无法从php中的任何HTTP请求获取参数

我无法从php // input获取请求参数,它总是返回空字符串。$ _POST也总是空的,我可以获取参数的唯一方法是使用$ _REQUEST,但我不确定这是否是解决此问题的正确方法。


var_dump(file_get_contents("php://input"));

返回:string(0)“”


var_dump($_POST);

返回:array(0){}


var_dump($_REQUEST);

返回:array(2){[“” ergerg“] => string(6)” ergerg“ [” regergerg“] => string(6)” ergreg“}


我在邮递员中的要求如下:


POST  /api/ajax.php?ergerg=ergerg&regergerg=ergreg


response

Status:200 OK

Time:192 ms

Size:236 B

ajax.php


<?php

include "config.php";


$rest_json = file_get_contents("php://input");

$obj = json_decode($rest_json);

$request = $obj->request;


if($request == 1){

    $userData = mysqli_query($con, "select * from rows");


    $response = array();


    while ($row = mysqli_fetch_assoc($userData)) {

        $response[] = $row;

    }

    echo json_encode($response);

}


if($request == 2){

    $platform   = $data->platform;

    $container1 = $data->container1;

    $container2 = $data->container2;

    $ZPU1       = $data->ZPU1;

    $ZPU2       = $data->ZPU2;

    $reportId   = $data->reportId;

    mysqli_query($con, "INSERT INTO 

                            rows(platform,container1,container2,

                                ZPU1,ZPU2,reportId) 

                    VALUES('" . $platform . "','" . $container1 . "','" . 

                            $container2 . "','" . $ZPU1 . "','" . $ZPU2 . 

                            "','" . $reportId . "')");

    echo "Insert successfully";

}


exit;

在配置文件中,只有与mysql数据库的连接:


config.php


<?php


$host = "localhost"; /* Host name */

$user = "root"; /* User */

$password = "password"; /* Password */

$dbname = "dbname"; /* Database name */


$con = mysqli_connect($host, $user, $password,$dbname);

// Check connection

if (!$con) {

  die("Connection failed: " . mysqli_connect_error());

}


子衿沉夜
浏览 243回答 1
1回答
打开App,查看更多内容
随时随地看视频慕课网APP