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Icebergness

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Posts posted by Icebergness

    Problem Feeding Sql Data Into Jquery Ui Dialog Box

    in Javascript Help

    Posted

    And the PHP file that the script is calling, just in case it matters :)

     

    <!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Transactions</title>
</head>
<body>
<h1>Transactions</h1>

<div id="content">

<?php

$client=$_GET["client"];
$stock=$_GET['stock'];

$mssql_server= "sql-primary";
$mssql_database = "FOUR_I_CORE";

include("../index_files/mssql_include.php");

$sql_count = "SELECT * FROM [TRA_CORE] WHERE [CLIENT REC NO] = '$client' AND [sTOCK REC NO] = '$stock' AND [QUANTITY] != 0";
$stmt_count = sqlsrv_query($connnection, $sql_count);

while($count = sqlsrv_fetch_array($stmt_count, SQLSRV_FETCH_ASSOC))

{

$quantity = $count['QUANTITY'];
$date = $count['EVENT DATE'];
$cost = $count['COST/PROCEEDS'];
$currency = $count['ORIGINAL CURRENCY'];

echo $cost . '<br>';

}

?>

</div>

</body>
</html>

    Problem Feeding Sql Data Into Jquery Ui Dialog Box

    in Javascript Help

    Posted

    Hi,

     

    I'm trying to make a table that display a list of what stocks each client holds. In simple terms:

     

    Client 1 holdings:

     

    | + | Apple |

    | + | Google |

    | + | Microsoft |

     

    I currently have it set up so that if I click on the '+' (which is an image), it opens up a jQuery UI dialog box with static information in:

     

    <script>
$(function() {

 $('div.dialog')
 .dialog(
 {
 autoOpen: false,
 modal: true
 }
);
 $('img.opener')
 .css("cursor","pointer")
 .click(function()
 {
		$('#' + this.id.replace(/opener/, 'dialog'))
		.dialog('open');
 return false;
}
);
});
</script>

     

    ...and the table...

     

    <?php
echo '<td><img id="opener' . $row_counter . '" class="opener" src="../images/procedural/plus-white.png" title="Click to show Transactions"></td>';
echo '<td align=left>' . $stock_sedol . '</td>';
echo '<td align=left>' . htmlentities($stock_short_name) . '</td>';

echo '<td><div href="trans.php?client=' . $client_ref . '&stock=' . $stock_id . '" id="dialog' . $row_counter . '" class="dialog" title="Transactions for ' . $stock_sedol . '">This is box ' . $row_counter . '</div></td>';
?>

     

    Now I want to be able to populate the table with data from my SQL server, so I have the following code that I found on the internet (which works in isolation), but I'm not sure how to integrate it in to the above code.

     

    <!doctype html>

<html lang="en">
<head>
<meta charset="utf-8" />
<title>jQuery UI Dialog - Animation</title>
<link rel="stylesheet" href="http://code.jquery.com/ui/1.9.1/themes/base/jquery-ui.css" />
<script src="http://code.jquery.com/jquery-1.8.2.js"></script>
<script src="http://code.jquery.com/ui/1.9.1/jquery-ui.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('#tire-specs th a').each(function() {
var $link = $(this);
var $dialog = $('<div></div>')
.load($link.attr('href') + ' #content')
.dialog({
autoOpen: false,
title: $link.attr('title'),
width: 600
});

$link.click(function() {
$dialog.dialog('open');

return false;
});
});
});
</script>

</head><body>


<table id="tire-specs">
<thead>
<tr>
<th>Size</th>
<th><a href="trans.php?client=12345&stock=67890" title="Transactions">Transactions</a></th>
<th>Max Load</th>
<th>Max Inflation Pressure</th>
<th>Tread Depth</th>
</tr>
</thead>
<tbody>
<tr>
<td>205/65R15</td>
<td>620 A A</td>
<td>1477 lbs.</td>
<td>44 psi</td>
<td>11/32"</td>
</tr>
</tbody>
</table>

</body></html>

     

    Any help you could give would be much appreciated.

     

    Dave

    Retrieving Data From Multiple Tables

    in Microsoft SQL - MSSQL

    Posted

    Hi,

     

    I work in a Stockbroker's and I'm trying to display our client's holdings from my MSSQL Database. To put it in to perspective...

     

    The page displays a single client. A client has multiple holdings. Each holding relates to a single stock, and each stock is linked to a single currency.

