PHP does not show output [duplicate] - php

Question

I have a PHP file that is needed to be run from the command line (via crontab). I need to pass type=daily to the file, but I don't know how. I tried:
php myfile.php?type=daily
but this error was returned:
Could not open input file: myfile.php?type=daily
What can I do?

Answer 1

The ?type=daily argument (ending up in the $_GET array) is only valid for web-accessed pages.
You'll need to call it like php myfile.php daily and retrieve that argument from the $argv array (which would be $argv[1], since $argv[0] would be myfile.php).
If the page is used as a webpage as well, there are two options you could consider. Either accessing it with a shell script and Wget, and call that from cron:
#!/bin/sh
wget http://location.to/myfile.php?type=daily
Or check in the PHP file whether it's called from the command line or not:
if (defined('STDIN')) {
$type = $argv[1];
} else {
$type = $_GET['type'];
}
(Note: You'll probably need/want to check if $argv actually contains enough variables and such)

Answer 2

Just pass it as normal parameters and access it in PHP using the $argv array.
php myfile.php daily
and in myfile.php
$type = $argv[1];

Answer 3

These lines will convert the arguments of a CLI call like php myfile.php "type=daily&foo=bar" into the well known $_GET-array:
if (!empty($argv[1])) {
parse_str($argv[1], $_GET);
}
Though it is rather messy to overwrite the global $_GET-array, it converts all your scripts quickly to accept CLI arguments.
See parse_str for details.
If you want the more traditional CLI style like php myfile.php type=daily foo=bar a small function can convert this into an associative array compatible with a $_GET-array:
// Convert $argv into associative array
function parse_argv(array $argv): array
{
$request = [];
foreach ($argv as $i => $a) {
if (!$i) {
continue;
}
if (preg_match('/^-*(.+?)=(.+)$/', $a, $matches)) {
$request[$matches[1]] = $matches[2];
} else {
$request[$a] = true;
}
}
return $request;
}
if (!empty($argv[1])) {
$_GET = parse_argv($argv);
}

Answer 4

Using the getopt() function, we can also read a parameter from the command line. Just pass a value with the php running command:
php abc.php --name=xyz
File abc.php
$val = getopt(null, ["name:"]);
print_r($val); // Output: ['name' => 'xyz'];

Answer 5

Parameters send by index like other applications:
php myfile.php type=daily
And then you can get them like this:
<?php
if (count($argv) == 0)
exit;
foreach ($argv as $arg)
echo $arg;
?>

Answer 6

Save this code in file myfile.php and run as php myfile.php type=daily
<?php
$a = $argv;
$b = array();
if (count($a) === 1) exit;
foreach ($a as $key => $arg) {
if ($key > 0) {
list($x,$y) = explode('=', $arg);
$b["$x"] = $y;
}
}
?>
If you add var_dump($b); before the ?> tag, you will see that the array $b contains type => daily.

Answer 7

You can use the following code to both work with the command line and a web browser. Put this code above your PHP code. It creates a $_GET variable for each command line parameter.
In your code you only need to check for $_GET variables then, not worrying about if the script is called from the web browser or command line.
if(isset($argv))
foreach ($argv as $arg) {
$e=explode("=",$arg);
if(count($e)==2)
$_GET[$e[0]]=$e[1];
else
$_GET[$e[0]]=0;
}
Edited. Using this I found a Small bug. If your parameter value contains an = it fails. I'm using this code now:
if(isset($argv))
foreach ($argv as $arg) {
$e=explode("=",$arg);
if(count($e)>=2)
$_GET[$e[0]]=substr($arg,strlen($e[0])+1,strlen($arg));
else
$_GET[$e[0]]=0;
}

Answer 8

You could use what sep16 on php.net recommends:
<?php
parse_str(implode('&', array_slice($argv, 1)), $_GET);
?>
It behaves exactly like you'd expect with cgi-php.
$ php -f myfile.php type=daily a=1 b[]=2 b[]=3
will set $_GET['type'] to 'daily', $_GET['a'] to '1' and $_GET['b'] to array('2', '3').

Answer 9

I strongly recommend the use of getopt.
If you want help to print out for your options then take a look at GetOptionKit.

Answer 10

Just pass it as parameters as follows:
php test.php one two three
And inside file test.php:
<?php
if(isset($argv))
{
foreach ($argv as $arg)
{
echo $arg;
echo "\r\n";
}
}
?>

Answer 11

There are four main alternatives. Both have their quirks, but Method 4 has many advantages from my view.
./script is a shell script starting by #!/usr/bin/php
Method 1: $argv
./script hello wo8844rld
// $argv[0] = "script", $argv[1] = "hello", $argv[2] = "wo8844rld"
⚠️ Using $argv, the parameter order is critical.
Method 2: getopt()
./script -p7 -e3
// getopt("p::")["p"] = "7", getopt("e::")["e"] = "3"
It's hard to use in conjunction of $argv, because:
⚠️ The parsing of options will end at the first non-option found,
anything that follows is discarded.
⚠️ Only 26 parameters as the alphabet.
Method 3: Bash Global variable
P9="xptdr" ./script
// getenv("P9") = "xptdr"
// $_SERVER["P9"] = "xptdr"
Those variables can be used by other programs running in the same shell.
They are blown when the shell is closed, but not when the PHP program is terminated. We can set them permanent in file ~/.bashrc!
Method 4: STDIN pipe and stream_get_contents()
Some piping examples:
Feed a string:
./script <<< "hello wo8844rld"
// stream_get_contents(STDIN) = "hello wo8844rld"
Feed a string using bash echo:
echo "hello wo8844rld" | ./script
// explode(" ",stream_get_contents(STDIN)) ...
Feed a file content:
./script < ~/folder/Special_params.txt
// explode("\n",stream_get_contents(STDIN)) ...
Feed an array of values:
./script <<< '["array entry","lol"]'
// var_dump( json_decode(trim(stream_get_contents(STDIN))) );
Feed JSON content from a file:
echo params.json | ./script
// json_decode(stream_get_contents(STDIN)) ...
It might work similarly to fread() or fgets(), by reading the STDIN.
Bash-Scripting Guide

Answer 12

if (isset($argv) && is_array($argv)) {
$param = array();
for ($x=1; $x<sizeof($argv);$x++) {
$pattern = '#\/(.+)=(.+)#i';
if (preg_match($pattern, $argv[$x])) {
$key = preg_replace($pattern, '$1', $argv[$x]);
$val = preg_replace($pattern, '$2', $argv[$x]);
$_REQUEST[$key] = $val;
$$key = $val;
}
}
}
I put parameters in $_REQUEST:
$_REQUEST[$key] = $val;
And it is also usable directly:
$$key = $val
Use it like this:
myFile.php /key=val

Answer 13

I found this vanilla/garden-cli on github. I think it answers all the needs for PHP CLI.

Answer 14

To bypass the complexity of passing to the file, it sounds like you could use sed to insert the line directly into the php file.
sed -i "i (backslash)type=daily myfile.php
or as I use it with variables:
sed -i "i (backslash)$type = "(backslash)"${daily}(backslash)"(backslash); ${path}"/myfile.php"