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使用PHP接受文件并获得其后缀名的方法

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Release: 2016-06-06 19:48:44
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这篇文章主要介绍了使用PHP接受文件并获得其后缀名的方法,作者着重提到了其中$_FILES全局变量的使用,需要的朋友可以参考下

HTML的form表单
用html的表单模拟一个文件上传的post请求,代码如下:

File Upload
Send this File:


注意:

要确保文件上传表单的属性是 enctype="multipart/form-data",否则文件上传不了


PHP
首先,需要解释一下PHP的全局变量$_FILES,此数组包含了所有上传的文件信息


思路
1、生成40位的随机字符串作为文件名
2、根据文件是图片还是语音转存到不同的文件位置
3、暂时不做文件大小和文件类型的校验

 

function processFile($files, $type) { $uploadName = null; foreach ($files as $name => $value) { $originalName = $value['name']; $arr = explode(".", $originalName); $postfix = $arr[count($arr) - 1]; $tmpPath = $value['tmp_name']; $tmpType = $value['type']; $tmpSize = $value['size']; } $newname = EhlStaticFunction::generateRandomStr(40).".".$postfix; switch ($type) { case 1 : // 处理声音文件 $destination = VIDEOUPLOADDIR.$newname; break; case 2 : // 处理图像文件 $destination = IMAGEUPLOADDIR.$newname; break; } move_uploaded_file($tmpPath, $destination); }

而获取所上传文件的后缀名则可以使用一下代码:

HTML


PHP

结果示例:

20158591330708.jpg (423×167)

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