PHP - Does The Destructor Work When Running On The Command Line Or As A Service

I am trying to execute a command line tool, but am having a problem adding a variable to the line.
Here is the code
Code: [Select]
$filename = "devices.txt";
$handle = fopen($filename, 'r');

while (!feof($handle))
{
$devices = fgets($handle, 4096);
$query = 'ipmitool -I lanplus -H 192.168.1.1 -U admin -P adminpass sdr list';
passthru($query);
}
flcose($handle);

On this line I don't want to have the IP manually added like so.
Code: [Select]
$query = 'ipmitool -I lanplus -H 192.168.1.1 -U admin -P adminpass sdr list';

Instead I want to be able to insert the $devices variable which contains the values from the text file so it would look something like this:
Code: [Select]
$query = 'ipmitool -I lanplus -H $devices -U admin -P adminpass sdr list';
Problem is the command line tool does not run this way. How do I properly add the $devices variable so the command line tool ipmitool will run with its value?

Thanks,

hey guys m using STDIN as input stream in php command line

my problem is that some time i get this error

Warning: fgets(): supplied argument is not a valid stream resource in D:\myquery
\test.php on line 16
$number=fgets(STDIN);
anyone knows solution

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

This topic has been moved to Other Web Server Software.

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, 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?

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?

Hello all

 

I'm new to this forum.

I've been struggling on this problem for a few days now.

I’m writing a script to automatically push code to a git server.
I’m using php and running the script via the command line.
After asking the user to enter the branch and commit message, the following code runs:
 

exec(‘git config user.name "’ . $userName . ‘"’);
exec(‘git config user.email "’ . $userEmail . ‘"’);

exec('git checkout -b ’ . $branch);
exec(‘git add --all’);
exec(‘git commit -m "’ . $message . ‘"’);

exec('git push origin ’ . $branch);

 

When running the last command, the script stops and asks for a user name, then a password.
How can I enter that username and password via the script?
I don’t want to have to type in the user name and password each time.
As I said, I've been stuck on this for 2 days now.

I tried other forums, searching the net.

I'm frustrated...

Thanks in advance.
Have a lovely day.

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.

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;

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? 

  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.

Okay,

On my original problem, I have been able to authenticate with the web service login service, and extract a cookie as follows from the web-service:

$s_Cookie = $agLoginService->agGetCookie();
$s_Cookie = AGSESSION=oxhbjcKtPw5sOymAD8m8KcKR6JL4i+TIE5mZxwOrXLwCRr8iTOvPxA==;

Now, I create data access soapclient:

$agDataService = new SoapClient($s_DataURL, $a_SoapOptions);   //yes I know I should build my own class wrapper around soapClient, but I am trying to make sure I know what to build before I do

next in the process I need to set the cookie for the data access service.

I tried using:
<b>$agDataService->__setCookie('cookie', $s_Cookie);</b>

but I don't think it's right because when I call any of the service's functions, I get no data back.

This is what their engineer had to say ( he is no help in php at all)

This is a question referring the http headers.  He needs to set the cookies as given in the examples on the wiki page.  The cookie is specified in a response http header as: Set-Cookie.  So the http headers he got back are correct.  The cookie he has to set is AGSESSION=oxhbjcKtPw5sOymAD8m8KcKR6JL4i+TIE5mZxwOrXLwCRr8iTOvPxA==;  There should be an api for setting the cookie if not there is one for setting the http header.  Set the http header name for the request (cookie - notice that is different than what you get back from an http response - read the rfc spec for http for all of the details) then set the value.  The semi colon at the end is the delimiter for http headers (in the rfc).  That is all he needs to do if he gets back some login error that means the cookies did not get set correctly.

Any ideas or help will be great. Thank you.

Hi all,

Perhaps, let me briefly explain what I would like to achieve.

Currently I am using INNER JOIN command to link 4 sql tables
1) tutor_overall_level_subject.sql
2) tutor_subject_level.sql
3) tutor_level.sql
4) tutor_subject.sql

In tutor_level.sql, there are 11 'level_id'
level id 1 == Pre-School
level id 2 == Lower Primary School
level id 3 == Upper Primary School
level id 4 == Lower Secondary
etc....


As you can see from the code below,
I am trying to retrieve 'lower primary' records only and place it in a table
Subsequently, 'upper primary' records and place it in another table

However when I executed the code, I receive a blank page. What could have gone wrong? Thanks
if ($row3['level_id'] == 2){}
if ($row3['level_id'] == 3){}

Code: [Select]
<?php
//Retrieve Lower Primary 
if ($row3['level_id'] == 2) {
echo'<table><tr>'; 
                echo '<td class="label">Lower Primary</td>';
$count = 0; 
while($row3 = mysqli_fetch_array($data3)) {

if ($count % 5 == 0) { 
echo "</tr><tr>"; $count = 0;
}
echo '<td>' . $row3['subject_name'] . '</td>';
$count++; 
} 
echo '</tr></table><br/>'; 
}

//Upper Lower Primary 
if ($row3['level_id'] == 3) {
echo'<table><tr>'; 
                echo '<td class="label">Upper Primary</td>';
$count = 0; 
while($row3 = mysqli_fetch_array($data3)) {

if ($count % 5 == 0) { 
echo "</tr><tr>"; $count = 0;
}
echo '<td>' . $row3['subject_name'] . '</td>';
$count++; 
} 
echo '</tr></table><br/>'; 
}
?>


//This is my complete code
Code: [Select]
<?php
$query3 = "SELECT sl.level_id, sl.subject_id, tl.level_name AS level_name, ts.subject_name AS subject_name " .
        "FROM tutor_overall_level_subject AS ols " .
"INNER JOIN tutor_subject_level AS sl USING (subject_level_id) " .
        "INNER JOIN tutor_level AS tl USING (level_id) " .
        "INNER JOIN tutor_subject AS ts USING (subject_id) " .
"WHERE ols.tutor_id = '" . $_GET['tutor_id'] . "'";


  $data3 = mysqli_query($dbc, $query3)
  or die(mysqli_error($dbc));

  if (mysqli_num_rows($data3) > 0) {

echo '<div id="panel4">';

if ($row3['level_id'] == 2) {
echo '<td class="label">Lower Primary</td>';
                echo'<table><tr>'; 
$count = 0; 
while($row3 = mysqli_fetch_array($data3)) {

if ($count % 5 == 0) { 
echo "</tr><tr>"; $count = 0;
}
echo '<td>' . $row3['subject_name'] . '</td>';
$count++; 
} 
echo '</tr></table><br/>'; 
}

echo '</div>'; //End of panel 4
?>