PHP - Need Help Retrieving Addrees Bar Url

i have had a difficult time trying to work this out.I need to do some pattern matching for certain urls and retrieve information from it.
For example,
$url=http://www.test.com/showpic.php?do=showpic&u=89165&a=34933

if $url contains the value showpic.php,then i need to retrieve the following
{
  $u=value of u(i.e. 89165 in this case)
  $a=value of a(i.e. 34933 in this case)
}
else do nothin..

the format of the url will always be the same as above if it contains showpic.php

Hi there

I am trying to use cURL to retrieve XML from a gateway. The code I am using works (ie. returns the HTTP response) for any URL that I use except the one that I need!  The URL is below, and when using that I get a blank response every time. I've tested and can telnet to that IP address on that port from my server.

I'm not sure if it is because of the format/structure of the URL? Could it have something to do with the port of the gateway? Any assistance would be much appreciated.

http://196.11.120.190:8080/mtnusa/client.jsp?command=<usareq NODE="tHTTP" USERNAME="HTTP" PASSWORD="1234" TRANSFORM="SMPP"><command><submit_sm><a_number>278200703520709</a_number><b_number>27824411926</b_number><service_type/><message>Test</message><registered_delivery/></submit_sm></command></usareq>

The function I am using (from http://davidwalsh.name) is:

/* gets the data from a URL */
function get_data($url)
{
   $ch = curl_init();
   $timeout = 5;
   curl_setopt($ch,CURLOPT_URL,$url);
   curl_setopt($ch,CURLOPT_RETURNTRANSFER,1);
   curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,$timeout);
   $data = curl_exec($ch);
   curl_close($ch);
   return $data;
}

$returned_content = get_data('http://196.11.120.190:8080/mtnusa/client.jsp?command=<usareq NODE="HTTP" USERNAME="HTTP" PASSWORD="1234" TRANSFORM="SMPP"><command><submit_sm><a_number>278200703520709</a_number><b_number>27824411926</b_number><service_type/><message>Test</message><registered_delivery/></submit_sm></command></usareq>');

In my script, users login with their Username & Password. However, I'd like to be able to echo the email address used on their account. I've tried adding the email to the session
I'm not having much luck... Here's a piece of the login code(untouched);

                $username = $_POST['name'];
                $passwd = $_POST['passwd'];
                $query = "SELECT name,passwd FROM users WHERE CONCAT('0x', hex(passwd)) = '{$salt}'";
		$result = mysql_query($query);
		$login_ok = false;
		if(mysql_num_rows($result) > 0) {
		    $login_ok = true;
		}
		if($login_ok)
		{
		    $row = mysql_fetch_array($result, MYSQL_NUM);
		    $_SESSION['user'] = $row;

I've also tried messing around with this piece below in a few different ways but still nothing.

<?php echo htmlentities($_SESSION['user']['email'], ENT_QUOTES, 'UTF-8'); ?>

Any help is greatly appreciated..

I have a basic table where I am trying to retrieve records that are filtered by a form request. Here is the code I have...

$ps = $pdo->prepare("SELECT * FROM `Products` Where `Vendor` LIKE concat('%',?,'%');
$ps->execute(array($_POST['Vendor']));;
echo "post=" . $_POST['vendor']; ////Displays correct data from form request
$count = $ps->rowCount();
echo "Count=" . $count; ////Count = 0 although I know for a fact that there is 1 record that should be in there

////Used for display of records
foreach ($ps as $row){
echo $row, PHP_EOL . "xxx<br>";
}

Where is the incorrect code? I am new to php and pdo. Thank you in advance

K

i have this script called view.php
Code: [Select]
<?php
  include 'db.inc';

  $file = clean($file, 4);

  if (empty($file))
     exit;

  if (!($connection = @ mysql_pconnect($hostName,
                                       $username,
                                       $password)))
     showerror();

  if (!mysql_select_db("Php_test", $connection))
     showerror();

  $query = "SELECT mime, data FROM file 
            WHERE id = $file";

  if (!($result = @ mysql_query ($query,$connection)))
     showerror();  

  $data = @ mysql_fetch_array($result);

  if (!empty($data["data"]))
  {
    // Output the MIME header
     header("Content-Type: {$data["mime"]}");
    // Output the image
     echo $data["data"];
   }
?>after i have db.inc
Code: [Select]
<?php

