php FILE POST上传没有保存
[英]php FILE POST upload without save
问题描述
I am using a plain text file in a PhP script. 
我在PhP脚本中使用纯文本文件。 Is there any way to process the file in the PhP script without saving it locally? 有没有办法在PhP脚本中处理文件而不在本地保存？ Everywhere I see simply uploading a file and saving it. 我到处都看到只是上传文件并保存。 I just need to pull some names off of the file and be done with it. 我只需要从文件中提取一些名称并完成它。 I have everything else working if I use a local copy of the file, so saving it, then deleting it works. 如果我使用文件的本地副本，我有其他一切工作，所以保存它，然后删除它的工作原理。 I was just wondering if there was a way to skip the saving a copy and just getting that information directly. 我只是想知道是否有办法跳过保存副本并直接获取该信息。
We upload the file here. 我们在这里上传文件。
<html>
<body>
<form action="test.php" method="post" enctype="multipart/form-data">
<label for="file">Filename:</label><input type="file" name="file" id="file" /> <br />
<input type="submit" name="submit" value="Submit" />
</form>
</body>
</html>
and in the script working with a copy of the file saved locally, I simply use 并且在脚本中使用本地保存的文件副本，我只是使用
$file = fopen($_FILES['file']['name'], "r");
6 个解决方案
解决方案1
19 已采纳 2012-02-01 18:26:55
If its text file (assuming size is not that huge) usually you could read it by 如果它的文本文件（假设大小不是那么大）通常你可以读它
$contents= file_get_contents($_FILES['file']['tmp_name']);
If you are not sure which type of upload it was, check the request method 如果您不确定它的上传类型，请检查请求方法
if(strtolower($_SERVER['REQUEST_METHOD'])=='post')
$contents= file_get_contents($_FILES['file']['tmp_name']);
elseif(strtolower($_SERVER['REQUEST_METHOD'])=='put')
$contents= file_get_contents("php://input");
else
$contents="";
解决方案2
4 2012-02-01 18:20:30
When the form is submitted, check the $_FILES['input_name']['tmp_name'] property. 提交表单时，请检查$_FILES['input_name']['tmp_name']属性。 This is the path to your file saved in a temporary /tmp system path. 这是保存在临时/tmp系统路径中的文件的路径。 You can proceed reading the name using, say, file_get_contents() and then simply forget the file. 您可以使用file_get_contents()继续读取名称，然后只是忘记该文件。 System will take care of removing it. 系统将负责删除它。
Just to stand out of the other answers, you could theoretically read the file without even uploading it using JavaScript, see http://www.html5rocks.com/en/tutorials/file/dndfiles/ . 只是为了脱离其他答案，理论上你可以在不使用JavaScript上传文件的情况下阅读文件，参见http://www.html5rocks.com/en/tutorials/file/dndfiles/ 。 Then submit only the data you need as part of AJAX request. 然后仅提交您需要的数据作为AJAX请求的一部分。
解决方案3
3 2012-02-01 18:21:56
The file has to be saved somewhere before reading in this case is the temporary directory; 在这种情况下读取文件是临时目录之前，必须将文件保存在某处; You can get the contents of the file from the temporary directory, then if you really need to you delete. 您可以从临时目录中获取文件的内容，然后如果您确实需要删除。 See the following code: 请参阅以下代码：
$file = file_get_contents($_FILES["ufiley"]["tmp_name"]);
unlink($_FILES["ufiley"]["tmp_name"]);
//contents would be stored in $file
解决方案4
1 2012-02-01 18:21:52
When you upload a file, before you save it locally it get's saved to a temporary file. 上传文件时，在本地保存文件之前，它会保存到临时文件中。 The location of which can be accessed by: 其位置可通过以下方式访问：
$_FILES['uploadedfile']['tmp_name']
You can choose not to save the file and fopen() the temporary file, as long as you do this within the same script that recieves the POST . 您可以选择不保存文件和fopen()临时文件，只要您在收到POST的同一脚本中执行此操作即可。
解决方案5
0 2012-02-01 18:19:59
不，它必须保存才能被读取，我通常的常规与你的相同，上传，阅读和删除。
解决方案6
0 2012-02-01 18:20:29
You could just use $_FILES["file"]["tmp_name"] which would be the temp file that was uploaded. 你可以使用$_FILES["file"]["tmp_name"] ，这将是上传的临时文件。 Once your script is done running, PHP should even clean up after itself and delete the temp file. 一旦脚本运行完毕，PHP甚至应该自行清理并删除临时文件。
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