transfer file using file_get_content
file_get_contents('http://url/to/upload/handler', false, $context);transfer file using file_get_content
file_get_contents('http://url/to/upload/handler', false, $context);transfer file using file_get_content
$filename = "/path/to/uploaded/file.zip";
$file_contents = file_get_contents($filename);
$content = "--".MULTIPART_BOUNDARY."\r\n".
"Content-Disposition: form-data; name=\"".FORM_FIELD."\"; filename=\"".basename($filename)."\"\r\n".
"Content-Type: application/zip\r\n\r\n".
$file_contents."\r\n";
// add some POST fields to the request too: $_POST['foo'] = 'bar'
$content .= "--".MULTIPART_BOUNDARY."\r\n".
"Content-Disposition: form-data; name=\"foo\"\r\n\r\n".
"bar\r\n";
// signal end of request (note the trailing "--")
$content .= "--".MULTIPART_BOUNDARY."--\r\n";take file data in variable php
I suppose you want to get the content generated by PHP, if so use:
$Vdata = file_get_contents('http://YOUR_HOST/YOUR/FILE.php');
Otherwise if you want to get the source code of the PHP file, its the same as
a .txt file:
$Vdata = file_get_contents('path/to/YOUR/FILE.php');take file data in variable php
ob_start();
include "yourfile.php";
$myvar = ob_get_clean();
reference :
https://stackoverflow.com/questions/1272228/how-do-i-load-a-php-file-into-a-variable
https://secure.php.net/manual/en/function.ob-get-clean.phptransfer file using file_get_content
$header = 'Content-Type: multipart/form-data; boundary='.MULTIPART_BOUNDARY;Copyright © 2021 Codeinu
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