PHP - Php Command Line: Enter Data Into Executed Function

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http://www.phpfreaks.com/forums/index.php?topic=350801.0

Hello everyone
I have a question?
I run a PHP script loads about 200 calls simultaneously on Linux environment,
so my question is what is best way to run
a command line As an example php-f example.php
or make it through the apache
example "curl http://localhost/example.php"

Thank you
woren

Hi guys,

I'm trying to run a php script from command line in linux.  I installed xampp for linux a.k.a lampp so typing

php /path/to/file.php

Just brings up a message saying I need to install php.

Whats the correct way to do this?

can anyone help me with an overwhelming problem ?

I want to use this command in the php curl code.

curl -v --data "WSCommunityStringRW?2=1200ve50set&Submit=Submit" http://10.2.3.111/Forms/SnmpCommunityString -u "admin:a1s2d3" --anyauth

hello dear php-experts


how to testrun a drive with command-line?

note: i had several blank screens during the last week

does this somewhat have to  do with the hard-drive?

which tests can be done?

Hi, suppose I want to write a PHP script that checks whether a valid email address exists or not. There are a number of GUI-based tools to verify this, but is there any command line tool to do the same so that I can use PHP to capture the output?

I would like to update the DB when a service ends and thought about doing so in the destructor, but I did not make much progress.  Then tried using pcntl_signal(), but also didn't get far.  Is doing so possible?  Thanks

/usr/lib/systemd/system/test123.service

[Unit]
Description=bla
After=syslog.target

[Service]
ExecStart=/usr/bin/php /var/www/testing/__destruct.php

[Install]
WantedBy=multi-user.target

 

/var/www/testing/__destruct.php

<?php
ini_set('display_errors', 1);

pcntl_signal(SIGTERM, function() {
    syslog(LOG_INFO, 'Caught SIGTERM');
    die;
});
pcntl_signal(SIGHUP, function() {
    syslog(LOG_INFO, 'Caught SIGHUP');
    die;
});
pcntl_signal(SIGUSR1, function() {
    syslog(LOG_INFO, 'Caught SIGUSR1');
    die;
});
pcntl_signal(SIGINT, function() {
    syslog(LOG_INFO, 'Caught SIGINT');
    die;
});
pcntl_signal(SIGABRT , function() {
    syslog(LOG_INFO, 'Caught SIGABRT ');
    die;
});
/*
pcntl_signal(SIGSTOP , function() {
    syslog(LOG_INFO, 'Caught SIGSTOP ');
    die;
});
*/
/*
pcntl_signal(SIGKILL , function() {
    syslog(LOG_INFO, 'Caught SIGKILL ');
    die;
});
*/

class Bla
{
    private $bla;
    public function __construct($bla) {
        syslog(LOG_INFO, '__construct');
        $this->bla = $bla;
    }

    public function __destruct()
    {
        syslog(LOG_INFO, '__destruct');
    }

    public function start()
    {
        $counter=0;
        $counter2=0;
        while(true) {
            $counter++;
            if($counter>10000000) {
                $counter2++;
                $counter=0;
                syslog(LOG_INFO, '.');
            }
            if($counter2>5) {
                die('all done!');
            }
        }
    }

}

$bla = new Bla('hello');
$bla->start();

 

So, I'm trying to read a file from a different filepath than the current working directory.    I have some simple PHP like this :

<?php
 
    $input_map = "readfile.php";
    $map_contents = file_get_contents($input_map);
    echo $map_contents."<br />\n";
?>

And this'll work fine because it's in the same directory but if I try to set $input_map = "/home/...", the browser will return an empty string. This script will work from the command line though. So how do I get file_get_contents() to read from the server's file structure instead of just the working directory?   Edit : For example, this will not work through the browser :

<?php
 
    $input_map = 
"/home/...";
    $map_contents = file_get_contents($input_map);
 
    echo $map_contents."<br />\n";
    var_dump($map_contents);
?>

Edited by MutantJohn, 21 January 2015 - 06:35 PM.

  No idea how to add a new line so it's easier to read in the terminal. Debugging this way is absolutely horrible.
Edited by Monkuar, 03 December 2014 - 09:06 AM.

Hi all,

In the process of moving websites off an old win 2008 server and onto a new win 2016 server. I discovered that dns_get_record is failing on the new server. I made a simple test page:

$dnsrecs = dns_get_record('google.com'); 

When i call this as a webpage, it comes back with dns query failed. It comes back instantly, so it's not like it's timing out. But, if i just call the webpage from the command line, it comes back with the dns record. Why would dns_get_record fail when used in a webpage, but work from the command line?  I can make outside connections since I was able to send an email to an smtp server. I thought maybe this is a permission issue, but seems odd I'd be able to send an email using php but not do a dns query.

Oh, and I have two versions of php installed on this server - 5.4.10 and 7.4.9. The old version is a direct copy from the 2008 server. 7.4.9 is a clean install. Both of them behave the same way. webpage fails, command line works. Also, nslookup works. I also tried switching to using google's name server, but it had no effect on anything.

Anyone have any suggestions? 

