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我使用下面的脚本来实时创建图像,并通过ajax和div显示它们,但是PHP图像头不会被读取,而是显示图像而不是我刚刚获得图像源代码。任何想法如何强制PHP读取标题作为图像?PHP和AJAX图像头不被读取
<?php
header('Content-Type: image/jpeg'); // JPG picture
$root = $_SERVER['DOCUMENT_ROOT'];
require("$root/include/mysqldb.php"); //mysql login details
require("$root/include/mysql_connect.php"); //mysql connect
if(!empty($small)) { $size = "50"; $folder = "thumnail_user_images"; } else { $size = "250"; $folder = "large_user_images"; }
$dice_image = $_GET["image"];
if (!file_exists("$root/$folder/$dice_image"))
{
require("$root/include/image_resizing_function.php"); //create image
$image = new SimpleImage();
$image->load("$root/raw_user_images/$dice_image");
$image->resizeToWidth($size);
$image->save("$root/$folder/$dice_image");
$image->output();
}
else {
readfile("$root/$folder/$dice_image");
}
这是阿贾克斯用换出图片:
<script type="text/javascript">
var bustcachevar=1
var loadedobjects=""
var rootdomain="https://"+window.location.hostname
var bustcacheparameter=""
function ajaxpage(url, containerid){
var page_request = false
if (window.XMLHttpRequest)
page_request = new XMLHttpRequest()
else if (window.ActiveXObject){
try {
page_request = new ActiveXObject("Msxml2.XMLHTTP")
}
catch (e){
try{
page_request = new ActiveXObject("Microsoft.XMLHTTP")
}
catch (e){}
}
}
else
return false
page_request.onreadystatechange=function(){
loadpage(page_request, containerid)
}
if (bustcachevar)
bustcacheparameter=(url.indexOf("?")!=-1)? "&"+new Date().getTime() : "?"+new Date().getTime()
page_request.open('GET', url+bustcacheparameter, true)
page_request.send(null)
}
function loadpage(page_request, containerid){
if (page_request.readyState == 4 && (page_request.status==200 || window.location.href.indexOf("http")==-1))
document.getElementById(containerid).innerHTML=page_request.responseText
}
function loadobjs(){
if (!document.getElementById)
return
for (i=0; i<arguments.length; i++){
var file=arguments[i]
var fileref=""
if (loadedobjects.indexOf(file)==-1){
if (file.indexOf(".js")!=-1){
fileref=document.createElement('script')
fileref.setAttribute("type","text/javascript");
fileref.setAttribute("src", file);
}
else if (file.indexOf(".css")!=-1){
fileref=document.createElement("link")
fileref.setAttribute("rel", "stylesheet");
fileref.setAttribute("type", "text/css");
fileref.setAttribute("href", file);
}
}
if (fileref!=""){
document.getElementsByTagName("head").item(0).appendChild(fileref)
loadedobjects+=file+" "
}
}
}
</script>
直接在浏览器中输入您的PHP处理程序的URL。如果浏览器显示你的图像,那么图像处理程序工作正常。然后,错误出现在您的JavaScript中。我也推荐你使用一个常用的JavaScript框架。通过这种方式,您可以避免代码中出现异步调用的潜在错误,并依赖经过良好测试的框架。 – CodeZombie 2012-03-11 22:22:52