A simple PHP File Upload example Program

This article introduces a simple PHP File Upload sample code for beginners. If you need to learn it, you can refer to it.

The Code is as follows:

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<? Php // Define the prompt Function Function alert ($ msg ){ Return '<script type = "text/javascript"> alert ("'. $ msg. '"); window. history. back (-1); </script> '; } // Define the allowed file types $ AllowType = array ('image/jpeg ', 'image/gif', 'image/jpg '); // Define the path, either absolute or relative. $ FilePath = './uploadFileDir /';   // Receives the form information. The file value written in the form is the name value in the form of the Static Page. $ File = $ _ FILES ['file'];   // Step 1: Determine whether the uploaded file is incorrect If ($ file ['error']! = 0 ){ Exit (alert ('file upload error ')); }   // Step 2: Determine the file size. Here, 102400 is the byte, and the conversion to kb is kb. If ($ file ['SIZE']> 102400 ){ Exit (alert ('file oversized ')); }   // Step 3: Determine the file type If (! In_array (mime_content_type ($ file ['tmp _ name']), $ allowType )){ Exit (alert ('file type error ')); }   // Step 4: Determine whether the path exists. If not, create If (! File_exists ($ filePath )&&! Mkdir ($ filePath, 0777, true )){ Exit (alert ('Directory creation error ')); }   // Step 5: Define the uploaded name and Path $ Filename = time (). '_'. $ file ['name'];   // Step 6: copy the file If (! Copy ($ file ['tmp _ name'], $ filePath. $ filename )){ Exit (alert ('file upload error, please try again later ')); }   // Step 7: delete a temporary file Unlink ($ file ['tmp _ name']);   // Prompt that the upload is successful Echo alert ('Congratulations, the file ['. $ filename.'] is uploaded successfully! '); ?>