问题:
答案 0 :(得分:1)
我的想法是从服务器获取路径
<form action="pageA.php" method="post" enctype="multipart/form-data">
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form>
pageA.php :
<?php
session_start();
$todir = "uploads/";
$file = $tooir . basename($_FILES["fileToUpload"]["name"]);
$_SESSION['img']=$todir;
$checkimg = 1;
$imageType = strtolower(pathinfo($file,PATHINFO_EXTENSION));
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo "File is an image - " . $check["mime"] . ".";
$checkimg = 1;
} else {
echo "File is not an image.";
$checkimg = 0;
}if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $file)) {
$up=1 ;
$_SESSION['img']=$todir;
$_SESSION['up']=$up;
echo "The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.";
header("location:./pageB.php");
} else {
up=o;// checking whether file is uploaded
echo "Sorry, there was an error uploading your file.";
}
}
?>
pageB:(在页面底部的顶部和session_start()
处写script
)
<?php
session_Start();
if($_SESSION['up'])
{
echo "<script>window.onload = function() {
var canvas=document.getElementById('myCanvas');
var context=c.getContext('2d');
var image=document.getElementById('draw');
ctx.drawImage(image,10,10);
};</script><img src=$_SESSION['img'] id=draw > <canvas id=myCanvas> </canvas>";
}
?>