PHP - Strange Sequence In Error Logs: Php Warning: Preg_match_all(): Unknown Modifier 'g'

Hi,
on my cpanel I can see some errors:
[Mon Mar 12 08:41:05 2012] [error] [client 46.120.43.163] File does not exist: /home/affilinn/public_html/504.shtml, referer: http://www.mysite.com/admin/
[Mon Mar 12 08:33:25 2012] [error] [client 46.120.43.163] File does not exist: /home/affilinn/public_html/404.shtml
[Mon Mar 12 08:33:25 2012] [error] [client 46.120.43.163] File does not exist: /home/affilinn/public_html/404.shtml
[Mon Mar 12 08:30:12 2012] [error] [client 46.120.43.163] File does not exist: /home/affilinn/public_html/404.shtml
[Mon Mar 12 08:30:12 2012] [error] [client 46.120.43.163] File does not exist: /home/affilinn/public_html/favicon.ico
[Mon Mar 12 08:30:12 2012] [error] [client 46.120.43.163] File does not exist: /home/affilinn/public_html/504.shtml, referer: http://www.mysite.com/admin/
[Mon Mar 12 08:29:55 2012] [error] [client 46.120.43.163] File does not exist: /home/affilinn/public_html/404.shtml
[Mon Mar 12 08:29:55 2012] [error] [client 46.120.43.163] File does not exist: /home/affilinn/public_html/favicon.ico



what is that?

Hey gang,

Is there any way to view the apache error.log file in PHP?

Hello,
I have php application, it is hosted on a shared server.
How can I view the errors in my application with the details of the error for my website users?

Thanks,

I have been scripting (some may say programming) in PHP for last 5 years but never faced this issue. Interesting!

Say for example, I created a custom function with one argument:

function myFunc($arg1) {
echo $arg1;
}

Now, if I call this function without supplying any argument it doesn't give any error/warning. Any other high level language like C++ or Java would produce error in this case. So, I had to modify my code as follows:


function myFunc($arg1) {
if (!isset($arg1)) {
echo "Missing argument 1";
}
echo $s;
}

But this is annoying. Now I have to add these EXTRA lines of code in every function and class methods I have created. Is there any shortcut/easy method that the script will always produce an error if its missing any required argument? I have tried with adding error_reporting(E_ALL); at the very beginning of my code. But it still didn't work out. By the way, I am using PHP 5.3.1 on Apache/2.2.14 (Unix).

Any suggestion will be appreciated.

im using 000webhost as a test site and when i run this code it redirects me to there err page but with no error message.
the sql query works fine in phpmyadmin and i added the rest of the code to try the php side.
i "think" the problem is the echo $rows"value's" as im unsure of what the $vars should be


<?php
include("config.php"); 

// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

$sql = "SELECT make, COUNT(*) AS total, SUM(IF(comments = \'pass\', 1, 0)) AS withComments FROM dsgi_serval GROUP BY make ORDER BY COUNT(*) DESC"; 
$result=mysql_query($sql);
echo "$sql";
echo "$result";

?>

<table><tr>
<td colspan="4"><strong>List data from mysql </strong> </td>
</tr>
<tr>
<td align="center"><strong>make</strong></td>
<td align="center"><strong>Total</strong></td>
<td align="center"><strong>Validated</strong></td>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td><?php echo $rows['make']; ?></td>
<td><?php echo $rows['total']; ?></td>
<td><?php echo $rows['withcomments']; ?></td>
</tr>


