Question

我试图对名为＆＃34; sendMail.php＆＃34;的文件执行ajax post请求。我不知道出了什么问题，我无法看到它。

js工作，它从输入中记录值，在chrome dev工具中我可以看到它被发送到PHP文件... chrome dev工具屏幕

我有点生疏，最近的PHP代码大约一年前。

下面你会找到我的PHP和js代码。

文件夹树（index.html是操作它的主文件）：

PHP：

Critical error: wait for JVM process failed
--> Wrapper Started as Console
Launching a JVM...
Unable to execute Java command.  Access is denied. (0x5)

Advice:
Access denied errors when attempting to launch the Java process are
usually caused by strict access permissions assigned to the directory
in which Java is installed.

JS：

if ($_SERVER['REQUEST_METHOD'] == 'POST'){
   header('Content-Type: application/json');
   $requestbody = file_get_contents('php://input','r');
   $jsonbody = json_decode($requestbody, true);
   $action = $jsonbody['action'];

    switch($action){
        case "sendMail":
            $name = $requestbody->{'name'};
            $gsm = $requestbody->{'gsm'};
            $mail = $requestbody->{'mail'};
            $msg = $requestbody->{'msg'};
            $response = json_encode( '{"naam":"'.$name.'","gsm":"'.$gsm.'","mail":"'.$mail.'"}');
            echo $response;
            break;
    }
}else{
    http_response_code(405);
    header('Content-Type: application/json');
    $response = '{"error":"request method is not allowed."}';
    echo($response);
}
?>

Answer 1

Ajax中没有method参数，因此请将其替换为type。

请尝试以下代码：

$.ajax({
        type:'POST',
        url:'mail/sendMail.php',
        dataType:'json',
        data: {
            "action":"sendMail",
            "name":  document.getElementById("mail_naam").value,
            "gsm": document.getElementById("mail_gsm").value,
            "mail":document.getElementById("mail_mail").value,
            "msg": document.getElementById("mail_msg").value
        }
    }).done(function(data) {
        console.log("succes");
        console.log(data);
        var result = JSON.parse(data);
        console.log(result);
    });

Answer 2

我猜测问题是$requestbody总是空的。您应该使用$_POST作为来源。

$requestbody = $_POST;

有点。

php://input好参考：Description

jQuery ajax

Answer 3

我添加了这个，这有效：

PHP

  if(empty($action) || $jsonbody){ $action = $_POST['action']; }

  switch ($action) {
      case "sendMail":    
          $name = $_POST["name"];
          $gsm =  $_POST['gsm'];
          $mail = $_POST['mail'];
          $msg =  $_POST['msg'];
          mailVraag($mail,$name,$msg);
          mailVraag("achiel@protoware.be","protoware",$msg . "tel:".$gsm ."email: ".$mail);

          $response = json_encode( '{"naam":"'.$name.'","gsm":"'.$gsm.'","mail":"'.$mail.'"}');
          echo $response;
          break;
  }

php ajax发布请求

3 个答案: