PHP - Get Id From Jobs And Insert It To Bookings

hello im having trouble with this code im not sure if you can help me but ima paste the code here and i will see what people say, the problem is this:

PHP Fatal error: require(): Failed opening required '../class/class_db_mysql.php' (include_path='.:') in /var/www/vhosts/dclxvi.co.uk/httpdocs/crons/cron_day.php on line 6

my code is this:

<?php
   include_once ('../Global/config.php');
global $_CONFIG;
if($_GET['code'] != $_CONFIG['code']) { die(""); }
define("MONO_ON", 1);
require "../class/class_db_{$_CONFIG['driver']}.php";
$db=new database;
$db->configure($_CONFIG['hostname'],
 $_CONFIG['username'],
 $_CONFIG['password'],
 $_CONFIG['database'],
 $_CONFIG['persistent']);
$db->connect();
$c=$db->connection_id;
$db->query("UPDATE fedjail set fed_days=fed_days-1");
$q=$db->query("SELECT * FROM fedjail WHERE fed_days=0");
$ids=array();
while($r=$db->fetch_row($q))
{
$ids[]=$r['fed_userid'];
}
if(count($ids) > 0)
{
$db->query("UPDATE users SET fedjail=0 WHERE userid IN(".implode(",", $ids).")");
}
$db->query("DELETE FROM fedjail WHERE fed_days=0");
$db->query("UPDATE users SET daysingang=daysingang+1 WHERE gang > 0");
$db->query("UPDATE users SET daysold=daysold+1, boxes_opened=0");
$db->query("UPDATE users SET mailban=mailban-1 WHERE mailban > 0");
$db->query("UPDATE users SET donatordays=donatordays-1 WHERE donatordays > 0");
$db->query("UPDATE users SET cdays=cdays-1 WHERE course > 0");
$db->query("UPDATE users SET bankmoney=bankmoney+(bankmoney/50) where bankmoney>0");
$db->query("UPDATE users SET cybermoney=cybermoney+(cybermoney/100*7) where cybermoney>0");
$q=$db->query("SELECT * FROM users WHERE cdays=0 AND course > 0");
while($r=$db->fetch_row($q))
{
$cd=$db->query("SELECT * FROM courses WHERE crID={$r['course']}");
$coud=$db->fetch_row($cd);
$userid=$r['userid'];
$db->query("INSERT INTO coursesdone VALUES({$r['userid']},{$r['course']})");
$upd="";
$ev="";
if($coud['crSTR'] > 0)
{
$upd.=",us.strength=us.strength+{$coud['crSTR']}";
$ev.=", {$coud['crSTR']} strength";
}
if($coud['crGUARD'] > 0)
{
$upd.=",us.guard=us.guard+{$coud['crGUARD']}";
$ev.=", {$coud['crGUARD']} guard";
}
if($coud['crLABOUR'] > 0)
{
$upd.=",us.labour=us.labour+{$coud['crLABOUR']}";
$ev.=", {$coud['crLABOUR']} labour";
}
if($coud['crAGIL'] > 0)
{
$upd.=",us.agility=us.agility+{$coud['crAGIL']}";
$ev.=", {$coud['crAGIL']} agility";
}
if($coud['crIQ'] > 0)
{
$upd.=",us.IQ=us.IQ+{$coud['crIQ']}";
$ev.=", {$coud['crIQ']} IQ";
}
$ev=substr($ev,1);
if ($upd) {
$db->query("UPDATE users u LEFT JOIN userstats us ON u.userid=us.userid SET us.userid=us.userid $upd WHERE u.userid=$userid");
}
$db->query("INSERT INTO events VALUES('',$userid,unix_timestamp(),0,'Congratulations, you completed the {$coud['crNAME']} and gained $ev!')");
}
$db->query("UPDATE users SET course=0 WHERE cdays=0");
$db->query("TRUNCATE TABLE `votes`");
?>

Hello there..
my webhost doesn't have Cron Jobs feature..
is there any alternative php script in able to run my php script every minute ??
i have a website containing mafia game.. it's coded in php.. 
i tried to use


Code: [Select]
<META HTTP-EQUIV="Refresh" CONTENT="60; URL=http://test.com/run_minute.php">but this will work only if the user is in the existing page where this code placed..


i want it something like, the http://test.com/run_minute.php will refresh every minute.. even thou the user is not on the page..


