【发布时间】:2012-03-25 14:00:06
【问题描述】:
我想做一个简单的图片上传,然后把第一张图片换成新上传的照片
php 代码:(为您的帮助发表评论)
<?php
//define a maxim size for the uploaded images in Kb
define ("MAX_SIZE","100");
//This function reads the extension of the file. It is used to determine if the file is an image by checking the extension.
function getExtension($str) {
$i = strrpos($str,".");
if (!$i) { return ""; }
$l = strlen($str) - $i;
$ext = substr($str,$i+1,$l);
return $ext;
}
//This variable is used as a flag. The value is initialized with 0 (meaning no error found)
//and it will be changed to 1 if an errror occures.
//If the error occures the file will not be uploaded.
$errors=0;
//checks if the form has been submitted
if(isset($_POST['Submit']))
{
//reads the name of the file the user submitted for uploading
$image=$_FILES['image']['name'];
//if it is not empty
if ($image)
{
//get the original name of the file from the clients machine
$filename = stripslashes($_FILES['image']['name']);
//get the extension of the file in a lower case format
$extension = getExtension($filename);
$extension = strtolower($extension);
//if it is not a known extension, we will suppose it is an error and will not upload the file,
//otherwise we will do more tests
if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif"))
{
//print error message
echo "";
$errors=1;
}
else
{
//get the size of the image in bytes
//$_FILES['image']['tmp_name'] is the temporary filename of the file
//in which the uploaded file was stored on the server
$size=filesize($_FILES['image']['tmp_name']);
//compare the size with the maxim size we defined and print error if bigger
if ($size > MAX_SIZE*1024)
{
echo "";
$errors=1;
}
//we will give an unique name, for example the time in unix time format
$image_name=time().'.'.$extension;
//the new name will be containing the full path where will be stored (images folder)
$newname="images/".$image_name;
//we verify if the image has been uploaded, and print error instead
$copied = copy($_FILES['image']['tmp_name'], $newname);
if (!$copied)
{
echo "";
$errors=1;
}}}}
//If no errors registred, print the success message
if(isset($_POST['Submit']) && !$errors)
{
echo "";
}
?>
-摄影部
<div id="photo1div">
<img id="photo1" src="" alt="No Image" width="251" height="146" />
</div>
-按钮(上传和浏览)
<form name="newad" method="post" enctype="multipart/form-data" action="">
<table>
<tr><td><table>
<tr>
<td><input type="file" name="image" /></td>
</tr>
<tr>
<td><input name="Submit" id="upload" type="submit" value="Upload image" onclick"load_images()" /></td>
</tr>
</table></td></tr>
</table>
</form>
-如果你有更好的方法我可以做到这一点(如 jquery 或 ajax) 请务必告诉我怎么做
Php 和所有的都在同一个文件中(.php 扩展名)
所以评论:
当我单击“浏览”并选择一个图像时,它会保存到名为“image”的目录文件夹中/然后当我单击“提交”时,我希望它将图像源 (photo1) 更改为我刚刚上传的图像。
我有什么办法可以做到这一点?
谢谢 -朱利安
【问题讨论】:
-
我的意思是,检查您已经提出的问题,并接受一些答案。