I have a PHP file that is executed via batch file very frequently for live updating. This is a sort of "sync" file for reading/writing to a MySQL database. I am able to get it functioning absolutely fine when executed manually via a web browser (and my echo debug lines output the expected information), yet when I run the same PHP file via the command line, it seems to just not be capable of opening any file for reading, so the equivalent variables that are correct when executed on a web browser are blank when echoed through the command line.

Is there a different way of handling reading of files when running a php script via the command line, or should it function exactly the same as when run via a browser?

For instance:

$k = "0";
$line = file("examplefile.log")[$k];
echo $line;

Sorry for the caps, but this is relatively time sensitive.  I am trying to make a register form, but when I click the submit button, nothing happens.  It doesn't add to the table, it doesn't bring me home, doesn't even display the errors if the PWD's don't match or the fields are blank.  Here's my code, thanks guys !  PS:  The DB name is phptest, and the table is called users. 
Code: [Select]
<?php
error_reporting(0);

require_once('connector.php');

$errors = array();

if ($_POST["submit"]) {

if (empty($_POST['username'])) { array_push($errors, 'You did not submit a username.');}
if (empty($_POST['email'])) { array_push($errors, 'You did not submit a email.');}
if (empty($_POST['password1'])) { array_push($errors, 'You did not submit a password.');}

$old_usn = mysql_query("SELECT id FROM users WHERE name = '".$_POST['username']."' LIMIT 1") or die (mysql_error());
if (mysql_num_rows($old_usn) > 0) { array_push($errors, 'This username is already registered.');}

$old_email = mysql_query("SELECT id FROM users WHERE email = '".$_POST['email']."' LIMIT 1") or die (mysql_error());
if (mysql_num_rows($old_email) > 0) { array_push($errors, 'This email is already registered.');}

if ($_POST['password1'] != $_POST['password2']) { array_push($errors,'You entered two different passwords');}

if(sizeof($errors) == 0) { 


$username = $_POST['username'];
$email = $_POST['email'];
$password = sha1 ($_POST['password1']);


mysql_query("INSERT INTO users (name, hashed_psw, email, joined)
VALUES ('{$username}', '{$password1}', '{$email}', NOW());") or die (mysql_error());

 

header ('Location: index.php?msg=1');

}
}
?>
<html>
<head>
<title>register</title>
</head>

<body>
<?php
foreach($errors as $e) {
echo $e;
echo "<br/>\n";
}


?>

<form action="register.php" method="post">
<h4>
Username:
<br />
<input name="username" type="text" value="" size="10" maxlength="16" />
<br />
<br />
Email:
<br />
<input name="email" type="text" value="" size="10" maxlength="100" />
<br />
<br />
Password:
<br />
<input name="password1" type="password" value="" size="10" maxlength="16" />
<br />
<br />
Confirm Password:
<br />
<input name="password2" type="password" value="" size="10" maxlength="16" />
<br />
<br />
<input name="submit" type="button" value="Register" />
</h4>
</form>

</body>
</html>

And heres the connector.php script:

Code: [Select]
<?php

mysql_connect("localhost", "***", "***") or die (mysql_error());
mysql_select_db("phptest") or die (mysql_error());

?>(yes, the asterisks have the name and pw, just put them just in caseys!

The company I work for is a not-for-profit organization, we provide medical services for people who can not afford to go to the regular doctor.  We need a way to keep track of our patients and their visits and a way to search the data.

I found an html form, with name, address, phone number, zip code,,,,,, and made a few changes to fit our needs.  But I do not know how to get that information into the database.

I also need a way to make sure the information is not a duplicate.

Since we are a not-for-profit organization, funds are kinda tight.  That is why I want to go with php instead of buying medical software.  Most of the medical software on the market cost more then we can afford.

I'am a computer tech here at the office, but my field is hardware, networking, and malware removal.  PHP programming is new to me.

So where do I start?

I have a html form that collect information such as [name], [address], [email] (amongst others) and once you hit the submit button all the data you send to "thank_you.php" which dumps all the data from fields to the predefined database, as wwell as it telly you thank you for registering.

What code do I need in order for "thank_you.php" to send [name] and [email] fields automatically in the background to other page on different server (such as www.otherdomain.com/storage.php)?

Is such a think possible?

Hi

I'm trying to resize pictures for thumbnails

I can do this correctly at linux command line with the following command

convert ideologies.jpeg -resize 300x200 ./Thumbnails/testphp.jpg

However trying to wrap this in PHP doesn't seem to work

// Upload file to server
        if(move_uploaded_file($_FILES["file"]["tmp_name"], $targetFilePath)){
            // Insert image file name into database
            $insert = $conn->query("INSERT into Threads (Title, ThreadBody, filename) VALUES ('$Title', '$BodyText', '$fileName')");
            if($insert){
                $statusMsg = "The file ".$fileName. " has been uploaded successfully.";
exec("convert {['fileName']} -resize 300x200 wankertestphp.jpg");
            }

If you look above I do successfully hit the "The File has been uploaded successfully" line so it is entering that If statement.

I've also checked the Threads table with a Select * from Threads and I can see the $fileName variable is entering there.  I've tried numerous different ways around the $fileName variable (removing the brackets...removing the brackets and the apostrophes")

 

Any help is appreciated.