PHP - Instant Change When Select Selectmenu

I'm currently developing an e-commerce website and I've been asked to integrate a live chat tool into the website.  I was just wondering if anyone here has done something similar and would reccomend a certain path go down?

Thanks for any help. 

Hi all,

Experimenting with this instant search buzz only I cant seem to replicate the function!
I think my PHP may be cocking up but Ive been staring for too long to see anything.

Could anybody take a quick glance and see if they can see anything?

Code: [Select]
<?php
if(!empty($_GET['q'])) {
    search();
}

function search() {
    $con = mysql_connect('localhost','login', 'pass');
    mysql_select_db('db', $con);

    $q = htmlspecialchars($_GET['q'],ENT_QUOTES);

    $sql = mysql_query("
        SELECT
            post_title as post
        FROM wp_posts
        WHERE post_title LIKE '%{$q}%' OR post_title LIKE '%{$q}%'
        ");

    
    $results=array();
    while($v = mysql_fetch_object($sql)){
        $results[] = array(
          'title'=>$v->title,
          'post'=>$v->post
        );
    }

  
    echo json_encode($results);
}
?>

Trying to make it so people need a active session in order to access the page after the log in page and if they dont then it redirects them back to the log in page.
My session works fine. I tested and made sure.
it saves the user_id lets me display the page

but how do I keep someone from simply going to the webpage with out loging in?
Just a simple if statment checking if lastactive is empty or not? is that secure?

Code: [Select]
<?php include_once("connect.php");


if(isset($_SESSION['user_id'])) {
 // Login OK, update last active
 
 $sql = "UPDATE users SET lastactive=NOW() WHERE id='".mysql_real_escape_string($_SESSION['user_id'])."'";
 mysql_query($sql);
 
}else{
   header("Location: index.php");
     exit();
 }
?>

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=343375.0

I am working with an API provided here http://ipndoc.paydotcom.com/ to get IPN call back for payment or refund

With the code he
 

$input = file_get_contents('php://input');
$json = @json_decode($input, true);

$secret = 'SECRETCODE';
$secret = substr(sha1($secret), 0, 32);

$notification = base64_decode($json['notification']);

$iv = base64_decode($json['iv']);

$decoded = openssl_decrypt(
$notification,
'AES-256-CBC',
$secret,
OPENSSL_RAW_DATA,
$iv
);

$data = json_decode($decoded, true);

$trxntype = $data['transactionInfo']['transactionType'];
$paymethod = $data['transactionInfo']['paymentService'];
$f_name = $data['customerInfo']['contactInfo']['firstName'];
$l_name = $data['customerInfo']['contactInfo']['lastName'];

The $f_name and $l_name are correct!
But now, the $ trxntype and $paymethod returns NULL instead of the payment details.

How can I fix this?

hirealimo.com.au/code1.php

this works as i want it:
Quote

SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1



but apparelty selecting * is not a good thing????

but if I do this:
Quote

SELECT priceID, price FROM price INNER JOIN vehicle....etc


it works but i lose the info from the vehicle table.

but how do i make this work:
Quote

SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc


so that i am specifiying which colums from which tables to query??

thanks

I have 2 queries that I want to join together to make one row  

Dear All,

I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box.

All is working fine except that the locality select box is not being populated.
I know that the problem is in the sql statement

WHERE country_id='$co'

because i am having an error that $co is an undefined variable.

All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement?

Thank you for your help in advance.

MySQL Tables indicated below:

CREATE TABLE countries(
country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT,
country_name VARCHAR(30) NOT NULL,
PRIMARY KEY(country_id),
UNIQUE KEY(country_name),
INDEX(country_id),
INDEX(country_name))
ENGINE=MyISAM;

CREATE TABLE localities(
locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
country_id INT(3) UNSIGNED NOT NULL,
locality_name VARCHAR(50),
PRIMARY KEY (locality_id),
INDEX (country_id),
INDEX (locality_name))
ENGINE=MyISAM;

Extract PHP script included below:

// connect to database
require_once(MYSQL);

if(isset($_POST['submitted']))

{

// trim the incoming data

/* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */
$trimmed=array_map('trim',$_POST);

// clean the data

$co=mysqli_real_escape_string($dbc,$trimmed['country']);

$lc=mysqli_real_escape_string($dbc,$trimmed['locality']);

}

?>

<form action="form.php" method="post">

<p>Country
<select name="country">
<option>Select Country</option>

<?php

$q="SELECT country_id, country_name
FROM countries";

$r=mysqli_query($dbc,$q) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($r))

{

$country_id=$row[0];
$country_name=$row[1];
echo '<option value="' . $country_id . '"';
if(isset($trimmed['country']) && ($trimmed['country']==$country_id))
echo 'selected="selected"';
echo '>' . $country_name . '</option>\n';

}

?>

</select>
</p>

<p>Locality
<select name="locality">
<option>Select Locality</option>

<?php

$ql="SELECT locality_id, country_id, locality_name
FROM localities
WHERE country_id='$co'
ORDER BY locality_name";

$rl=mysqli_query($dbc,$ql) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($rl))

{

$locality_id=$row[0];
$country_id=$row[1];
$locality_name=$row[2];
echo '<option value="' . $locality_id . '"';
if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id))
echo 'selected="selected"';
echo '>' . $locality_name . '</option>\n';

}

// close database connection
mysqli_close($dbc);

?>

</select>
</p>

<p><input type="submit" name="submit" value="Submit" /></p>

<input type="hidden" name="submitted" value="TRUE" />

</form>

Hi,

I'm not quite sure how to do this, so i thought i'd ask you guys from some assistance.

Basically i am inserting Author's into a table successfully using an array. The reason for this is that i have multiple authors being added and there is no limit as to how many. An example of what i mean can be seen he

http://www.prima.cse.salford.ac.uk:8080/~ibrarhussain/test.html

You can click on "Add author" to add however many necessary..

Anyway, i have got the insert working, however when i edit i want to be able to see all the authors that have been added but obviously i don't know how many there are..

Typically i would like to see something like this:

http://www.prima.cse.salford.ac.uk:8080/~ibrarhussain/edit.jpg

So i would click on an edit link and it would pre populate the text boxes. I don't have a problem with doing this, but how can i show the correct amount of input textboxes based on how many authors exist for that specific record?

Can someone offer some advice please?

The input elements are like so:

Quote

<input type="text" name="author[]" id="author1"/>
<input type="text" name="author[]" id="author2"/>
<input type="text" name="author[]" id="author3"/>
...
...
...
<input type="text" name="author[]" id="author10"/>



Some records may have 1 author some may have 10, so how can i do this?

Thanks again..

I'm using a PHP helpdesk script, but when a customer submits a ticket, they are able to assign a category. I want to be able to have the category function so I can move tickets around, but only allow customers to assign to one. (Basically I want to hide the Select Box on the new ticket page)

This is the code which displays the select box:
Quote

<td style="text-align:right" width="150"><?php echo $hesklang['category']; ?>: <font class="important">*</font></td>
<td width="80%"><select name="category">


Can you tell me what I can do to "hide" the category select box without getting rid of it? (The page requires a variable for category)

Thanks

Hi guys,

I'm having a issue with a little bit of coding involving SELECT the problem is I want it to select from the database and say limit it to 5 so basicly it prints 5 results from the database but its only printing 1 result

Here is the code i'm using

Code: [Select]
$q=mysql_query("SELECT * FROM papercontent LIMIT 5",$c);
$content=mysql_result($q,0,0);

print "
<center><table width = '40%' border = '1'><tr><th>
<center><u>Latest Announcements</u></center>

$content
</tr></td></table>
";

Thanks

I'm trying to run a readout of a db which runs fine as an individual script. When I embed it in PHP inside an html div container, it bails when it encounters the first ">" and simply outputs the PHP characters from there through the "?>". The rest of the html runs fine before and after. For example, this line

echo "<select>";

would output

";

and any other php script up to the ?> end, after which it renders html fine. If I run the script as a separate php file, it runs as expected. Any help would be appreciated. Thanks.

I am trying to get an id from a table where clientid is equal to a session var. I know the session is set because if I echo it, it shows up, and I know there is a row in the db that matches because I am looking at it right now.

The problem is, the query is not selecting any data. I need to select the clientid from the table so I can set another session var.

Please please help my brain is not coping! Thank you!

<?
session_start();
$uid = $_SESSION['logid'];

include('../connectdb.php');
mysql_select_db("maindb", $con);

$result = mysql_query("SELECT * FROM projects WHERE clientid = $uid");

while($row = mysql_fetch_array($result))
  {
echo $row['id'];
  }

?>

I'm down to just echoing the row to problem solve but nothing comes up.