PHP - Only Display Current And Up And Coming Event

Hi,

I'm wanting to find rows whose date is within the next week of the current month of the current year. The format of the date is, for example: 2010-10-28

Any ideas guys?

Thanks lots!

This topic has been moved to PHP Applications.

http://www.phpfreaks.com/forums/index.php?topic=330163.0

Hi i have this drop down list current the year is 2010 and downwards but i want to change the list to 2010 upwards u can notice on the 50-- so shows current year minus so current is 2010 to 61 how can i change 2010 to 2030 or sunfin??

echo '<select name="year_of_birth">',"\n";
    $year = date("Y");
    for ($i = $year;$i > $year-50;$i--) {
      if($i == $thisYear) { 
        $s = ' selected'; 
      } else { 
        $s='';
      }
      echo '<option value="' ,$i, '"',$s,'>' ,$i, '</option>',"\n";
    }
  echo '</select>',"\n";

I have a listener which executes a HTTP request to a remote API before the User entity is persisted and uses the response to set one of the entity's properties.  It will also listen for update and remove and will make the appropriate HTTP request to the API but will not modify the entity.  All works as desired... Almost.  If when persisting the entity, I have some error, the remote API and my application become out of sync.  I wish to change my application to perform a second call to the API if an error occurs and reverse the previous call.  My thoughts on how to implement a

Place a try/catch block when executing the query.  Don't like this approach. Add an ExceptionListener which somehow retrieves the entity and makes the applicable changes.  Maybe part of the solution, but too complicated to be the full solution. When adding, updating, or deleting a user from the remote API under UserListener's three methods, adding a callback which gets executed upon a PDOException.  I think this is the best approach and expanded my thoughts below.

 

<?php

namespace App\EventListener;

use Doctrine\Persistence\Event\LifecycleEventArgs;
use App\Service\HelpDeskClient;
use App\Entity\AbstractUser;

final class UserListner
{
    private $helpDeskClient;

    public function __construct(HelpDeskClient $helpDeskClient)
    {
        $this->helpDeskClient = $helpDeskClient;
    }

    public function prePersist(AbstractUser $user, LifecycleEventArgs $event)
    {
        $this->helpDeskClient->addUser($user);   //$user will be updated with the HTTP response
    }

    public function preUpdate(AbstractUser $user, LifecycleEventArgs $event)
    {
        $this->helpDeskClient->updateUser($user);
    }

    public function preRemove (AbstractUser $user, LifecycleEventArgs $event)
    {
        $this->helpDeskClient->deleteUser($user);
    }
}

Okay, how do I actually do this?  Was thinking of modifying UserListner as follows:

//...
use Symfony\Component\HttpKernel\KernelEvents;

final class UserListner
{
    // ...
    
    public function prePersist(AbstractUser $user, LifecycleEventArgs $event)
    {
        $this->helpDeskClient->addUser($user);
        
        $event->getObjectManager()->getEventManager()->addEventListener(KernelEvents::EXCEPTION, function($something) use($user) {
            // Use $this->helpDeskClient to reverse the changes
        });
    }
    
    // Similar for update and remove
}

 

But when trying this approach, I get a PDOException, but my callback never gets excecated.  I've also tried replacing KernelEvents::EXCEPTION with '\PDOException' (note the quotes) with no success.

Any ideas what I should be doing differently?

Maybe some totally different approach?  I suppose I could make the request to the API after the DB query is complete for updating and deleting, but not for adding.

Hey,

I'm using a script which allows you to click on a calendar to select the date to submit to the database.

The date is submitted like this: 2014-02-08

Is there a really simple way to prevent rows showing if the date is in the past?

Something like this:
if($currentdate < 2014-02-08 || $currentdate == 2014-02-08) {

}

Thanks very much,
Jack

My error:  Code: [Select]
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 'key FROM applications WHERE key = '0188f'' at line 1
Code:


function randomNum()
{
$random = substr(md5(rand(5000,300000)), -5);
$query = mysql_query("SELECT key FROM applications WHERE key = $random") or die(mysql_error());

//make sure it doesn't already exist
if(mysql_num_rows($query) > 0)
{
randomNum();
}
else
{
return $random;
}
}


I simply cannot seem to figure out what is happening...I've tried many solutions, but all seem to fail.

Okay, so, I'm trying to pull the info from my `shop` table where the `users` table matches at x, y, z, and plane.

Instead, I'm getting this:

Resource id #22     

Where is this coming from? I only have 1 row in my shop table, and it has:
id = 7
name = admin shop
x = 1
y = 0
z = admin
plane = batai

I'm standing in the shop right now (as my character, so my users table has the exact same x, y, z, plane)...

Any help would be appreciated!


