥去额外的'？'

从Javascript传递php变量时出错

2 个答案: