1

我正在尝试将 json 格式的值发送到 php，但是当我调用该函数时，php 上的 $_POST 返回 null。此过程类似于在底部https://www.w3schools.com/js/js_json_php.asp找到的 w3schools php json 示例，但仍然没有得到相同的结果。json 中字符串旁边的值是其中包含值的变量。这是我所做的：

var args = {'name': name, 'birthday': birthday, 'policy': policy, 'cl': cl, 'claim': claim, 'incoming': incoming, 'theDate': theDate,
    'xora': xora, 'treatment': treat, 'session': sess, 'inpat': inpat, 'outpat': outpat, 'daycase': daycase,
    'radioBtn': radioBtn, 'admDate': admDate, 'invDate': invDate, 'invNum': invNum, "nomisma": nomisma,
    'provSelect': provSelect, 'specSel1': specSelect1, 'prescSel': prescSelect, 'specSel2': specSelect2, 'amount': amount,
    'deduct': deduct, 'dedColl': dedColl, 'copay': copay, 'copayColl': copayColl, 'totalAm': totalAm, 'comms': comms,
    'diagnosiDesc': diagnosiDesc, 'diagnosiCode': diagnosiCode, 'nonAmount': nonAmount, 'reason': reason, 'categs': category };
var json = JSON.stringify(args);

if (window.XMLHttpRequest) {
    // code for IE7+, Firefox, Chrome, Opera, Safari
    xmlhttp = new XMLHttpRequest();
} else {
    // code for IE6, IE5
    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function () {
    if (this.readyState === 4 && this.status === 200) {
        //alert("Saved and Continue");
        window.open("PHP/SaveAndCont.php?q="+json); // Test
    }
};

xmlhttp.open("POST","PHP/SaveAndCont.php",true);
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xmlhttp.send("q="+json);

这是我的 php 文件：

include('Connection.php');
header("Content-Type: application/json; charset=UTF-8");
$parse = $_POST['q']; // undefined index q
$obj = json_decode($parse, false);
var_dump($parse, $obj); // NULL
4

2 回答 2

2

将内容类型标头设置为application/json.

在带有原始请求的 PHP 中，您不会在 $_POST 中填充变量，而是需要通过原始请求获取 json 来获取它：

检查请求类型是否为 post 如果不退出 PHP

<?php
if ($_SERVER['REQUEST_METHOD'] !== 'POST') {
    die("wrong type of request"); 
}
 // get raw json post object and decode it
 $json = json_decode(file_get_contents("php://input"));
 var_dump($jsonString);

JS：

var args = {'name': name, 'birthday': birthday, 'policy': policy, 'cl': cl, 'claim': claim, 'incoming': incoming, 'theDate': theDate,
    'xora': xora, 'treatment': treat, 'session': sess, 'inpat': inpat, 'outpat': outpat, 'daycase': daycase,
    'radioBtn': radioBtn, 'admDate': admDate, 'invDate': invDate, 'invNum': invNum, "nomisma": nomisma,
    'provSelect': provSelect, 'specSel1': specSelect1, 'prescSel': prescSelect, 'specSel2': specSelect2, 'amount': amount,
    'deduct': deduct, 'dedColl': dedColl, 'copay': copay, 'copayColl': copayColl, 'totalAm': totalAm, 'comms': comms,
    'diagnosiDesc': diagnosiDesc, 'diagnosiCode': diagnosiCode, 'nonAmount': nonAmount, 'reason': reason, 'categs': category };
var json = JSON.stringify(args);

if (window.XMLHttpRequest) {
    // code for IE7+, Firefox, Chrome, Opera, Safari
    xmlhttp = new XMLHttpRequest();
} else {
    // code for IE6, IE5
    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function () {
    if (this.readyState === 4 && this.status === 200) {
        //alert("Saved and Continue");
        alert(xmlhtml.responseText); // this should print the response
    } else if(this.status !== 200) {
        alert(this.status); // maybe it'll be 404 or 500 if so then correct the url in xmlhttp.open it depends on your server configuration but it needs to be accessed via http://localhost/ or host defined by you
    }
};

xmlhttp.open("POST","php/saveandcont.php", true); // this should be path that can be opened via browser
xmlhttp.setRequestHeader("Content-type", "application/json");
xmlhttp.send(json);
于 2018-03-06T20:37:58.357 回答
-1

$parse 和 $obj NULL 变量都意味着 $_POST['q'] 没有定义。

我认为这个字符串中的问题：

xmlhttp.open("POST","PHP/SaveAndCont.php",true);

第二个参数是一个 URL：https://www.w3schools.com/xml/ajax_xmlhttprequest_send.asp 但是您将它设置为文件的路径。尝试将其替换为

xmlhttp.open("POST","SaveAndCont.php",true);

或者

xmlhttp.open("POST","http://fullserverurl/SaveAndCont.php",true);
于 2018-03-06T20:16:30.710 回答