我有一個HTML5應用程序將畫布圖像保存到服務器,然後提供一個鏈接到在新窗口中打開的圖像。保存的圖像的HTML5 URL顯示以前創建的圖像時點擊
這個工作正常,我第一次保存它,但如果我創建並保存一個新的圖像,然後點擊鏈接,它會顯示以前創建的舊圖像。
點擊刷新將強制它顯示新的,但我想知道是否有辦法確保它顯示正確的圖像,以便我不必刷新頁面?
下面是我用來保存圖像。
<script>
function saveImageAs (imgOrURL) {
if (typeof imgOrURL == 'object')
imgOrURL = imgOrURL.src;
window.win = open (imgOrURL);
setTimeout('win.document.execCommand("SaveAs")', 500);
}
</script>
<script type="text/javascript">
//****************************************************************
// Save canvas content into image file. //
//****************************************************************
function saveViaAJAX()
{
document.getElementById('saveimage').style.visibility="hidden";
document.getElementById("debugFilenameConsole").innerHTML="Please wait while your image is been generated";
var testCanvas = document.getElementById('canvas');
var canvasData = testCanvas.toDataURL("image/jpg");
var postData = "canvasData="+canvasData;
var debugConsole= document.getElementById("debugConsole");
debugConsole.value=canvasData;
//alert("canvasData ="+canvasData);
var ajax = new XMLHttpRequest();
ajax.open("POST",'savecanvas.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 <a target='_blank' href='myimage.php'> MyImage.jpg"+ajax.responseText+"</a><br>Reload this page to start a new image or click on the link above to open the file.";
}
}
ajax.send(postData);
}
</script>
而且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;
// Save file. This example uses a hard coded filename for testing,
// but a real application can specify filename in POST variable
$fp = fopen('MyImage.jpg', 'wb');
fwrite($fp, $unencodedData);
fclose($fp);
}
header("Content-Type: image/jpg");
header("Expires: 0");
header("Cache-Control: must-revalidate, post-check=0, pre-check=0");
header("content-disposition: attachment; filename=MyImage.jpg");
imagejpeg($img, null, 100);
?>
嘗試'imgOrURL = imgOrURL.src + +新的Date()的getTime( )' – mplungjan