简单PHP上传图片、删除图片实现代码_php实例

WBOY
Release: 2016-05-17 09:25:50
Original
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上传图片:

复制代码 代码如下:

if (!empty($_FILES["img"]["name"])) { //提取文件域内容名称,并判断
$path=”uppic/”; //上传路径
if(!file_exists($path))
{
//检查是否有该文件夹,如果没有就创建,并给予最高权限
mkdir(“$path”, 0700);
}//END IF
//允许上传的文件格式
$tp = array(“image/gif”,”image/pjpeg”,”image/jpeg”);
//检查上传文件是否在允许上传的类型
if(!in_array($_FILES["img"]["type"],$tp))
{
echo “<script>alert(‘格式不对');history.go(-1);</script>”;
exit;
}//END IF
$filetype = $_FILES['img']['type'];
if($filetype == ‘image/jpeg'){
$type = ‘.jpg';
}
if ($filetype == ‘image/jpg') {
$type = ‘.jpg';
}
if ($filetype == ‘image/pjpeg') {
$type = ‘.jpg';
}
if($filetype == ‘image/gif'){
$type = ‘.gif';
}
if($_FILES["img"]["name"])
{
$today=date(“YmdHis”); //获取时间并赋值给变量
$file2 = $path.$today.$type; //图片的完整路径
$img = $today.$type; //图片名称
$flag=1;
}//END IF
if($flag) $result=move_uploaded_file($_FILES["img"]["tmp_name"],$file2);
//特别注意这里传递给move_uploaded_file的第一个参数为上传到服务器上的临时文件
}//END IF
//这里再将$img的值写入到数据库中对应的字段

删除图片:
复制代码 代码如下:

unlink(“uppic/”.$img); //当然,变量的值是从数据库中读取出来的,PHP删除图片比ASP简洁多了
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