<?php
}
?>
</table>
<?php
mysql_close();
?>

Hello! I have a strange error on my PHP script and i dont how to fix it. If someone can help me, please help me then! Here is my error: Code: [Select]
logout(); } else { $iq = mysql_query("SELECT * FROM users WHERE username='{$signin_username}' AND password='{$signin_password}' AND suspended='0' LIMIT 1;"); $ir = mysql_fetch_array($iq); $_SESSION['me'] = $ir; } } } } else { die("The configuration did not recieve appropriate variables to accept your request."); } if ($set['next_clearup'] < time ()) { $next_clearup = time () + 60 * 60 * 24; mysql_query ('' . 'UPDATE settings SET set_value=\'' . $next_clearup . '\' WHERE set_name=\'next_clearup\' LIMIT 1;'); mysql_query ('UPDATE users SET ads_clicked=\'\' WHERE ads_clicked!=\'\''); } } ?>
Warning: include(THDIRindex.php) [function.include]: failed to open stream: No such file or directory in C:\xampp\htdocs\Upload\index.php on line 16

Warning: include() [function.include]: Failed opening 'THDIRindex.php' for inclusion (include_path='.;\xampp\php\PEAR') in C:\xampp\htdocs\Upload\index.php on line 16
And here is the PHP file the error is in:
Code: [Select]
<?php

session_start();
include_once('lib/lib.php');
include_once('lib/configuration.php');

$ddir = THDIR.$do->get_file_url();
include($ddir);

if(file_exists(HEADER)) { include_once(HEADER); }
if($contents) { print $contents; }
if(file_exists(FOOTER)) { include_once(FOOTER); }
?>
Help ASAP if you can!

I Am getting along with php better than I was previously. But this 68 year old brain still refuses to learn very fast!

Here is the error I'm receiving when I'm trying to open the db:

Parse error: syntax error, unexpected T_VARIABLE in /home/taft65/public_html/memProtest.php on line 197


<?php
                            error_reporting(E_ERROR | E_PARSE | E_CORE_ERROR);
                            $host = "localhost";
                            $dbname="database;"
       Failing ------>$username = "user";
                            $password="drDedf#hj";


I understand you do not need to declare varibles in PHP, Correct? I checked the db to ensure that I'm calling the correct value. NuSpherePhpEd to validate the code. I also check it with DSV PHP Editor. Both come up with the same error. I'm also using MyPhpAdmin to create the database and tables.

I know also to place this calling info in another folder and include it by calling it with a php include statement. I just have it within the code to quickly test it.

Thank you for any assistance.
Bob...

Hey all,

I keep getting this error:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 2

When I use the script below. I'm finding it a bit confusing because everything about it continues to work, it's just it gives me an error. When the script runs, the outcome is

"Success! Your dog now looks far more energetic!
Looks like this food is all used up."

Followed by the error, which cuts the remainder of the page off. Does anybody know why it's doing this?

$dogyay = $_POST['dogid'];

$checkenergy = "SELECT energy FROM dogs WHERE id=$dogyay";   
$energylevel = mysql_query($checkenergy) or die(mysql_error());
$row = mysql_fetch_array($energylevel) or die(mysql_error());

if($row['energy'] >= 100)
{ echo "<b>Oops!</b> Looks like your dog is full right now...";}

else{ 

echo "<b>Success! Your dog now looks far more energetic!</b><br><br>";

$sql11="UPDATE dogs SET energy=energy + 50 WHERE id=$dogyay";
$result11=mysql_query($sql11);

$sql12="UPDATE items SET uses = uses - 1 WHERE itemid=$id";
$result12=mysql_query($sql12);

$checkuses = "SELECT uses FROM items WHERE itemid=$id";   
$useslevel = mysql_query($checkuses) or die(mysql_error());
$row = mysql_fetch_array($useslevel) or die(mysql_error());

if($row['uses'] == 0)
{ echo "Looks like this food is all used up.<bR><br>";

mysql_query("DELETE FROM items WHERE itemid='$id'") 
or die(mysql_error());}


Thanks !