Thank you.
More power..

This topic has been moved to Linux.

http://www.phpfreaks.com/forums/index.php?topic=306518.0

This topic has been moved to Miscellaneous.

http://www.phpfreaks.com/forums/index.php?topic=333230.0

Basically I am setting up my own cricket management game, but there are 3 areas that I am slightly stuck on and would appreciate help:

1) When a use registers, I need to create 11 players for their team. For this, I am going to have one text files with first names and one text file with surnames. How can I open each file and select a name and then match them together?

2) I need to create the list of games to be executed at particular times through a cron job. How can I go through all the teams in a table and match them together in games, ensuring everyone plays everyone? If a new player registers, he will be put into a random league and so the games for that league will have to be redone

3) How can I set up a cron job to execute the above games? Each team should play once a day and so if there are 24 teams in a table, there will be 12 games that occur through this cron job. I assume cron jobs are the best thing for this.

Any help would be appreciated.
Thanks.

Can anyone tell me why this is not INSERTing? My array data is coming out just fine.. I've tried everything I can think of and cannot get anything to insert.. Ahhhh!

<?php
  $query = "SELECT RegionID, City FROM geo_cities WHERE RegionID='135'";
  $results = mysqli_query($cxn, $query);
  $row_cnt = mysqli_num_rows($results);
  echo $row_cnt . " Total Records in Query.<br /><br />";
  if (mysqli_num_rows($results)) {
      while ($row = mysqli_fetch_array($results)) {
          $insert_city_query = "INSERT INTO all_illinois SET state_id=$row[RegionID], city_name=$row[City] WHERE id = null" or mysqli_error();
          $insert = mysqli_query($cxn, $insert_city_query);
          if (!$insert) {
              echo "INSERT is NOT working!";
              exit();
          }
          
          echo $row['City'] . "<br />";
          
          echo "<pre>";
          echo print_r($row);
          echo "</pre>";
      }
      //while ($rows = mysqli_fetch_array($results))
      
      } //if (mysqli_num_rows($results))
      
      else {
          echo "No results to get!";
      }
?>

Here is my all_illinois INSERT table structu

CREATE TABLE IF NOT EXISTS `all_illinois` (
  `state_id` varchar(255) NOT NULL,
  `city_name` varchar(255) NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

Here is my source table geo_cities structu

CREATE TABLE IF NOT EXISTS `1` (
  `CityId` varchar(255) NOT NULL,
  `CountryID` varchar(255) NOT NULL,
  `RegionID` varchar(255) NOT NULL,
  `City` varchar(255) NOT NULL,
  `Latitude` varchar(255) NOT NULL,
  `Longitude` varchar(255) NOT NULL,
  `TimeZone` varchar(255) NOT NULL,
  `DmaId` varchar(255) NOT NULL,
  `Code` varchar(255) NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

I'm missing something here. I have a form, and when the submit is pressed, the relevant post data inserts into table one, then I want the last insert id to insert along with other form data into a second table. The first table's still inserting fine, but I can't get that second one to do anything. It leapfrogs over the query and doesn't give an error. EDIT: I forgot to add an error: I get: 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 'usage, why VALUES ('14', '', '123', '','1234', '', '')' at line 1

query:INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('test 14', 'asdfa', 'asdf', 'adf','asdf', '', '', '', '123', '', '')


Code: [Select]
if (empty($errors)) {

require_once ('dbconnectionfile.php');

$query = "INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company)
 VALUES ('$description12', '$sn', '$description4', '$cne','$description5', '$description6', '$description7', '$description8',
'$description9', '$description10', '$description11')";

$result = @mysql_query ($query);


if ($result) {

$who_donated=mysql_insert_id();

$query2 = "INSERT INTO tbl_donation (donor_id, donor_expyear, donor_cvv, donor_cardtype, donor_authorization, amount, usage, why)
 VALUES ('$who_donated', '$donate2', '$donate3', '$donate4','$donate5', '$donate6', '$donate7')";
$result2 = @mysql_query ($query2);

if ($result2) {echo "Info was added to both tables! yay!";}

echo "table one filled. Table two was not.";
echo $who_donated;

//header ("Location: http://www.twigzy.com/add_plant.php?var1=$plant_id");

exit();
} else {
echo 'system error. No donation added';

Hello, I'm having a bit of a problem here, all help to this issues would be much appreciated

I am trying to use text boxes to insert numbers into the database based on what is inputed.
If I have a string, like this for example:
$variable = 09385493;

And I want to insert it into the database like this:
mysql_query("INSERT INTO integers(number)
VALUES ('$variable')");

When checking the integers table in my database, looking at the number field, the $variable that was inserted is outputted as
9385493

Notice the number zero was taken out of the front of the number. If the number is double 0's (009385493), both of those zero's would disappear, too.

Thanks

i need the insert code

I am wondering about the following problem:
I have two sessions, one for the user ($_SESSION['user_id']) and second one for the products inside cart ($_SESSION['cart']). I need to insert data into table racun.
I did it this way and it works, but i want to know if this can be done better?
Code: [Select]
while ($row=mysql_fetch_array($query)){
$id = $_SESSION['korisnik_id'];
$idp =  $row['product_id'];
$kol = $_SESSION['cart'][$row['product_id']]['quantity'] ."<br>";
$insert = "INSERT INTO racun (product_id, quantity, korisnik_id) VALUES ('$idp', '$kol', '$id') ";
$result = mysql_query($insert);
  }

hey developers,   i have problem to get last insert id. btw.. i try to put that code on my mvc creation.   on register_model page i write like this:  public function numbering(){
        $sth = $this->db->prepare('SELECT * FROM member_data');
        $result = $this->db->lastInsertId();
        return $result;
    }
  and for the controllers page, i write the code like this:   $c = $this->model->numbering();
         foreach ($c as $key => $value) {
$data = $result('id');             
}          
        
$abc['member_id'] = $data;     but i didn't get the results,.. can you help me?

i am trying to update a column in the database. I can update int columns but what about text? i want to update it so that the new text is added to it while keeping the old one. Here is what i have(which isnt working):

mysql_query("UPDATE ".DB_PREFIX."posts
            SET post_content = '$old_content' + '".$_POST['test']."'
            WHERE post_id = '$double_post_id'") or die(mysql_error());

i have also tried:

SET post_content = post_content + '".$_POST['test']."' 

but that didnt work either. What should i be doing?

Just a quick question I have a form, couple of drop down boxes, input fields etc.. at the end is the submit button. now my question is should my insert into database sqls go before the submit button inside the form?
out side the form? after the button inside the form?

If by any chance you could give me some advice on the best way to layout my sql that would be great. They are big and go into three tables, at the moment I have them as three seperate queries but I think they must be able to go into one. can someone please help me with the layout and syntax.

here is what the queries look like at the moment:


if ($IBselect=$_POST ['IBselect']);
{
//CUSTOMER
$enterCust="INSERT INTO customer(username, password, title, firstName, lastName, address, town, country, postCode, phone, email, dateCust) 
                        VALUES ('$_POST[username]', '$_POST[password]', '$_POST[title]', '$_POST[firstName]', '$_POST[lastName]', '$_POST[address]', '$_POST[town]', '$_POST[country]', '$_POST[postCode]', '$_POST[phone]', '$_POST[email]', 'DATE: Auto CURDATE()', CURDATE()";
$enterCust_query=mysql_query($enterCust)or die(mysql_error());

