PHP - 2 Exec() Functions, One Works The Other Doesn't

Hello everyone,

I'm having this problem which is really annoying, tried to solve it but couldn't, I write that code in PHPMyAdmin and it works great, but it doesn't work in the website it self

ok long story short, there are three tables, hotels, cities, countries
hotels include in addition to hotel info, 2 columns (city_id) and (country_id)
Cities include id and name and also countries include id and name

what I was trying to do, that when a person inputs a city or country name in the search form, it should get the hotels that exists in this city or country, but unfortunately it shows all the hotels in all cities and countries, although the pagination code for number of pages works just fine, it count the number of hotels in that city or country and show the number of pages correctly

so here is the code for both

for hotel search

Code: [Select]
class hotelManager
{


public function getHotel($where)
{
$where = isset($_POST['where']) ? $_POST['where'] : "";
$dbObj = new DB();
$sql = "select * from hotels where city_id = (select id from cities where name = '$where' ) or country_id = (select id from countries where name = '$where' )";
$result = MYSQL_QUERY($sql);
$arr = array();
echo "<table>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td valign=\"top\" width=\"120px\">";
$rowid = $row['id'];
$imageqry=mysql_query("SELECT * FROM `hotelphotos` where hotel_id='$rowid' LIMIT 1");
$image=mysql_fetch_array($imageqry);
$imagename=$image['attachmentName'];
echo "<img src=\"foxmaincms/webroot/files/small/$imagename\"/>";
echo "</td>";
echo "<td valign=\"top\">";

echo "<table>
<tr>
<td valign=\"top\">
<a href=\"hotels.php?id=".$row['id']."\" class=\"titleslink\">".$row['name']."</a>
</td>
</tr>
<tr>
<td class=\"text\" valign=\"top\">
".$row['location']."
</td>
</tr>
</table>";

echo "</td>";
echo "</tr>";
}
echo "</table>";

for hotel pagination

Code: [Select]
<?php
  include("includes/hotelsManager.php");
  $hotelObj = new hotelManager();
  $where = isset($_POST['where']) ? $_POST['where'] : "";
  if(isset($_POST['where']))
  {
    
$hotelObj -> getHotel($where);


$per_page = 9;

//Calculating no of pages
$sql = "select * from hotels where city_id = (select id from cities where name = '$where' ) or country_id = (select id from countries where name = '$where' )";
$result = MYSQL_QUERY($sql) or die("<br />No Hotels found in this city, please check the city name and try again");
$count = mysql_num_rows($result);
$pages = ceil($count/$per_page)
?>


<div id="loading" ></div>
<div id="maincontent" ></div>
<ul id="pagination">
thank you in advance

Hi,

My site was working fine, but I just switched servers/web hosts and now I'm getting the following errors (in blue). The thing is, I don't get the error if I try to run the query loop after just using 'mysql_select_db' once, but if I use it again to select a different database and then run the query loop, I start getting the error.

Code: [Select]
mysql_select_db ("database1", $mysql_con);

$mysql_query = mysql_query ("SELECT ID FROM someTable1", $mysql_con);
$someVar1 = 0;

while ($mysql_array = mysql_fetch_array ($mysql_query))
{
$someVar1 ++;
}


////// The above code gives me no error. But If I then try to select a different database and run a loop in the same way (in the same file), I start getting the below error...


mysql_select_db ("database2", $mysql_con);

$mysql_query = mysql_query ("SELECT ID FROM someTable2", $mysql_con);
$someVar2 = 0;

while ($mysql_array = mysql_fetch_array ($mysql_query))
{
$someVar2 ++;
}
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/snubby5/public_html/coozymcmillan.com/view.php on line 86

So I looked it up and somewhere it told me to add some extra code to handle errors better:

Code: [Select]
if ($mysql_query === FALSE) {
    die (mysql_error ()); // TODO: better error handling
}
But when I do that, I get a different error:

Table 'database1.someTable2' doesn't exist

It doesn't seem to recognize that I switched databases, and it says 'database1.someTable2' doesn't exist (because it doesn't), when really it should be checking for 'database2.someTable2' (which exists).

So to reiterate, I can run the query loop fine after selecting database1, but after that if I select database2, it either gives me the first error, or If I add the 'die' code it gives me the second error.


Thanks much!

PS: I can get it working if I put everything into one database, but I have multiple websites (which sometimes need to call eachothers' databases) and it would be a huge hassle to cram them all into one database unless I have to.

Hi, this query runs fine when I run it from PHPMyAdmin:
 

UPDATE `tran_term_taxonomy` SET `description` = (SELECT keyword from `good_keywords` ORDER BY RAND() LIMIT 1,1) WHERE `tran_term_taxonomy`.`taxonomy` = 'post_tag' AND `tran_term_taxonomy`.`description` = "" LIMIT 1

However, when I run the same query in a PHP file on my server, the page doesn't load at all. The message I get is: www.somesite.com is currently unable to handle this request. HTTP ERROR 500. This is my PHP code:

<?php 
include("/database/connection/path/db_connect.php"); 

$result4 = mysqli_query($GLOBALS["___mysqli_ston"], "UPDATE `tran_term_taxonomy` SET `description` = (SELECT keyword from `good_keywords` ORDER BY RAND() LIMIT 1,1) WHERE `tran_term_taxonomy`.`taxonomy` = 'post_tag' AND `tran_term_taxonomy`.`description` = "" LIMIT 1");
echo $result4;

?>

So how do I make this query work please? Thanks for your guidance.

Hello,
The following is my situation where I seem to get a 500 error code from the linux server:

i have an 'index' file like this:
Code: [Select]
<?php
require("includes/config.php");

$a = $_REQUEST['a'];

switch ($a)
{
   case "home":
      include("frontpage/main.php");
   case "user-process":
      include("user-process.php");
}
?>
config.php is something like this:
Code: [Select]
<?php

require(includes/classes/session.class.php);
require(includes/classes/user.class.php);
require(includes/classes/db.class.php);
...
?>
Now if we fall into the case "home" it works fine. Instead, if we fall into user-process it writes to the logs file Fatal Error: Class User does not exist bla bla bla. Why doesn't it exist ? every class is included in the config.php file then index.php includes first config.php ( which has all the classes) and then includes the requested page.

I also have a .htaccess file which is as follows:
Code: [Select]
RewriteEngine On
RewriteRule ^([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?/?([^/\.]+)?$ index.php?a=$1&b=$2&c=$3&d=$4&e=$5&f=$6&g=$7 [NC,L]
which is used to access in a SEO friendly way the pages that users request.

I'm trying to update every record where one field in a row is less than the other. The code gets each row i'm looking for and sets up the query right, I hope I combined the entire query into one string each query seperated by a ; so it's like
UPDATE `table` SET field2= '1' WHERE field1= '1';UPDATE `table` SET field2= '1' WHERE field1= '2';UPDATE `table` SET field2= '1' WHERE field1= '3';UPDATE `table` SET field2= '1' WHERE field1= '4';UPDATE `table` SET field2= '1' WHERE field1= '5';
this executes properly if i run the query in phpMyAdmin, however when I run the query in PHP, it does nothing... Any advice?

Hi guys!

I have a php script that works and outputs info:

<?php
echo exec("uptime");
?>

and another that doesn't output anything:

<?php
echo exec("MP4Box");
?>

Why?

I'm trying to open a batch file in php with these contents:
Code: [Select]
@echo off
java -classpath rscd.jar;lib/mina.jar;lib/xpp3.jar;lib/slf4j.jar;lib/xstream.jar;lib/hex-string.jar; org.rscdaemon.server.Server
pause

My PHP code is simply
Code: [Select]
$run = "C:\\Users\\Zorian\\Desktop\\EasyRSC\\Server\\run-win.bat";
echo exec("cmd.exe /c " . $run);

It should open up a java applet, instead it just echos:
Code: [Select]
Press any key to continue . . .

I figure I’m doing something basic wrong.
I have this php code.

$cmd = “at $wtime $wdate <<< “sftp -a -r -P xxx xxx@xxx.xxx.net:”.$file.” /mnt/TRFR” ;
exec ($cmd);
echo $cmd;

the “at” command is never generated atq returns blank. But if I copy the output of the echo from the web page into bash the “at” job gets created.

 

What am I missing here?

Edited April 9, 2020 by Henry_Who

Hello, I was just seeing if someone could push me in the right direction. I have a form that uploads items to a database and it also includes uploading images to a directory as a part of that. That all works, but I was thinking for security it'd be best not to leave that directory as 0777 for permissions. In researching this I found the exec() function. I'm still figuring out the relation of who the permissions are associated with and why. All I know is if the directory has 0777 for permissions my script works just fine. I was just seeing if using exec to change the permissions during the process of uploading the images and then change back would be cumbersome and leave open other problems (there won't be any user-based input being passed to that) or if that actually would be the way to go for this.

Thanks!
-Frank

i have a bash script that i was running with cron jobs:


#! /bin/bash
 start=1000
 homedir="/home/alin/NetBeansProjects/Fragger2"

 numberofprocess=`expr $start + $2`

 i=$start

 currentnumber=`ps ax |  grep index | grep -c php`

case "$1" in
  start)

        while [ $i -lt $numberofprocess ]
                do
                   if [ `ps axo cmd | grep  index | grep php | grep -c $i` -lt 1 ] ; then

                   
                   /server/php/bin/php $homedir/index.php $i &
                   
                   fi;
                   
                   i=`expr $i + 1`
                   
        done
       ;;
   stop)
        while [ $currentnumber -ne 0 ]
        do
                kill -9 `ps ax | grep php | grep index | awk {'print $1'} | head -n 1`;

                currentnumber=`ps ax |  grep index | grep -c php`
        done
        
      ;;

  *)
echo "Usage : " $0 "start|stop"
        ;;

esac
exit 0



and it was working fine,

now i'm tring to make it run from a page .


if(isset ($_POST['stop'])){
    $conn->update("update `links` set `proccess`=0 ");
    exec("/home/alin/NetBeansProjects/Fragger2/Trans stop 0");
}


if(isset ($_POST['start'])){
    if(!preg_match("/^[0-9]+$/", $_POST['proccess']))
            echo "proccess should be only a number!";
    
    exec("/home/alin/NetBeansProjects/Fragger2/Trans start ".$_POST['proccess']);
}


and i don't know why it is not working,

i have set for this folder all the permision for this folder  "chmod -R 777 /home/alin"

it doesn't stop running and i can't figure out why

i have to kill the procces from the comand line to stop

Hi,

I have a java file that needs to be executed from my php script, the command I have in it works perfectly when I execute directly in the command line, but when I execute it from php it returns this: array(0) { }

I have tested that the exec works by using antiword, which does return results as expected, but its not working for the code below.
But I would much rather use apache tika.

Also your thoughts on whether this is good practice please. I am using this to extract text from CV's and store that in the db to be able to search through. I need to extract from: pdf, doc, rtf and docx. this is the only tool I have found that can handle them all.

your help on the exec problem and any suggestions will be much appreciated.

Code: [Select]
<?php
   exec('java -jar /etc/tika-app-1.0.jar -t /var/www/html/cvdatabase/400001.doc', $output);
   var_dump($output);
?>

Hello i am trying to execute a linux program for every line of a textarea

Code: [Select]
        $text = $_POST['links'];
       
        $text = htmlspecialchars($_POST['links']);
        $text = nl2br($text);
       
        foreach(explode("\n",$text) as $row)
        {
                $result = NULL;
                $command = 'the command';
                exec($command,&$result);
                echo $result[0]."<br>";
        }
but this always print the result for last line of text area
for example if there are 3 lines it print 2 blanks and 1 result...
whats wrong??

 have a executable binary to which i pass some dynamic values as hostname,host id,server name,server id and a text field value.

Now i call this binary using exec or shell_exec functions in PHP to stimulate Command executions as a admin.

But i am able to invoke the binary and not able to use psexec service of windows.

I get logs and DB entries successfully but i also have a xml push operation in psexec call that is required.

the format is like ::

exec($url,$output,$return_var);

url is the command path with parameters .
output is an out param and return is the code on execution

this is executed in PHP and i want this binary to generate the same output as done from CLI .

Currently i guess it's a permission issue with shell that disallows psexec call.

How do i invoke the binary with psexec access outside PHP on command shell

I have set access privilege of Powershell  to "unrestricted" so that any script can be executed.

Still it hangs up somewhere in th script causing problems in copying XML file.

Powershell psexec is required in the process and i need to track down the root cause so that php side scripting of binary is successfull.

Please let me know about your ideas in fixing this.

I am using google and trying examples ,but i havent got any feasible solution.

Hope some expert advice can guide me out of this.

Hi all.

I tried to execute command below
` echo '<br/>'.$data. " >=> " . exec("java -jar encrypt.jar ".$data." ".$privateKey, $key);`
it's look like we run `java -jar encrypt.jar 192.168.101.1BBB10052 ssdsds`
I can run this command on the server by ssh but when I use PHP code, it's nothing.
so I need to enale the exec function or different command on this version?

Hey guys i am trying to create a small script using exec() to convert PDF files into images ... The script worked well in the past and for some reasons it staped working.

exec('"C:\\Program Files\\ImageMagick-6.7.1-Q16\\convert.exe" -verbose -density 150 D:\\Inetpub\\www.goulet.ca\\Pub\\Web\\new\\media\\other\\'.$value.' -quality 100 -sharpen 0x1.0 D:\\Inetpub\\www.goulet.ca\\Pub\\Web\\new\\media\\pdf_flipbook\\'.$value.'\\img.png', $out);

the $out var returns an empty array