1

我有以下javascript函数

function ajax_runs3(value){
    var ajaxRequest;  // The variable that makes Ajax possible!
    try{
        ajaxRequest = new XMLHttpRequest();
    } catch (e){
        // Internet Explorer Browsers
        try{
            ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
        } catch (e) {
            try{
                ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
            } catch (e){
                // Something went wrong
                alert("Your browser broke!");
                return false;
            }
        }
    }
    ajaxRequest.onreadystatechange = function(){
        if(ajaxRequest.readyState == 4){
            document.myForm.time.value = ajaxRequest.responseText;
        }
    }

    var runs3= value;
    ajaxRequest.open("POST","runs3.php"+ runs3, true);
    ajaxRequest.send(null); 
}

还有PHP文件

<?php
$servername = "localhost";
$username = "USER";
$password = "PASS";
$dbname = "labi8575_inventory";
$conn = mysql_connect($servername, $username, $password);
if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}
mysql_select_db('labi8575_inventory');  
$runs3 = $_POST["runs3"];
$sql = mysql_query("UPDATE demo SET runs3 = '$runs3'");
$retval = mysqli_query( $sql, $conn );
?>

问题是我无法将 var runs3 从 javascript 函数传递到 php 文件。我还尝试根据以下主题（使用 ajaxRequest.open 向 php 发送变量）解决方案，例如 ajaxRequest.open("POST", "runs3.php?variable="+runs3) 或 AjaxRequest.open("POST" , "runs3.php?myvar=runs3", true); 但就我而言，它不起作用。你知道我的情况有什么问题吗？感谢您的关注。

4

2 回答 2

2

POST 请求不使用 URL 作为参数！它是使用 in-url 参数的 GET 方法...

解决方案：

var runs3= value;
ajaxRequest.open("POST","runs3.php", true); //We open the url
ajaxRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); //IMPORTANT!! We add this header to tell to PHP that it is a "form" which sent requesdt
ajaxRequest.send("value=" + encodeURIComponent(runs3)); //Then we send DATA HERE  (encodeURIComponent encodes data to prevents URL-specific characters (for example '&'))

然后你在 PHP 中得到 runs3 值$_POST["value"]

这是“常规”方式。

但是如果您想要更灵活的请求格式，您也可以将数据发送为 JSON：

var runs3 = {"val" : value};
ajaxRequest.open("POST","runs3.php", true); //We open the url
ajaxRequest.setRequestHeader("Content-type", "application/json");
ajaxRequest.send(JSON.stringify(runs3));

和 PHP 方面：（在这里解释： Reading JSON POST using PHP）：

$request = file_get_contents('php://input'); //raw request data
$object  = json_decode($request, true);      //we convert it to associative array, by JSON

print_r($object); //Should return Array[1] {  "val" => YOUR_VALUE};

不是“常规”方式，但您发送的数据具有更好的灵活性（因为您不发送字符串，而是发送原始数据：对象/数组......）

于 2016-04-12T08:49:11.857 回答
0

尝试这个。你的功能（ajax_runs3）

function ajax_runs3(value){
    var ajaxRequest;  // The variable that makes Ajax possible!
    try{
        ajaxRequest = new XMLHttpRequest();
    } catch (e){
        // Internet Explorer Browsers
        try{
            ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
        } catch (e) {
            try{
                ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
            } catch (e){
                // Something went wrong
                alert("Your browser broke!");
                return false;
            }
        }
    }
    ajaxRequest.onreadystatechange = function(){
        if(ajaxRequest.readyState == 4){
            document.myForm.time.value = ajaxRequest.responseText;
        }
    }

    var runs3= value;
    //ajaxRequest.open("POST","runs3.php"+ runs3, true);
   // ajaxRequest.send(null); 

    var url = "runs3.php";
    var params = "runs3="+value;
    ajaxRequest.open("POST", url, true);

    //Send the proper header information along with the request
    ajaxRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    ajaxRequest.setRequestHeader("Content-length", params.length);
    ajaxRequest.setRequestHeader("Connection", "close");

    ajaxRequest.onreadystatechange = function() {//Call a function when the state changes.
        if(ajaxRequest.readyState == 4 && ajaxRequest.status == 200) {
            alert(ajaxRequest.responseText);
        }
    }
    ajaxRequest.send(params);
}
于 2016-04-12T08:56:52.383 回答