Question

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); 
}

以及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功能传递到php文件。我还尝试了根据以下主题（Using an ajaxRequest.open to send a variable to php）解决方案，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。你知道我的案子有什么问题吗？谢谢你的兴趣。

Answer 1

POST请求不使用参数URL ！这是使用in-url params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中获得$_POST["value"]

中的runs3值

这是“常规”方式。

但是如果你想要一个更灵活的请求格式，你也可以发送数据为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:(在此解释：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，但您可以更灵活地使用您发送的数据（因为你不发送字符串，而是原始数据：对象/数组......）

Answer 2

试试这个。你的功能（ajax_runs3）

#include <algorithm>
#include <iterator>

std::partition_copy(array, array + size,
                    std::back_inserter(real),
                    std::back_inserter(imag),
                    [&toggle](int t){ return t != toggle; });

通过Ajaxrequest.open

2 个答案: