Detailed explanation of move_uploaded_file() function in PHP-PHP Tutorial-php.cn

This article mainly introduces the PHP move_uploaded_file() function, which actually moves the uploaded file to a new location. Friends who need it can refer to it. I hope it can help everyone.

move_uploaded_file()Function moves the uploaded file to a new location.

If successful, return true, otherwise return false.

Syntax

move_uploaded_file(file,newloc)

##Parameters Description file Required. Specifies the files to be moved. newloc Required. Specifies the new location of the file.

Description

This function checks and ensures that the file specified by file is a legal upload file (that is, uploaded through PHP's HTTP POST upload mechanism) . If the file is legal, it is moved to the file specified by newloc.

If file is not a legal uploaded file, no operation will occur and move_uploaded_file() will return false.

If file is a legal uploaded file but cannot be moved for some reason, no action will occur and move_uploaded_file() will return false and a warning will be issued.

This kind of check is particularly important if the uploaded file may cause its content to be displayed to the user or other users of this system.

Tips and Notes

Note: This function is only used for files uploaded via HTTP POST.

Note: If the target file already exists, it will be overwritten.

Security Supplement

Introduction from w3c, let’s talk about the problems I encountered.

Generally speaking, we will write the save file like this:

$fileName = $_SERVER['DOCUMENT_ROOT'].'/Basic/uploads/'.$_FILES['file']['name']; move_uploaded_file($_FILES['file']['tmp_name'],$fileName )

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First explain, the meaning of these two codes: save the file directly, and at the same time file name Also for the file name uploaded by the user

Okay, now the risk is here:

① Save the file directly.

This means that the file will not be identified in any way. If a user uploads a piece of background code and saves it as a jpg suffix or other, if the administrator accidentally maps it to php and then accesses the background, - - Result It is conceivable that if he deletes all databases in the background, the entire website will directly GG. In short, saving files directly is very risky.

②Use the same file name as the user file name.

The above code will report an error if the user uses a Chinese file name.

As soon as the file name is involved, encoding is involved. If the file name is English + numbers, it is fine. If it contains Chinese, it will be a big problem and it must be re-encoded.

I think reliable storage should be like this:

①The files uploaded by users must be identified.

File recognition, this part has many functions. I think it is good to use MIME type, which is also difficult to forge.

② Change the file name.

I think it’s best to change the file name to a time format like “201803264104421”, or you can also correspond the file name to the database.

Supplement:

There are two parameters. The first parameter is the temporary file name after you upload it, which is automatically generated by the system. Usually the style is:

$_FILE["file"]["tmp_name"];

where file is the name of your front-end file upload form.

The second parameter is the new file name containing the path. For example:

"upload/1.jpg";

In this way, the file you uploaded will be moved to the subdirectory named upload in the current directory, and the file name will be saved as :1.jpg.

move_uploaded_file() function example

Use the move_uploaded_file() function to upload files to the server.

"; echo "Please ensure that if safe_mode is on that the " . "UID PHP is using matches the file."; exit; } else { echo "The file has been successfully uploaded!"; } ?>

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move_uploaded_file File upload failure cases and solutions

Today I am implementing a method to upload avatar image files when users register. A problem occurred when using the PHP script: The PHP script code is as follows:

 0) { echo "Return Code: " . $_FILES["file"]["error"] . "
"; } else { echo "Upload: " . $_FILES["file"]["name"] . "
"; echo "Type: " . $_FILES["file"]["type"] . "
"; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb
"; echo "Temp file: " . $_FILES["file"]["tmp_name"] . "
"; if (file_exists("upload/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { if(is_uploaded_file($_FILES['file']['tmp_name'])){ $stored_path = ROOT.'/upload/'.basename($_FILES['file']['name']); if(move_uploaded_file($_FILES['file']['tmp_name'],$stored_path)){ echo "Stored in: " . $stored_path; }else{ echo 'Stored failed:file save error'; } }else{ echo 'Stored failed:no post '; } } } ?>

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When I execute the above script, the script outputs "Stored failed: file save error", which is very strange. Obviously something went wrong. I saw the error in the php_error_log file: insufficient permissions. I finally found the error: the destination directory where we store the images does not have permissions for the user who executes PHP. The user who executes the PHP script and The user who wrote the script code and created the picture folder is not the same user, so I only need to change the file permissions to 777.

PHP Development Learning File Upload (move_uploaded_file)

Function: Move the uploaded temporary file to the upload directory. Upload has been created in the root directory! ! !

上传文件：
 
  * 1：超过了文件大小，在php.ini文件中设置
 * 2：超过了文件的大小MAX_FILE_SIZE选项指定的值
 * 3：文件只有部分被上传
 * 4：没有文件被上传
 * 5：上传文件大小为0 */ $error=$upfile["error"];//上传后系统返回的值 echo "================
"; echo "上传文件名称是：".$name."
"; echo "上传文件类型是：".$type."
"; echo "上传文件大小是：".$size."
"; echo "上传后系统返回的值是：".$error."
"; echo "上传文件的临时存放路径是：".$tmp_name."
"; echo "开始移动上传文件
"; //把上传的临时文件移动到upload目录下面(upload是在根目录下已经创建好的！！！) move_uploaded_file($tmp_name,"upload/".$name); $destination="upload/".$name; echo "================
"; echo "上传信息：
"; if($error==0){ echo "文件上传成功啦！"; echo "
图片预览:
"; echo ""; //echo " alt=\"图片预览:\r文件名:".$destination."\r上传时间:\">"; }elseif ($error==1){ echo "超过了文件大小，在php.ini文件中设置"; }elseif ($error==2){ echo "超过了文件的大小MAX_FILE_SIZE选项指定的值"; }elseif ($error==3){ echo "文件只有部分被上传"; }elseif ($error==4){ echo "没有文件被上传"; }else{ echo "上传文件大小为0"; } }else{ echo "请上传jpg,gif,png等格式的图片！"; } } ?>

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Execution result:

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