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '["rid"]., SELECT refid1 FROM oto_members WHERE id='13', SELECT refid2 FR' at line 5

With this script area:

$affiliate1=('.$_REQUEST["rid"].');
            $affiliate2=("SELECT refid1 FROM oto_members WHERE id='".$_REQUEST["rid"]."'");
            $affiliate3=("SELECT refid2 FROM oto_members WHERE id='".$_REQUEST["rid"]."'");
            $affiliate4=("SELECT refid3 FROM oto_members WHERE id='".$_REQUEST["rid"]."'");
            $affiliate5=("SELECT refid4 FROM oto_members WHERE id='".$_REQUEST["rid"]."'");
            $affiliate6=("SELECT refid5 FROM oto_members WHERE id='".$_REQUEST["rid"]."'");
            $affiliate7=("SELECT refid6 FROM oto_members WHERE id='".$_REQUEST["rid"]."'");
            $affiliat8e=("SELECT refid7 FROM oto_members WHERE id='".$_REQUEST["rid"]."'");
            $affiliate9=("SELECT refid8 FROM oto_members WHERE id='".$_REQUEST["rid"]."'");
            $affiliate10=("SELECT refid9 FROM oto_members WHERE id='".$_REQUEST["rid"]."'");
            
            
            $qry="INSERT INTO ".$prefix."members(firstname,lastname,email,address,city,state,postcode,country,telephone,username,password,refid1,refid2,refid3,refid4,refid5,refid6,refid7,refid8,refid9,refid10,geo,paypal_email,joindate,mtype,groupid,cb_id,status,signupip)

             VALUES('".$_REQUEST["firstname"]."','".$_REQUEST["lastname"]."','".$_REQUEST["email"]."','".$_REQUEST["address"]."','".$_REQUEST["city"]."','".$_REQUEST["state"]."','".$_REQUEST["postcode"]."','".$_REQUEST["country"]."','".$_REQUEST["telephone"]."','".$_REQUEST["username"]."','".md5($_REQUEST["password"])."',
            
            $affiliate1,
            $affiliate2,
            $affiliate3,
            $affiliate4,
            $affiliate5,
            $affiliate6,
            $affiliate7,
            $affiliate8,
            $affiliate9,
            $affiliate10
            
            ,'".$_REQUEST["geo"]."','".$_REQUEST["paypal_email"]."',NOW(),".$_REQUEST["mtid"].",$eogroup,'".$_REQUEST["cb_id"]."','$memberstatus','$signupip')";

I'm a newbie, can you please help?

I'm getting a mysql error "Unknown column 'E0000001' in 'where clause'"

The id is in the URL: ...user-profile.php?id=E0000001

My Query

$query = mysqli_query($con, "SELECT * FROM UserList WHERE UserID=".$id) or die (mysqli_error($con));

Thank you

Edited September 18, 2019 by ACBMSE

Hi there

I'm trying to insert audi files' records into mysql database this is how the records insertion looks like:

Code: [Select]
$fileName = $_FILES['uploaded']['name'];//name
$tmpName  = $_FILES['uploaded']['tmp_name'];//temp location
$fileSize = $_FILES['uploaded']['size'];//size of the file
$fileType = $_FILES['uploaded']['type'];//type of file
$error    = $_FILES['uploaded']['error'];//verifys errprs
$ext = substr($fileName, strrpos($fileName, '.') +1); //this will get the extention out of the file name e.g. .mp3



//check that a file is passed by and no errors
if(isset($fileName) && $error == 0 && $fileSize != 0){

//condition to accept only certain file types/extentions
if($ext == "mp3" || $ext == "wma" || $ext == "wav"){

//get file content
$fp      = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);

if(!get_magic_quotes_gpc())
{
    $fileName = addslashes($fileName);
   
}


        //query to retrieve user's id
        $userid = mysql_query("select userID from user where username = '$username'");
//to get id by username
$row = mysql_fetch_assoc($userid);
$userid = $row['userID'];

$query = "INSERT INTO tracks (trackName, userID, tag, price, file, fileName, fileSize, fileType)
    VALUES ('$tname','$userid','$tag','$price','$content','$fileName','$fileSize' , '$fileType)";
   
mysql_query($query) or die (mysql_error());

when i send it i get this error:  Unknown column 'application' in 'field list' i did set data type for the fileType field to varchar.  when i remove the fileType all together the record is inserted successfully  into database though. ?!! can anyone help please

I have searched this forum as well as over 200 other forums and have not found the answer that is specific to my question. I have shortened my code drastically to assist in resolving this quickly -   I have a search form that has criteria for the search criteria with "virtual" "columns" in an array but it's not working. If I search one column at a time it works just fine but when I try to search 8 columns with one select I get the following error: SELECT Error: Unknown column 'achievements' in 'where clause'.   When a user selects search in Achievements, I need it to look at all 8 columns that are associated with achievements and bring back the results that match - the same as if the user selects search in Associations, I need it to look at all 5 columns and bring back the results that match.   My shortened code is as follows:

        <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
    <html xmlns="http://www.w3.org/1999/xhtml">
    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>Search</title>
    </head>
    
    <body>
    
    <form name="search" action="" method="POST">
    
      <p>Search:</p>
      <p>
        
        Achievements/Associations: <input type="text" name="find1" /> in 
      <Select NAME="field1">
        <Option VALUE="achievements">Achievements</option>
        <Option VALUE="associations">Associations</option>
      </Select>
      <br><br>
      
      Secondary Education: <input type="text" name="find2" /> in 
      <Select NAME="field2">
        <Option VALUE="edu1sectype">Highest Certificate Attained</option>
        <Option VALUE="edu1secname">Highest Grade Passed</option>
        <Option VALUE="edu1secinst">Name of High School</option>
        <Option VALUE="edu1secdate">Date Completed</option>
        <Option VALUE="edu1secinsttyp">Type of Institution</option>
        <Option VALUE="subjects">Subjects</option>
      </Select>
      <br><br>
         
      <input type="hidden" name="searching" value="yes" />
      <input type="submit" name="search" value="Search" />
      </p>
    </form>
    
    <?php
    $searching = $_POST['searching'];
    
    $find1 = $_POST['find1'];
    $field1 = $_POST['field1'];
    
    $find2 = $_POST['find2'];
    $field2 = $_POST['field2'];
    
    if ($searching =="yes") 
     { 
     
     echo "<br><b>Searched For:</b>  $find1 $find2<br>"; 
     echo "<br><h2>Results</h2><p>"; 
     
     //If they did not enter a search term we give them an error 
     
     
     // Otherwise we connect to our Database 
    include_once "connect_to_mysql.php";
    mysql_select_db("table_name") or die(mysql_error());
     
     // We preform a bit of filtering 
    $find = strtoupper($find);
    $find = strip_tags($find);
    $find = trim($find);
    $find = mysql_real_escape_string($find);
    $field = mysql_real_escape_string($field);
    
    $data = mysql_query("SELECT * FROM table_name
    
    WHERE upper(".$field1.")  LIKE '%$find1%'
    AND upper(".$field2.")  LIKE '%$find2%'
    ") or die("SELECT Error: ".mysql_error());
    
    $result = mysql_query("SELECT * FROM table_name 
     WHERE upper($field1) LIKE '%$find1%'
    AND upper($field2) LIKE '%$find2%'
    ") or die("SELECT Error: ".mysql_error());
    $num_rows = mysql_num_rows($result);
    
    echo "There are $num_rows records:<br>";
     echo '<center>';
    										
    									    echo "<table border='1' cellpadding='5' width='990'>";
                                            
                                            // set table headers
                                            echo "<tr><th>Reference</th>
    										          <th>First Name</th>
    										          <th>Last Name</th>
    												  </tr>";
    
    //get images and names in two arrays
    
    $name= $row["name"];
    $surname= $row["surname"];
    $achieve1 = $row["achieve1"];
    $achieve2 = $row["achieve2"];
    $achieve3 = $row["achieve3"];
    $achieve4 = $row["achieve4"];
    $achieve5 = $row["achieve5"];
    $achieve6 = $row["achieve6"];
    $achieve7 = $row["achieve7"];
    $achieve8 = $row["achieve8"];
    $assoc1 = $row["assoc1"];
    $assoc2 = $row["assoc2"];
    $assoc3 = $row["assoc3"];
    $assoc4 = $row["assoc4"];
    $assoc5 = $row["assoc5"];
    $edu1sectype = $row["edu1sectype"];
    $edu1secinst = $row["edu1secinst"];
    $edu1secname = $row["edu1secname"];
    $edu1secdate = $row["edu1secdate"];
    $edu1secinsttyp = $row["edu1secinsttyp"];
    $subject1 = $row["subject1"];
    $subject2 = $row["subject2"];
    $subject3 = $row["subject3"];
    $subject4 = $row["subject4"];
    $subject5 = $row["subject5"];
    $subject6 = $row["subject6"];
    $subject7 = $row["subject7"];
    $subject8 = $row["subject8"];
    $compsoft1name = $row["compsoft1name"];
    $compsoft2name = $row["compsoft2name"];
    $compsoft3name = $row["compsoft3name"];
    $compsoft4name = $row["compsoft4name"];
    $compsoft5name = $row["compsoft5name"];
    $compsoft6name = $row["compsoft6name"];
    
    $achievements = array('achieve1', 'achieve2', 'achieve3', 'achieve4', 'achieve5', 'achieve6', 'achieve7', 'achieve8');
    
    $associations = array('assoc1', 'assoc2', 'assoc3', 'assoc4', 'assoc5');
    
    $subjects = array('subject1', 'subject2', 'subject3', 'subject4', 'subject5', 'subject6', 'subject7', 'subject8' );
    
    $compsoft = array('compsoft1name', 'compsoft2name', 'compsoft3name', 'compsoft4name', 'compsoft5name', 'compsoft6name');
    
    while ($row = mysql_fetch_array($result))
    {
    
     echo "<tr>";
    												echo "<td ALIGN=LEFT>" . $row['id'] . "</td>";
                                                    echo "<td ALIGN=LEFT>" . $row['name'] . "</td>";
                                                    echo "<td ALIGN=LEFT>" . $row['surname'] . "</td>";
    												echo "</tr>";
                                            }
                                            
                                            echo "</table>";
                                   
     
     
     //This counts the number or results - and if there wasn't any it gives them a little message explaining that 
     $anymatches=mysql_num_rows($data);
    if ($anymatches == 0) {
    echo "Sorry, but we can not find an entry to match your query";
    }
     }
     
    ?>
    
    </body>
    </html>

Any assistance will be greatly appreciated as I have been working on this website for the past 4 months which has totalled over 150 pages and this is one of the last pages left to program and it's taken 6 days to get to this search page to this point.

Just a heads up, my OOP skills are in development so sorry for the roughness of things...(any tips always appreciated!)

I'm basically making my own little repository site that will have snippets (for quick copy paste anywhere) (I know there are other services, but mine is a little different and needs to be private).

I'll try to just post the necessary code:

HTML trying to pass:
Code: [Select]
<p><br /><span class="something">Hey buddy</span></p>
Javascript/JQuery:
Code: [Select]
//send info
$.ajaxFileUpload({
url:'./upload_zip.php',
secureuri:false,
fileElementId:'form_zip',
dataType: 'json',
data: {...
'description': htmlEntities($('#modularContent form textarea[name="description"]').val())...},
success:function(data, status){
var html = ...
html+= ...'<div class="description"><p>'+data.description+'<br /><br />';...
$('#list').append(html);
...
$('#modularOverlay,#modularContent').remove();
},
error:function(data, status, e){
alert(e);
} 