//CARD 
$enterCard="INSERT INTO card(cardNumber,  name, expDate, cardID) 
                        VALUES ('$_POST[cardNumber]','$_POST[name]', ' $_POST[expDate]', '$_POST[cardID]')";
$enterCard_query=mysql_query($enterCard);

//BOOKING
$RCenterBook=mysql_query("INSERT INTO booking (custNumber, cardID, rideName, seatNo1, seatNo2, price, price2, dateBook, ID_time_tbl) 
                VALUES ('$custNumber', '$_POST[cardID]', '$rideName', '$_SESSION[IBextract]', '$_SESSION[IBextract2]', '$IBendPrice1', '$IBendPrice2', 'DATE: Auto CURDATE()', CURDATE()',  '$_SESSION[IB_slot_Time]'");
$IBenterBooking_query=mysql_query($IBenterBook); 
        if (!mysql_query($enterCard, $enterCust, $IBenterBook))
        {
        die("Error:" .mysql_error());
        }
        echo "1 record added IB";
}


Thanks =)

Not being able to upload images in the directory with a path name in the database?

1) This is my code for insert into the database and directory:

<?php


$host="localhost";
$username="root";
$pass="";
$db="registration";

$conn=mysqli_connect($host,$username,$pass,$db);
if(!$conn){
	die("Database connection error");
}

// insert query for register page
if(isset($_POST['ronel'])){
     $images = $_FILES['file']['name'];
  $target_dir = "uploads/";
  $target_file = $target_dir . basename($_FILES["file"]["name"]);

    // Select file type
  $imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));

  // Valid file extensions
  $extensions_arr = array("jpg","jpeg","png","gif","pdf");

  // Check extension
  if( in_array($imageFileType,$extensions_arr) )
  {

$details=$_POST['details'];
$location=$_POST['location'];
$checkbox=$_POST['checkbox'];
$injured=$_POST['injured'];    
$agegender=$_POST['agegender'];   
$contact=$_POST['contact'];
$empid=$_POST['empid'];
$dept=$_POST['dept'];    
$organization=$_POST['organization'];     
$summary=$_POST['summary']; 
$name=$_POST['name'];
$outcome=$_POST['outcome'];    
$cause=$_POST['cause'];     
$action=$_POST['action'];
$reportedname=$_POST['reportedname'];
$position=$_POST['position'];    
$organisation=$_POST['organisation'];
$reportedcontact=$_POST['reportedcontact'];
$reporteddept=$_POST['reporteddept'];    
$status="Pending";
$comment=$_POST['comment'];
	
	
	 $query="INSERT INTO `proposals` (`details`,`location`,`date`,`time`,`checkbox`,`injured`,`agegender`,`contact`,`empid`,`dept`  ,`organization`,`summary`,`image`,`outcome`,`cause`,`action`,`reportedname`,`position`,`organisation`,`reportedcontact`,`reporteddept`,`status`,`comment`) VALUES ('$details','$location', current_timestamp(),current_timestamp(),'$checkbox','$injured','$agegender','$contact','$empid','$dept' ,'$organization','$summary','$name','$outcome','$cause','$action','$reportedname','$position','$organisation','$reportedcontact','$reporteddept','$status','$comment')";
	
	$res=mysqli_query($conn,$query);
	if($res){
		$_SESSION['success']="Not Inserted successfully!";
		header('Location:');
	}else{
		echo "<script>alert('Proposal not applied!');</script>";
	}
      // Upload file
     move_uploaded_file($_FILES['file']['tmp_name'],$target_dir.$image);
	}
}
date_default_timezone_set("Asia/Kolkata");
?>

2) Here is the input file:

