Javascript 如何使用php和ajax保存画布图像数据?

Javascript 如何使用php和ajax保存画布图像数据?,javascript,php,jquery,ajax,html,Javascript,Php,Jquery,Ajax,Html,我在vps上遵循了本教程: testSave.php <?php if (isset($GLOBALS["HTTP_RAW_POST_DATA"])) { // Get the data $imageData=$GLOBALS['HTTP_RAW_POST_DATA']; // Remove the headers (data:,) part. // A real application should use them according to needs such as

我在vps上遵循了本教程:

testSave.php

<?php
if (isset($GLOBALS["HTTP_RAW_POST_DATA"]))
{
  // Get the data
  $imageData=$GLOBALS['HTTP_RAW_POST_DATA'];

  // Remove the headers (data:,) part.
  // A real application should use them according to needs such as to check image type
  $filteredData=substr($imageData, strpos($imageData, ",")+1);

  // Need to decode before saving since the data we received is already base64 encoded
  $unencodedData=base64_decode($filteredData);

  //echo "unencodedData".$unencodedData;
 $randnum = rand(1111111111,9999999999);
  // Save file. This example uses a hard coded filename for testing,
  // but a real application can specify filename in POST variable
$tmpfname = tempnam("http://123.xx.xx.xx/test/tmp/", "FOO");
$fp = fopen(http://123.xx.xx.xx/test/test . uniqid() .".png","wb");
  fwrite( $fp, $unencodedData);
  fclose( $fp );
}
?>

JS

function saveViaAJAX()
{
    var testCanvas = document.getElementById("testCanvas");
    var canvasData = testCanvas.toDataURL("image/png");
    var postData = "canvasData="+canvasData;
    var debugConsole= document.getElementById("debugConsole");
    debugConsole.value=canvasData;

    //alert("canvasData ="+canvasData );
    var ajax = new XMLHttpRequest();
    ajax.open("POST",'testSave.php',true);
    ajax.setRequestHeader('Content-Type', 'canvas/upload');
    //ajax.setRequestHeader('Content-TypeLength', postData.length);


    ajax.onreadystatechange=function()
    {
        if (ajax.readyState == 4)
        {
            //alert(ajax.responseText);
            // Write out the filename.
                document.getElementById("debugFilenameConsole").innerHTML="Saved as<br><a target='_blank' href='"+ajax.responseText+"'>"+ajax.responseText+"</a><br>Reload this page to generate new image or click the filename to open the image file.";
        }
    }

    ajax.send(postData);
}
函数saveViaAJAX() { var testCanvas=document.getElementById(“testCanvas”); var canvasData=testCanvas.toDataURL(“image/png”); var postData=“canvasData=”+canvasData; var debugConsole=document.getElementById(“debugConsole”); debugConsole.value=canvasData; //警报(“canvasData=“+canvasData”); var ajax=new-XMLHttpRequest(); open(“POST”,“testSave.php”,true); setRequestHeader('Content-Type','canvas/upload'); //setRequestHeader('Content-TypeLength',postData.length); ajax.onreadystatechange=function() { if(ajax.readyState==4) { //警报(ajax.responseText); //写出文件名。 document.getElementById(“debugFilenameConsole”).innerHTML=“另存为

重新加载此页面以生成新图像,或单击文件名以打开图像文件。”; } } 发送(postData); } 问题是,当用户单击“通过ajax发送”时,服务器目录()中没有发送/生成图像。结果是:目录中的所有内容都保持不变

应该在“另存为”下生成图像链接

有什么解决办法吗

p.S.
是否还有其他方法可以使用php来回传图像链接?

问题在于您的AJAX。参考Url此处给出了使用php和ajax保存画布的代码。

很好的尝试,您的代码将覆盖文件而不是生成文件。谢谢你!