在 PHP 中读取 JavaScript 个变量

Reading JavaScript Variables in PHP

我正在寻找一种方法来读取 PHP 中的图像 src 或读取 PHP 中的 JavaScript 变量。

计划: 我在 JavaScript 中有一个网络摄像头脚本,它使用 base64 Canvas 方法拍摄图像,但我需要它来使用 PHP.

更新 MySQL 记录

尝试了很多方法后; cookies,提交表格,ajax。我决定在这里做一个 post 是最好的主意。

(!) 谨慎使用。这是一个概念证明。在 PHP 中,你永远不应该只接受和计算客户端数据而不检查它是否存在恶意代码并清理输入。 (!)

您可以使用 AJAX 将您的 base64 源代码放入 formData 将其发送到 PHP。

Javascript:

// ajax.js
var xhr = new XMLHttpRequest(); // initialize a new XHR
var formData = new FormData(); // initialize form data
var myFile = 'data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAASwAAACWCAYAAABkW7XSAAAEaElEQVR4Xu3UgQkAMAwCwXb/oS10i4fLBHIG77YdR4AAgYDANViBlkQkQOALGCyPQIBARsBgZaoSlAABg+UHCBDICBisTFWCEiBgsPwAAQIZAYOVqUpQAgQMlh8gQCAjYLAyVQlKgIDB8gMECGQEDFamKkEJEDBYfoAAgYyAwcpUJSgBAgbLDxAgkBEwWJmqBCVAwGD5AQIEMgIGK1OVoAQIGCw/QIBARsBgZaoSlAABg+UHCBDICBisTFWCEiBgsPwAAQIZAYOVqUpQAgQMlh8gQCAjYLAyVQlKgIDB8gMECGQEDFamKkEJEDBYfoAAgYyAwcpUJSgBAgbLDxAgkBEwWJmqBCVAwGD5AQIEMgIGK1OVoAQIGCw/QIBARsBgZaoSlAABg+UHCBDICBisTFWCEiBgsPwAAQIZAYOVqUpQAgQMlh8gQCAjYLAyVQlKgIDB8gMECGQEDFamKkEJEDBYfoAAgYyAwcpUJSgBAgbLDxAgkBEwWJmqBCVAwGD5AQIEMgIGK1OVoAQIGCw/QIBARsBgZaoSlAABg+UHCBDICBisTFWCEiBgsPwAAQIZAYOVqUpQAgQMlh8gQCAjYLAyVQlKgIDB8gMECGQEDFamKkEJEDBYfoAAgYyAwcpUJSgBAgbLDxAgkBEwWJmqBCVAwGD5AQIEMgIGK1OVoAQIGCw/QIBARsBgZaoSlAABg+UHCBDICBisTFWCEiBgsPwAAQIZAYOVqUpQAgQMlh8gQCAjYLAyVQlKgIDB8gMECGQEDFamKkEJEDBYfoAAgYyAwcpUJSgBAgbLDxAgkBEwWJmqBCVAwGD5AQIEMgIGK1OVoAQIGCw/QIBARsBgZaoSlAABg+UHCBDICBisTFWCEiBgsPwAAQIZAYOVqUpQAgQMlh8gQCAjYLAyVQlKgIDB8gMECGQEDFamKkEJEDBYfoAAgYyAwcpUJSgBAgbLDxAgkBEwWJmqBCVAwGD5AQIEMgIGK1OVoAQIGCw/QIBARsBgZaoSlAABg+UHCBDICBisTFWCEiBgsPwAAQIZAYOVqUpQAgQMlh8gQCAjYLAyVQlKgIDB8gMECGQEDFamKkEJEDBYfoAAgYyAwcpUJSgBAgbLDxAgkBEwWJmqBCVAwGD5AQIEMgIGK1OVoAQIGCw/QIBARsBgZaoSlAABg+UHCBDICBisTFWCEiBgsPwAAQIZAYOVqUpQAgQMlh8gQCAjYLAyVQlKgIDB8gMECGQEDFamKkEJEDBYfoAAgYyAwcpUJSgBAgbLDxAgkBEwWJmqBCVAwGD5AQIEMgIGK1OVoAQIGCw/QIBARsBgZaoSlAABg+UHCBDICBisTFWCEiBgsPwAAQIZAYOVqUpQAgQMlh8gQCAjYLAyVQlKgIDB8gMECGQEDFamKkEJEDBYfoAAgYyAwcpUJSgBAgbLDxAgkBEwWJmqBCVAwGD5AQIEMgIGK1OVoAQIGCw/QIBARsBgZaoSlAABg+UHCBDICBisTFWCEiDwAIGUVl1ZhH69AAAAAElFTkSuQmCC';

xhr.open('POST', '/requestHandler.php') // init your request
xhr.onreadystatechange = function () {
// check if the server was able to compute the XHR
    if(xhr.readyState === XMLHttpRequest.DONE && xhr.status === 200) { 
        // handle the response
        console.log(xhr.responseText) // log the PHP output
    }
}

// a form data element needs a descriptor/name and a value.
formData.append('myfile', myFile);
xhr.send(formData); // send your xhr to the server

PHP:

// requestHandler.php
<?php
if('$_POST') {
    echo $_POST['myfile']; // this is your image source
}

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