// These are the DBMS credentials
$hostName = "localhost";
$username = "root";
$password = "oxioxi";

// Show an error and stop the script
function showerror()
{
   if (mysql_error())
      die("Error " . mysql_errno() . " : " . mysql_error());
   else
      die("Could not connect to the DBMS");
}

// Secure the user data by escaping characters 
// and shortening the input string
function clean($input, $maxlength)
{
  $input = substr($input, 0, $maxlength);
  $input = EscapeShellCmd($input);
  return ($input);
}

?>and i have my page witch part of it is  this
Code: [Select]
   <?php
  include 'db.inc';
  
  $query = "SELECT id, name, mime FROM file";

  if (!($connection = @ mysql_pconnect($hostName, 
                                    $username,
                                    $password)))
     showerror();

  if (!mysql_select_db("php_test", $connection))
     showerror();
        
  if (!($result = @ mysql_query ($query, $connection)))
     showerror();
?>
    <h1>Image database</h1>


<?php 


  if ($row = @ mysql_fetch_array($result))
  {
?>

    <table>
    <col span="1" align="right">
    <tr>
       <th>Short description</th>
       <th>File type</th>
       <th>Image</th>

    </tr>
<?php
   do 
   {
?>
    <tr>
       <td><?php echo "{$row["name"]}";?></td>         
       <td><?php echo "{$row["mime"]}";?></td>
       <td><?php echo "<img src=\"view.php?file={$row["id"]}\">";?></td>
    </tr>
<?php
   } while ($row = @ mysql_fetch_array($result));
?>
    </table>
<?php
  } // if mysql_fetch_array()
  else
     echo "<h3>There are no images to display</h3>\n";
?>i have do it right but for some reason it doesnt displays the pictures the type and the name are displayed normally but not the picture!! instead of them there are small icons of not existing pictures.. please i want help i really need to do it

ok im back quicker than i thought....

i got my drop box sorted and i got it reloading the page. so it all works correctly.

but how do i get the page to display information regarding the film i have selected in the drop box.? i have no code for this at the mo.

also i would like the drop box to display the selected item at top of box when it refreshes

code for drop box:

Code: [Select]
<FORM>
<?php $result = mysql_query( "SELECT * FROM movie_info ORDER BY title ASC ")
;
echo "<select name= Film onChange='submit()' >film name</option>";
while
($nt=mysql_fetch_array($result)){
?>
<?php
  echo "<option value='$nt[id]'>$nt[title] </option>";
}
?>
</select>
</FORM>



any help would be great

I have a simple query and need to print the result.
My code does not print a resulr but the query is valid and returns a result.
Any help appreciated.
Code: [Select]
$query = ("SELECT * FROM wp_book WHERE Author_ID = '$Author_ID'");

            $result = mysql_query($query) or trigger_error("Query: $query\n<br />MySQL Error: " . mysql_error());
            echo mysql_error();
            if (!$result) {
                echo "<br />";
                echo"No Result";
            }
            echo"Result";
           

            while($row = mysql_fetch_assoc($result)){
           
            echo $row['WorkingTitle'];

            }

I've searched all over and have found some scripts but none of them work anymore.

Is there a way to fetch Google images using PHP? There's no Google Image API (only the regular Google API)

Hello, I have a few questions.  First, this is what I am trying to accomplish - I am trying to take values that are entered into forms.  And then store them into a database.  My plan was to take them into an array, and create a loop that wrote the values to the database.  So, the first step I thought would be to learn how to take info from a form, then display it onto the screen.  Theoretically, if I could do that, I could just learn the MySQL commands to write to the database, and paste them instead of printing to the screen.  So, I gave it a go.  It didn't work.  The way the 'Key' is chosen with arrays in PHP was different than I thought.  I gave my sample code to a friend and asked for his help.  He gave me a working copy back, but signed off before I could ask questions 
This is what I am working with:
Code: [Select]
<html>
<body>
<form name = "barInfo" method = "post">