$q1 = mysql_query("SELECT * FROM `shop` WHERE `x`='".$x["x"]."' AND `y`='".$x["y"]."' AND `z`='".$x["z"]."' AND `plane`='".$x["plane"]."'")  or die("qry failed: ".mysql_error());


Variables:
$x = $x = $player->username; (so it's pulling my username, which it is doing correctly ==> i echoed out $x and it came up with my username)

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
</head>

<?php // Connect to database==================================================

include("connect_db.php");

// sending query ===========================================================

$result = mysql_query("SELECT * FROM members Order By last") or die(mysql_error());  

if (!$result) {
    die("Query to show fields from table failed");
}

echo "<br />";
echo "<p>MEMBERS LIST</p>";

echo "<table border='10' cellpadding='3' cellspacing='2'>";
echo "<tr><th>First Name</th><th>Last Name</th>><th>Phone</th><th>Email</th></tr>";

// keeps getting the next row until there are no more to get ================

while($row = mysql_fetch_array( $result )) {

// Print out the contents of each row into a table ==========================

echo "<tr><td>";

echo $row['first'];

echo "</td><td>";

echo $row['last'];

echo "</td><td>";

echo $row['phone'];

echo "</td><td>";

echo $row['email'];

echo "</td></tr>"; 

} 

echo "</table>";

?>
</body>
</html>


This is the output.


MEMBERS LIST


>
  First         Last        Phone        Email
  data          data        data         data
ect...............................................................


I cannot find where the > (greater than) sign is coming from.

There is a space after MEMBERS LIST but I could not get it to work in this display.

Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\xampp\htdocs\page\verify_security.php on line 170     earlier the code has been running fine and now it is showing error to me , can you tell something more about it ???

Hi,

I just started learning PHP... Below is my first programme infact..

Out put is not coming when i type some thing in input boxes...

Can somebody suggest where is the error...

Index.php

<html>
<head>
<title>Welcome to our site</title>
</head>
<body>

<form method="POST" action="submit.php">
<label for ="business_name">Business name</label>
<input type="text" id="business_name" \><br/>
<label for ="contact_person">Contact person</label>
<input type="text" id="contact_person" \><br/>
<label for ="Telephone">Telephone</label>
<input type="text" id="Telephone" \><br/>
<label for ="FAX">FAX</label>
<input type="text" id="FAX" \><br/>
<label for ="Mobile">Mobile</label>
<input type="text" id="Mobile" \><br/>
<label for ="Addres_line_1">Addres_line_1</label>
<input type="text" id="Addres_line_1" \><br/>
<label for ="Addres_line_2">Addres_line_2</label>
<input type="text" id="Addres_line_2" \><br/>   
<label for ="Addres_line_3">Addres_line_3</label>
<input type="text" id="Addres_line_3" \><br/>
<label for ="Addres_line_4">Addres_line_4</label>
<input type="text" id="Addres_line_4" \><br/>
<label for ="City">City</label>
<input type="text" id="City" \><br/>
<label for ="Pin_code">Pin_code</label>
<input type="text" id="Pin_code" \><br/>
<label for ="State">State</label>
<input type="text" id="State" \><br/>
<label for ="email">email</label>
<input type="text" id="email" \><br/>
<input type="submit" value="Please submit the details" name="submit"/>
</form>


</body>
</html>   

submit.php



<head><title>Your submission details</title></head>
<body>
<H3>YOU HAVE SUBMITTED THE FOLLOWING DETAILS</H3>
<?php

$business_name= $_POST["business_name"];
$contact_person = $_POST["contact_person"];
$Telephone = $_POST["Telephone"];
$fax = $_POST["FAX"];
$mobile = $_POST["Mobile"];
$address1 = $_POST["Addres_line_1"];
$address2 = $_POST["Addres_line_2"];
$address3 = $_POST["Addres_line_3"];
$address4 = $_POST["Addres_line_4"];
$city = $_POST["City"];
$pincode = $_POST["Pin_code"];
$state = $_POST["State"];
$email = $_POST["email"];

echo "Your business name".$business_name.'<br/>';
echo "Contact person".$contact_person.'<br/>';
echo "Telephone no.".$Telephone.'<br/>';
echo "Fax No.".$fax.'<br/>';
echo "Mobile No.".$mobile.'<br/>';
echo "Address line one".$address1.'<br/>';
echo "Address line two".$address2.'<br/>';
echo "Address line three".$address3.'<br/>';
echo "Address line four".$address4.'<br/>';
echo "Your City".$city.'<br/>';
echo "Your Pincode".$pincode.'<br/>';
echo "Your State".$state.'<br/>';
echo "Your email".$email.'<br/>';
?>
<a href=index.php>Home</a>
</body>
</html>

Hi everyone reading this,

I have used PHP before as a hobby programmer but recently my boss asked me to make a webpage to accept worker's hours and store them in a database. So I realize that in the last few years object oriented style has been added into PHP. I have learnt OOP (C++) about a decade ago and understand the fundamentals of it.

If I am writing this website code, am I going to be better off spending the time to get familiar with the new OOP stuff in PHP and use that or would it be better to just use an old sequential style of programming? I know this may seem like apples and oranges and that it varies based on opinion, but maybe there can be reasons for me to use OOP over sequential.