<form class="form-horizontal" method="post" action="" enctype="multipart/form-data">
            <input type="hidden" name="ronel" value="">
  <div class="form-group">
  <label style="position:absolute; left:63%; top:425px;" for="inputEmail" class="col-lg-3"><b>Upload Images Here :</b></label><br><br>
  <div class="col-lg-9">
  <input style="position:absolute; left:78%; top:420px;" type="file" name="file" enctype="multipart/form-data" class="form-control" name="incident_reference" onchange="document.getElementById('inc_ref').src = window.URL.createObjectURL(this.files[0]); document.getElementById('inc_ref').className +='_active'; document.getElementById('inc_ref_span').className += '_hidden'">
  </div><iframe id="inc_ref" class="form-group" width="220px" height="130px" style="position:absolute; left:78%; top:32%;"></iframe></div>
</form>

3) error code:

Notice: Undefined index: name in /opt/lampp/htdocs/create-nearmiss.php on line 57
ONGC TRIPUR

 

Edited March 28, 2020 by Ronel
change of var

Okay,
I'm hoping one of you can help me.
I have a mysql database that I have configured through phpmyadmin.
I have an android app that simply makes and sends a mysql query
I can get it to successfully return values when using Select statements but when I use INSERT INTO, it returns " Error Query is invalid" BUT BUT BUT, when I use the same string and enter it through the sql tab in myphpadmin it works fine !

So here is the string ( the semicolons at the end of each field name are so I can use something common to split the string up when the data arrives back on the phone)
Code: [Select]
randomkey||||||INSERT INTO table4  (`geolat;` , `geolong;` , `mode;` , `destgeolat;` , `destgeolong;` , `cellphone;` , `email;` , `carrego;` , `colour;` ,  `rating;` , `comment;`)
VALUES (0.0,0.0,'driver' ,-43.54779,172.62472, , '' ,'' , 'text' , 'ratingleftblank' , 'commentblank' )
the index4.php script is as follows
Code: [Select]
?php
/*
 * Written By:
 * James
 */


/************************************CONFIG****************************************/
//DATABSE DETAILS//
$DB_ADDRESS="mysql1.openhost.net.nz";
$DB_USER="bling44";
$DB_PASS="sadlyinept";
$DB_NAME="bling44";

//SETTINGS//
//This code is something you set in the APP so random people cant use it.
$SQLKEY="randomkey";
/************************************CONFIG****************************************/

//these are just in case setting headers forcing it to always expire and the content type to JSON
header('Cache-Control: no-cache, must-revalidate');
header('Content-type: application/json');


if(isset($_POST['tag'])){ //checks ifthe tag post is there
$tag=$_POST['tag'];
$data=explode("||||||",$tag); //split the SQL statement from the SQLKEY
if($data[0]==$SQLKEY){ ///validate the SQL key
$query=$data[1];
$link = mysql_connect($DB_ADDRESS,$DB_USER,$DB_PASS); //connect ot the MYSQL database
mysql_select_db($DB_NAME,$link); //connect to the right DB
if($link){
$result=mysql_query($query); //runs the posted query (NO PROTECTION FROM INJECTION HERE)
if($result){
if (strlen(stristr($query,"SELECT"))>0) { //tests if its a select statemnet
$outputdata=array();
while ($row = mysql_fetch_assoc($result)){
$outputdata[]=$row; //formats the result set to a valid array
}
echo json_encode(array("VALUE",$tag,array_merge($outputdata))); //sends out a JSON result with merged output data
} else {
echo json_encode(array("VALUE",$tag,array_merge(array(array("AFFECTED_ROWS ".mysql_affected_rows($link)))))); //if the query is anything but a SELECT it will return the array event count
}
} else echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR QUERY IS INVALID"))))); //errors if the query is bad
mysql_close($link); //close the DB
} else echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR Database Connection Failed"))))); //reports a DB connection failure

} else {
echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR BAD CODE SUPPLIED"))))); //reports if the code is bad
}
}
?>
So to reiterate.
I can search the DB but can't INSERT INTO, unless I go through the myphpadmin interface.
Any ideas are very much appreciated