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通过Ajaxrequest.open从Javascript到PHP文件获取变量_Javascript_Php_Ajax_Ajax Request - Fatal编程技术网

通过Ajaxrequest.open从Javascript到PHP文件获取变量

通过Ajaxrequest.open从Javascript到PHP文件获取变量,javascript,php,ajax,ajax-request,Javascript,Php,Ajax,Ajax Request,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{ ajaxRequ

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 );
?>

POST请求不使用参数的URL!它是在url参数中使用的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 '&'))
然后在PHP中得到
$\u POST[“value”]

这是“常规”方式

但如果您想要更灵活的请求格式,还可以将数据作为JSON发送:

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));
PHP方面:(解释如下:):

不是“常规”方式,但您在发送数据时具有更好的灵活性 (因为您不发送字符串,而是发送原始数据:对象/数组…

试试这个。您的函数(ajax\u runs3)


请开始使用
mysqli*
PDO
,因为
mysql.*
在PHP7中已被弃用和删除。您正在混合使用mysqli和mysql
$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};
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);
}