...
function htmlEntities(str) {
    return String(str).replace(/&/g, '&amp;').replace(/</g, '&lt;').replace(/>/g, '&gt;').replace(/"/g, '&quot;');
}
//found via: http://css-tricks.com/snippets/javascript/htmlentities-for-javascript/

PHP:
Code: [Select]
<?php
//Start object and modify initial values
public function __construct($title,$description,$url,$snippet){
...
if($this->snippet == 'yes'){
$this->handle_snippet();
}
}
public function handle_snippet(){
$code = htmlentities($this->description, ENT_QUOTES | ENT_IGNORE, 'UTF-8',false);
$this->description = $code;
}
//returning value
public function getJSON($action){
//temp fix for js breaking if  remove_tag_ids is empty/null
if(empty($this->remove_tag_ids)){
$this->remove_tag_ids = 'null';
}
if($this->snippet == 'yes'){
$this->description = stripslashes($this->description);
}
switch($action){
case 'delete':
//not important
default:
$json = array(...'description'=>$this->description...);
break;
return $json;
}
?>

POST Value: (via firebug)
Code: [Select]
description: <p><br /><span class="something">Hey buddy</span></p>
JSON Return value: (via firebug)
Code: [Select]
description: &lt;p&gt;&lt;br /&gt;&lt;span class=&quot;something&quot;&gt;Hey buddy&lt;/span&gt;&lt;/p&gt;
DB value:
Code: [Select]
&lt;p&gt;&lt;br /&gt;&lt;span class=\&quot;something\&quot;&gt;Hey buddy&lt;/span&gt;&lt;/p&gt;
As far as error, I have no idea what the error is haha. It does not get through to the error code cause the plugin tosses an error b4 that (using an ajaxUpload plugin http://www.phpletter.com/Our-Projects/AjaxFileUpload/, (this does not affect the upload, it uploads just fine, but it tosses an error for some reason)).

Anyone have any insights/ideas?
Thanks for any and all help,
Justin

I don`t get it, waht is wrong?!
Code: [Select]
<?php
require_once 'auth.php';
if (!isset($_SESSION['SESS_VERIFY'])) {
header("location: access-denied.php");
exit();
}
if ($_SESSION['lang'] == 'Ro') {

// setare data romania
date_default_timezone_set('Europe/Bucharest');

$today = getdate();


$zi = $today['mday'];
$luna = $today['mon'];
$lunastring = $today['month'];
$an = $today['year'];
$data = $zi.$luna.$an;
$data = (string)$data;
$ora = date('H:i:s'); 

$msg = array();
$err = array();

$luni = array (
1=>'Ianuarie',
2=>'Februarie',
3=>'Martie',
4=>'Aprilie',
5=>'Mai',
6=>'Iunie',
7=>'Iulie',
8=>'August',
9=>'Septembrie',
10=>'Octobrie',
11=>'Noiembrie',
12=>'Decembrie');

// comun
const SQL_ERR = 'SQL statement failed with error: ';
const ADD_MODEL = 'ADAUGA UN MODEL NOU';
.
.many constants..
.
}
elseif ($_SESSION['lang'] == 'It') {...

Thank you!

Hello again,

I have some form data, which I then search through for particular code data like so:
$html2 = $_POST['fname'];
preg_match_all("/<bla>(.*)<\/bla>/", $html2, $matches40);

So the above searches for all the data between <bla>XXXXXX</bla> from $POST

Which I then print to my page using:
 (Only so I can see while developing)
print_r($matches40);

This displays HTML output like so:
Code: [Select]
Array ( [0] => Array ( [0] => Hello [1] => My [2] => Name [3] => Is [4] => Tom ) [1] => Array ( [0] => Hello [1] => My [2] => Name [3] => Is [4] => Tom ) )
What I am trying to do is again use the preg_match_all function to look through the array output and find data that I want to remove. E.g. If one of the variables from $matches40 is 'Tom' I want to find and replaces this with 'Ben'.

I spent a day searching Google but to not success. Any help?