For instance, if I write this code to be flexible enough that I can do a basic "worker times" table first, then I might be able to add additional functionality later, like worker details and qualifications, all of which could be brought up with the click of a button. Will it generally be easier to add onto code using OOP?

Hello all,

I am trying to create a PHP CLi Script that will help me setup new config and server files / directories for every new web project i get instead of me having to manually create everything with each project. The script accepts the project name, and y/n for whether i want to create subdomains for css, jss and image files. After it has that, it must create a new http config file under sites-available (in Linux), by pasting the config file text.

Problem : If I output regular text to the file, it displays exactly in the format that I write the string variable. But when I use a printf, or try to use variables directly in the config text, it starts formatting the text in the config file in a strange way. I just want the variables to just be replaced and the text to appear in the exact format below. The text displays perfectly as i typed below in the config file if I don't use any variables.


$subDomainConfig .=
'<VirtualHost *:80>
ServerAdmin webmaster@%1$s
ServerName  %1$s.%2$s
ServerAlias %1$s.%2$s

# Indexes + Directory Root.
DirectoryIndex index.html index.php
DocumentRoot /var/www/%2$s/public_html/

# CGI Directory
ScriptAlias /cgi-bin/ /usr/lib/cgi-bin/
<Location /cgi-bin>
Options +ExecCGI
</Location>


# Logfiles
ErrorLog  /var/www/%2$s/logs/apache/error.log
CustomLog /var/www/%2$s/logs/apache/access.log combined
</VirtualHost>
'.PHP_EOL;
$httpConfigFileText = sprintf($subDomainConfig, $subDomain, $shortProjectName);


The above code outputs text in this way in the file :

<VirtualHost *:80>
   ServerAdmin webmaster@images
   ServerName  images.mysite

   ServerAlias images.mysite


   # Indexes + Directory Root.
   DirectoryIndex index.html index.php
   DocumentRoot /var/www/mysite
/public_html/

   # CGI Directory
   ScriptAlias /cgi-bin/ /usr/lib/cgi-bin/
   <Location /cgi-bin>
      Options +ExecCGI
   </Location>


   # Logfiles
   ErrorLog  /var/www/mysite
/logs/apache/error.log
   CustomLog /var/www/mysite
/logs/apache/access.log combined
</VirtualHost>   

if any body copy and paste anything from word to editor some unwanted css also coming with that pasting. when we request the data same css also coming with that. so how to clean data  when we request
$desc =$_request['contents'];


how to solve this issue. please help me.

hello,
I have a search that matches users ISBN with two databases....code is below


$query_search_exact_match = mysql_query("SELECT 
   nvc_site.title, 
   nvc_site.id,
   nvc_site.description, 
   nvc_site.search_text, 
   nvc_site.image, 
   nvc_site.date, 
   nvc_site.price,
   nvc_site.location_city,
   nvc_site_ads_extra.name,
   nvc_site_ads_extra.value,
   nvc_site_ads_extra.classified_id
   
FROM   nvc_site,nvc_site_ads_extra
WHERE  name = 'ISBN%3A' AND live=1") or die(mysql_error());



then i take the ISBN that the user entered and match that with the isbn's in the DB


while ($fetch_extra = mysql_fetch_array($query_search_exact_match))
{

$value = ereg_replace( "[^0-9]", "", $fetch_extra['value'] );
$to_find_isbn = mysql_real_escape_string(ereg_replace( "[^0-9]", "",$_POST['szs']));
if($value == $to_find_isbn)
{
        echo "Match Found";
        } 
        else
        {
         echo "No Match Found";
         }
      //ELSE doesn't work here.....it displays both the if and else at the same time
}

i need it to display no match found if there was no match found.....I am soo lost right now!!

please help


thank you

Hi I have a form that is popluated from the database now I am trying to get the switch case to be based on the option the user selects is this possible to make a case a variable popluated by the drop down if so can someone show me what am i missing?
the value I am trying to get from the form to the switch case is the ( $val->slug) but the value not been set help please

here the form
Code: [Select]
                                       <form method="post">
                                           <select name="deal-locations" onChange="this.form.submit()">
                                             <?php
                                              while(list($key,$val) = each($locations)) {
                                             ?><option value="<?php echo $val->slug; ?>"><?php echo $val->slug; ?></option><?php
                                                 }
                                              ?>
                                           </select>
                                           <input type="submit" value="submit">
                                       </form>
Here the switch case
Code: [Select]
switch ($_POST['deal-locations'])
   {
       case " $val->slug": echo "<h3>deal-locations: ASC</h3>";
           $args = array(
                 's' => $_GET['s'],
                 'post_type' => 'deals',
                 'deal-locations' => ' $val->slug',
                 'where' => ' $val->slug',
                 'paged' => 'paged'
                );
           break;
}