Establishment name: <input  type = "text" name = "EstablishmentName">
<br>
Street Address: <input type = "text" name = "StreetAddress">
<br>
City: <input type = "text" name = "City">
<br>
State: <input  type = "text" name = "State">
<br>
Zip: <input type = "text" name = "Zip">
<br>
<input  type="submit" name="Submit"  value="Submit">
</form>

<?php 

if (!empty($_POST['EstablishmentName']))
{
<br>
print "Now, lets see if this shit works...";
<br>
print "Establishment Name:" . $_POST['EstablishmentName'];
<br>
print "Street Address:" . $_POST['StreetAddress'];
<br>
print "City:" . $_POST['City'];
<br>
print "State:" . $_POST['State'];
<br>
print "Zip:" . $_POST['Zip'];
<br>
}

?>
</body>
</html>
Ok, first off I get a parse error or something like that when I try it.  My book has '<br>' thrown all over the place inside '<?php>' things.  How can I do an endl; type thing in PHP?  Obv <br> does not work.,
Code: [Select]
print "\n";Does nothing, either....?
Second, the reason my code didn't display anything, but his did, was because I had a loop that displayed $details[$x], and 1 was added to $x every time it looped.  Am I right in saying that during the html part of the code, whatever value is assigned to 'name = ', is the key pointing to that value in my array?  If that is true, how would I make that become an integer and use it with a loop.  Do I have to put:
Code: [Select]
name = "1", name = "2", etc...?  Thanks in advance for the help!  Sorry if I am not referring to terms correctly, I just started looking at HTML/PHP last week...

Code: [Select]
$query = mysql_query("SELECT a.*, b.*
         FROM friendlist a
INNER JOIN friendlist b
ON (a.friendemail=b.friendemail)
INNER JOIN users c
ON (b.friendemail = c.EmailAddress)
WHERE a.email = 'asdf@gmail.com'
AND c.Username LIKE '%carol%'
GROUP BY a.id
ORDER BY count(*) DESC");
Code: [Select]
while ($showfriends = mysql_fetch_array($query)) {
     echo $showfriends['Username'];
}
and I would get nothing. It produces the correct number of <div> so i know it's getting through, but it's having trouble displaying the entries?

Hello,

I am having a problem with retrieving information from my database. I am kinda new in programming industry, and with the error message displayed, I can't pinpoint what went wrong in the code. Here is the PHP code:

<?php
include("dbconnection.php");

if(isset($_POST["btnSubmit"]))
{
$id = $_POST["id"];
$territory = $_POST["territory"];
$job_title = $_POST["job_title"];
$area_of_work = $_POST["area_of_work"];
$employer = $_POST["employer"];
$location = $_POST["location"];
$job_title = $_POST["job_title"];
$date_posted = $_POST["date_posted"];
$closing_date = $_POST["closing_date"];
$department = $_POST["department"];
$gender = $_POST["gender"];
$first_name = $_POST["first_name"];
$last_name = $_POST["last_name"];
$title = $_POST["title"];
$telephone = $_POST["telephone"];
$address_1 = $_POST["address_1"];
$address_2 = $_POST["address_2"];
$address_3 = $_POST["address_3"];
$country = $_POST["country"];
$city = $_POST["city"];
$postal_code = $_POST["postal_code"];
$website = $_POST["website"];
$email_address = $_POST["email_address"];
$cg_comment = $_POST["cg_comment"];
$date_emailed = $_POST["date_emailed"];
$mailing_comments = $_POST["mailing_comments"];
$telesales_comments = $_POST["telesales_comments"];

$query = "UPDATE contacts SET territory= '".$territory."', job_title = '".$job_title."', area_of_work = '".$area_of_work."', employer = '".$employer."', location = '".$location."', job_title = '".$job_title."', employer = '".$employer."', date_posted = '".$date_posted."', department = '".$department."', closing_date = '".$closing_date."', gender = '".$gender."', first_name = '".$first_name."', last_name = '".$last_name."', title = '".$title."', telephone = '".$telephone."', address_1 = '".$address_1."', address_2 = '".$address_2."', address_3 = '".$address_3."', country = '".$country."', city = '".$city."', postal_code = '".$postal_code."', website = '".$website."', email_address = '".$email_address."', cg_comment = '".$cg_comment."', date_emailed = '".$date_emailed."', mailing_comments = '".$mailing_comments."', telesales_comments = '".$telesales_comments."' WHERE id = '".$id."'";
mysql_query($query) or die(mysql_error());

echo "<script>
alert('You have successfully updated a record');
window.location = 'view.php';
</script>";
}

