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Javascript 使用dataURL发送变量_Javascript_Php - Fatal编程技术网

Javascript 使用dataURL发送变量

Javascript 使用dataURL发送变量,javascript,php,Javascript,Php,嗨,我有这段代码,它可以将一个dataURL发送到php并保存到服务器 JS: PHP 有人知道为什么吗?试试这个: type:"POST", url: "php/save.php", data: { img_val: canvas.toDataURL("image/png"), random: random }, success: function(){ $\u

嗨,我有这段代码,它可以将一个dataURL发送到php并保存到服务器

JS:

PHP


有人知道为什么吗?

试试这个:

        type:"POST",
        url: "php/save.php",
        data: {
            img_val: canvas.toDataURL("image/png"),
            random: random
        },
        success: function(){

$\u POST
包含
img\u val
random

这是正确的,在发布的代码片段中,数据URL没有名称是的,这很有效,谢谢!我会在几分钟内接受。
//Show the image
echo '<img src="'.$_POST['img_val'].'" />';

//Get the base-64 string from data
$filteredData=substr($_POST['img_val'], strpos($_POST['img_val'], ",")+1);

//Decode the string
$unencodedData=base64_decode($filteredData);

//Save the image
file_put_contents('img.png', $unencodedData);
?>
            type:"POST",
            url: "php/save.php",
            data: $('#img_val').val(canvas.toDataURL("image/png"))+"&random="+random,
            success: function(){
<?php
//save.php code

$random = $_REQUEST['random'];

//Show the image
echo '<img src="'.$_POST['img_val'].'" />';

//Get the base-64 string from data
$filteredData=substr($_POST['img_val'], strpos($_POST['img_val'], ",")+1);

//Decode the string
$unencodedData=base64_decode($filteredData);

//Save the image
file_put_contents($random.'img.png', $unencodedData);
?>
        type:"POST",
        url: "php/save.php",
        data: {
            img_val: canvas.toDataURL("image/png"),
            random: random
        },
        success: function(){