I have this code, provided by someone else:

<?php

$request = new HttpRequest();
$request->setUrl('https://api.ltl.xpo.com/rating/1.0/ratequotes');
$request->setMethod(HTTP_METH_POST);

$request->setQueryData(array(
  'format' => 'XML'
));

$request->setHeaders(array(
  'cache-control' => 'no-cache',
  'content-type' => 'application/xml',
  'authorization' => 'Bearer 5ce5e6e3304fd4e069b1c00f1c1f9bab'
));

$request->setBody('<?xml version="1.0" encoding="UTF-8"?>
<GetXpoLtlRateQuoteRqst xmlns="http://www.ltl.xpo.com/API/LTLRatingAPI">
<shipmentInfo>
    <paymentTermCd>P</paymentTermCd>
    <accessorials>
        <accessorialCd>DNC</accessorialCd>
        <accessorialCd>GUR</accessorialCd>
    </accessorials>
    <commodity>
        <grossWeight>
            <weight>500</weight>
            <weightUom>lbs</weightUom>
        </grossWeight>
        <nmfcClass>100</nmfcClass>
        <hazmatInd>false</hazmatInd>
    </commodity>
    <shipmentDate>2018-11-19</shipmentDate>
    <shipper>
        <address>
	<acctInstId>117258878</acctInstId>
            <postalCd>93727</postalCd>
        </address>
    </shipper>
    <consignee>
        <address>
            <postalCd>92027</postalCd>
        </address>
    </consignee>
</shipmentInfo>
</GetXpoLtlRateQuoteRqst>
');

try {
  $response = $request->send();

  echo $response->getBody();
} catch (HttpException $ex) {
  echo $ex;
}

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It fails with: Fatal error: Class 'HttpRequest' not found in /home/lakoshva/public_html/freight_test.php on line 3

Doesn't there have to be something like "require once xxxxxxxxxxxxxclass.php"?

Thanks

I am not sure if this join is going to be possible but I am asking here in the hope that it will be.

I have an order_detail table:

order_id
product_id
product_qty
product_price

So, some data might look like

1      25      2      299
2      26      1      199

And a coupon table (I have left out the other fields that aren't necessary for this question)

voucher_id
product_id
rep
purchased
order_id

And some data:

#12675      25      14      2018-11-05      1
#12676      25      14      2018-11-05      1
#12677      26      12      2018-11-07      2

I want to show a report of each voucher/coupon sale without duplicating data. My current query ends up showing the same coupon number twice.


 

      $this->db->query("SELECT `serial`, `rep`, `purchased` FROM `vouchers` AS `v`
        INNER JOIN `order_detail` AS `od` ON od.`order_id` = v.`order_id`
        WHERE cast(`purchased` as date) BETWEEN :first_day AND :last_day
        ");
        
        $this->db->bind(":first_day", $first_day);
        $this->db->bind(":last_day", $last_day);
        $results = $this->db->arraySet();
        return $results;

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Anyone know why the input "ScreenWidth" has a value of nothing instead of 1920?  I have tried " as well as '.

I have a form/page that I'm using to book in stock based on a poNumber that generates the stock items linked to that poNumber/supplier. Currently for each line (there may be 10 items for example on order) the user needs to add qty rec'd, qty outstanding, receipted by and delivery note number in the form for each line.

I want to short cut some of the data entry by having a single entry for receipted by and delivery note number but for this to populate the rows for each entry once submitted.

The code I have at the moment is as follows: -

<div class="container">	
	<form class="form" method="post" action="transit_temp.php">	
	
	<?php
	
	$poNumber = isset($_POST['poNumber']) ? $_POST['poNumber'] : false;
	
	$poNumber = $mysqli->real_escape_string($poNumber);
	
	$index=0;

	$query = 'SELECT * FROM goodsInTrans WHERE poNumber = "'.$poNumber.'"';
			
				if (!$result = $mysqli->query($query)) {
				$err
				= "QUERY FAIL: "
				. $query
				. ' ERRNO: '
				. $mysqli->errno
				. ' ERROR: '
				. $mysqli->error
				;
				trigger_error($err, E_USER_ERROR);
				}		
	
				while ($row = $result->fetch_object())
				
				{ 

				?>
					<div class="form-group row">
						<div class=""><input type="hidden" 				class="form-control" 	name="data[<?php echo $index; ?>][supplier]" 			value="<?php echo $row->supplier ?>"/></div>
						<div class="col-xs-3"><input type="text" 		class="form-control" 	name="data[<?php echo $index; ?>][sku]"	 				value="<?php echo $row->sku 

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…

I need to search for an order in my WooCommerce back office by email address.  I can see the email address in wp_users but don't know how to put a name to the email address.  Is there a plugin or mysql code that will accomplish that?

Thanks,

  <?php if ( ! isset($_GET["company"]) ): ?>

            <p>No Company Info has been passed in!</p>

        <?php else: ?>

            <?php
            // Let's run the query
            $query = $conn->prepare("SELECT frm_id, frm_companyname, frm_contact, frm_email, frm_rep, frm_date, frm_return, frm_timefrom, frm_timeto, frm_comment1, frm_comment2, frm_comment3, frm_comment4, frm_comment5, frm_todo1, frm_todo2,  frm_todo3,  frm_todo4,  frm_todo5, frm_updated  FROM siteform WHERE frm_companyname = ?");
            $query->bind_param("s", $_GET['company']);
            $query->execute();
            $results = $query->get_result();
            ?>
     
            <?php while ($row = $results->fetch_assoc()):$style = "";
  //if ($row['frm_id'] < "5"){
    //$style2 = "style='background:#FF0000;'";
  //} 
                  
  //if ($row['frm_id'] > "5"){
    //$style2 = "style='background:#FFF000;'";
  //} 
         //if ($row['frm_id'] =="5"){
    //$style2 = "style='background:#ddd000;'";
  //} 
       $data_db = $row['frm_date'];
       $current_date = date('Y-m-d', strtotime(" Today"));
        $current_date2 = date('Y-m-d', strtotime(" -1 month"));
        $current_date3 = date('Y-m-d', strtotime(" -2 month"));
        $current_date4 = date('Y-m-d', strtotime(" +1 day"));
        $current_date5 = date('Y-m-d', strtotime(" +1 month"));
         $current_date6 = date('Y-m-d', strtotime(" +2 month"));
    
if ($data_db == $current_date){
    $style = 

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…

Hello,
I'm working on a code that can record vocal messages and store them into a database to retrieve them later on, but my PHP script isnt working and I don't have enough expertise in web dev to correct it, can someone help me please ?
index.html
upload.php
app.js

This is a php script, which basically does the send mail function via terminal shell, I am studying it and the curious thing is that it worked in my tests only in the Ubuntu 14.04 versions with php5 and postfix. But I could not get it to work on Ubuntu 18.04 or Debian 9. I would like to ask for help to change the code in order to make it work in those versions. Thank you.

javali.php

I have some code which stores session detail in a database, but when I issue the command 'session_regenerate_id();' I get

PHP Catchable fatal error:  session_regenerate_id(): Failed to create(read) session ID

<?php
    //Session
    class pdoSession {
        //DB Object
        private $db;

        public function __construct($pdoDB){
            // Instantiate new Database object
            $this->db = $pdoDB;

            // Set handler to overide SESSION
            session_set_save_handler(
                array($this, "_open"),
                array($this, "_close"),
                array($this, "_read"),
                array($this, "_write"),
                array($this, "_destroy"),
                array($this, "_gc")
            );

            // Start the session
            session_start();
        }

        //Open
        public function _open(){
            // If successful
            if($this->db){
                return true;
            }
            // Return False
            return false;
        }

        //Close
        public function _close(){
            // Close Session
            return true;
        }

        //Read
        public function _read($id){
            $stSession = $this->db->prepare("SELECT `data` FROM `sessions` WHERE `SID` = :id");
            $stSession->bindValue(':id', $id, PDO::PARAM_STR);

            if($stSession->execute()){
                $row = $stSession->fetch(PDO::FETCH_ASSOC);
          

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…

Hello all,

I have a line in my code which checks to see if the email value entered in a submitted form already exists in the database. If so,  a value of 0 or 1 is assigned to a string and the string value is displayed.

$select = mysqli_query($db_connection, "SELECT `email` FROM `teachers_table` WHERE `email` = '".$_POST['email']."'") or exit(mysqli_error($db_connection));
	if(mysqli_num_rows($select)) {
    $email_exists = '1';
	}  
	else{
		$email_exists = '0';
	}
		echo $email_exists;

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This part seems to work as intended - at least the value of 0 or 1 is displayed correctly depending on whether the email exists in the database or not. The next part of my code calls another function, (which in turn calls another validation function) which displays errors.

The validation worked but I added to it:

	if($email_exists == '1') {
      $errors[] = "EMAIL Exists IN OUR DATABASE";
    }

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However, when I do this, I get an  Undefined variable: email_exists in... notice / error.