$query = "SELECT * FROM contacts WHERE id = '".$_GET["id"]."'";
$result = mysql_query($query, $connection);

if(mysql_num_rows($result) > 0)
{
$territory = mysql_result($result,0, "territory");
$job_title = mysql_result($result, 0, "job_title");
$area_of_work = mysql_result($result, 0, "area_of_work");
$employer = mysql_result($result, 0, "employer");
$status = mysql_result($result, 0, "status");
$location = mysql_result($result,0, "location");
$department = mysql_result($result,0, "department");
$date_posted = mysql_result($result,0, "date_posted");
$closing_date = mysql_result($result,0, "closing_date");
$gender = mysql_result($result,0, "gender");
$first_name = mysql_result($result,0, "first_name");
$last_name = mysql_result($result,0, "last_name");
$title = mysql_result($result,0, "title");
$telephone = mysql_result($result,0, "telephone");
$address_1 = mysql_result($result,0, "address_1");
$address_2 = mysql_result($result,0, "address_2");
$address_3 = mysql_result($result,0, "address_3");
$city = mysql_result($result,0, "city");
$country = mysql_result($result,0, "country");
$postal_code = mysql_result($result,0, "postal_code");
$website = mysql_result($result,0, "website");
$email_address = mysql_result($result,0, "email_address");
$cg_comment = mysql_result($result,0, "cg_comment");
$date_emailed = mysql_result($result,0, "date_emailed");
$mailing_comments = mysql_result($result,0, "mailing_comments");
$telesales_comments = mysql_result($result,0, "telesales_comments");
}


?>

The error I am getting is:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\invent-asia\edit_client.php on line 268

Line 268 in the code is:

Quote

if(mysql_num_rows($result) > 0)
{ - This one is line 68
   $territory = mysql_result($result,0, "territory");


Immediate response is well appreciated. Thank you very much!

Hi, i have a transaction table which includes following fields.
 
    tran_ID,account_number,account_type,transaction_type,amount,Date

I want to know how these data can be retrieved and display in another table. The table is filled with several records.

Thanks,
Heshan

Hi All, small question here. I was just playing around with a small script I made to obtain an ip-address, remove the dots and than glue the parts together. Not for any use but for practice. But I am having difficulties to accessing the array values. But maybe that because its the first time i am using explode() and print_r. If someone has a spare minute I would be pleased to be enlightened.

<?php
             $ip=$_SERVER['REMOTE_ADDR'];
             echo $ip.'<br />';
             $cleaned_ip = substr($ip,0,6);
             echo $cleaned_ip.'<br />';
             $cleaned_ip2 = print_r(explode('.', $ip, 6)).'<br />';
             // this is the point where I dont know how to get the array values.
             // echo $cleaned_ip2[0].[1].[2];   that doesnt work. ;(
         ?>

Hello all,

I'm currently working on a project and it is the first time I have had to store/retrieve an array with PHP and MySQL. Basically its a website that has a list of discounts/coupons on it. People can register and submit their discounts and such. I am storing an array for who is eligible for the discount, as well as which categories the discount falls under.

I did some googling and there seems to be tons of thoughts on the best way to do this. Some say to use implode/explode, some say to serialize/unserialize, I was just wondering if there is a common, concrete way to do this?

I will need to be able to search the array for its contents. So if children are eligible for the discount, I need to be able to store that in a database, retrieve it later, and search it for only "Children" so I can display the discounts available only to children.

If someone could please advise the best way to do this and post some examples, it would be greatly appreciated. I learn best with examples that are explained properly.

I would like to display weather conditions on my website and store them in mysql database.
I wonder if it is possible to load web page into php for parsing so the required info could be found and used?
Is it doable with php and what functions could you recommend for this functionality?
How can I have program running in the background on the server which would be triggered every so often to perform this task?
Another question is if I want to trigger some action once a week like Tuesday at noon, is there function in php which could be used to check what day of the week and time it is ?
I am new to website design so any help will be greatly appreciated.