I don't understand why this is coming up as not defined if the function is being called after I have defined this value.  The error message is displayed even though I can see the value of $email_exists (i.e. the 1 or 0) output in the screen.

I've tried removing the quotes, changing to single quotes and using TRUE and FALSE but this didn't appear to change anything.

 When I use '=' instead of '==' i.e:

if($email_exists = '1') {
      $errors[] = "EMAIL Exists IN OUR DATABASE";
    }

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...the not defined error goes away but the logic doesn't work, in that $email_exists seems to be always 1 and the error EMAIL Exists IN OUR DATABASE is always displayed.

Any help, as always, much appreciated.

Many thanks,

Adam,

Hi everyone,

I am learning and I want to make search with more filters. I have one table where I have three columns with foreign keys which are related to other tables. So, I have 4 tables. First table containt user details, second table contain category, third table contain institution, and fourth table containt city.

First table look like on picture below named (users)



Second table (named categories)



Third table (named institutions)



Fourth table (named city)



I have this HTML form:

<form action="izvjesce.php" method="GET">
										<div class="form-group col-lg-6">
										<select  class="form-control" name="status" required>
											<option  value="Aktivan" checked><b>Aktivan</b></option>
											<option  value="Neaktivan"><b>Neaktivan</b></option>
										</select>
										</div>
										<div class="form-group col-lg-6">
										<select  class="form-control" name="spol">
											<option  value=""><b>Spol</b></option>
											<option  value="Muški"><b>Muški</b></option>
											<option  value="Ženski"><b>Ženski</b></option>
										</select>
										</div>
										<div class="form-group col-lg-6">
										<select  class="form-control" name="pos_mj" >
											<option  value="">Izaberi mjesec</option>
											<option  value="siječanj">Januar/Siječanj</option>
											<option  

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…

my view is not passing data to the controller. and i am new to codeigniter and i am confused..anyone please help

this is my view

<?php require_once(APPPATH.'views/web/header.php');?>
<?php

$ses=$this->session->userdata('webuser');
/*print_r($ses);
die;*/


?>
<style>
.select2-container {
width : auto !important;
}
.select2
{
width : 100% !important;
}
.select2-container--bootstrap
{
width : auto !important;
}
.numcolor{
color:black;
}
.backcolor{
background:#32c5d2!important;
}
.activecolor{
font-weight:bold;
}
.nav-pills>li.active>a, .nav-pills>li.active>a:focus, .nav-pills>li.active>a:hover{
	background:grey!important;
}

<?php if(strtolower($this->uri->segment(3))=="parentuser"){?>
.one{
background:#32c5d2!important;

}	
.onenumcolor{
color:black;
font-weight:bold;
}

.twonumcolor,.threenumcolor,.fournumcolor{
color:#32c5d2!important;
}
.two>a:hover{
background:black!important;

}
.three>a:hover{
background:black!important;

}
.four>a:hover{
background:black!important;


}
<?php }elseif(strtolower($this->uri->segment(3))=="student"){?>
.one,.two{
background:#32c5d2!important;
color:white;
}
.twonumcolor{
color:black;
font-weight:bold;
}
.onenumcolor{
color:white!important;

}
.threenumcolor,.fournumcolor{
color:#32c5d2!important;
}
.one>a:hover{
background:black!important;

}
.three>a:hover{
background:black!important;

}
.four>a:hover{
background:black!important;


}
<?php	}?>
				

				
</style>

<div class="page-content-wrapper">
	<!-- BEGIN CONTENT BODY -->

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…

I'm fairly novice with PHP, but I can usually figure things out. But I'm stuck on working around this legacy code that we use on our sales form.
 I have to shoehorn in another Card processing merchant that does things a little differently with their API.

This is processing the response from a cURL API where I receive a SOAP XML response. I'm having some issues with the actual response as well, so I have it sampled in what appears to be a proper xml format, then attempting to parse out the message objects.

The process:
1. I have the response XML converting into a "SimpleXMLElement" with the simplexml_load_string
2. Then the XML is "converted to an array", I think, with the $array = (array)$xml;
3. Then, the code should be able to look into the array values and process a Success or Failed based on the Status Value
4. Then the $data is json encoded and is passed to the JavaScript file for further processing of the notification and filling in the Auth code field of the sales form.

The Problem(s):
1. You'll see in the PHP code that I have 2 testing sections to process the response. The first uses the $array and the second skips the array and just uses the $xml. One is commented out for testing the other, of course.
2. When using the $xml section to get the values, I get real close to what I need until it get json encoded, then the output just says Object Object in the JS Alert. It appears that the code is finding the proper value of the key, but it's not …