     

    So what I need is something like:

     

    Bob Marley's Holdings

     

    Holding 1: Apple | USD | 1.6549

    Holding 2: Google | EUR | 1.2525

     

    The code I have so far is: 

    <?php

$client_ref = "50000";

$holding_sql = "SELECT * FROM [HOL_CORE] INNER JOIN [sTO_CORE] on [HOL_CORE].[sTOCK REC NO] = [sTO_CORE].[sTOCK ID] WHERE [HOL_CORE].[CLIENT REC NO] = '$client_ref' AND [HOL_CORE].[QTY HELD] != '0'";
$holding_result = sqlsrv_query($connnection, $holding_sql);
while($holding = sqlsrv_fetch_array($holding_result, SQLSRV_FETCH_ASSOC))

{

    $stock_currency = $holding['CURRENCY'];
 $stock_name = $holding['SHORT NAME'];

 echo $stock_name . ' | ' . $stock_currency . ' | ' . '<br>';

}

?>

     

    Now I need to get the exchange rate from a table called [CUR_CORE] for each stock, where [CURRENCY ABBREV] = $stock_currency. Does anyone know how I would go about retrieving this?

     

    Cheers!

     

    Dave

    Problems inserting NULL into database

    in MySQL Help

    Posted

    Your $sql assumes that there should be quotes around $isin (the general case). Remove those.

    $isin may need quotes around its value (the specific case). Add them if it should have them. As in

    $isin = "'" . $user_isin . "'";

    But $user_isin is a string, right? Not a number? If it is a number then it shouldn't have quotes at all.

     

    Hi Requinix,

     

    Thanks for clearing that up for me, makes a lot more sense to me now :) You understood me correctly, yes, and your suggestion worked perfectly, so thank you very much! You've saved me countless hours of smashing my head against a wall!

     

    Cheers,

    Dave

    Problems inserting NULL into database

    in MySQL Help

    Posted

    Hi,

     

    This one has been doing my head in for hours!.

     

    I have a script that, to put it simply, inserts either NULL or a string in to a MySQL database. However, it's acting rather weird with different values.

     

    Here is my SQL script...

    <?php

$sql = "INSERT INTO stocks (s_isin) VALUES ('$isin')";

mysql_query($sql);

?>

     

    And then the variables being fed in to it are...

    <?php 
if ($user_isin_dummy == "1")
// ISIN is Dummy
	{
		$isin = "NULL";	
	}

else
// ISIN is live
	{
		$isin = $user_isin;
	}
?>

     

    In it's current state, the "else" clause works fine, but the "if" statement inserts the string "NULL" in to the database.

     

    If I remove the single quotes from "$isin" in the SQL code, the "else" works only if the string begins with a number. If it starts with a letter, it simply doesn't insert at all. However, the "if" now correctly inserts a null value.

     

    How can I make it so that a proper null value is inserted on the "if" statement, and also for any string to be inserted on the else clause?

     

    Cheers,

    Dave

    Finding percentage in string

    in Regex Help

    Posted

    Your regex is malformed. Plus these type of questions we have a specially dedicated Regex forum for.

     

    Use this instead:

     

    ([\d.]+)%

     

    Sorry :( now that you've pointed the sub-forum out, I feel an idiot for missing it

     

     

    $text = "Bob the builder 5% 17/05/12 and 33.456% on another day, but fdsfd% is not valid";
preg_match_all('#[\d|.]+%#', $text, $matches);
print_r($matches);

     

    Output

    Array
(
    [0] => Array
        (
            [0] => 5%
            [1] => 33.456%
        )
)

     

    Thank you for your help :) this was exactly what I was looking for!

    Finding percentage in string

    in Regex Help

    Posted

    Hi,

     

    I'm trying to extract a percentage from data sent through a form. In the below example, I'm trying to dig out the "5%" part, however, this can often be multiple digits, sometimes including decimals as well. Basically I need the numbers and decimals preceding the % symbol, so "5%", "123.456%" and so on.

     

    <?php

$text = "Bob the builder 5% 17/05/12";

$n = preg_match_all('/ (.*?)%/s', $text, $match); 
if ($n)
	{ 
    			$search=$match[0]; 
    			for ($k=0;$k<$n;$k++)
    				{ 
            				$accpay = $search[$k];
    				} 
	}

?>

     

    This example gives me the result "the builder 5%". Does anyone know how I can further restrict this? It works fine if 5% is at the start of the string, but the string can be given in any order unfortunately.

     

    Cheers

    Date > A future date

    in PHP Coding Help

    Posted

    try

    <?php
  $creststartdate = '15/6/2012';                 // UK date, unfortunate a format
                                                 // not recognised by strtotime()
  date_default_timezone_set('GMT');
                                                 
  list($d, $m, $y) = explode('/', $creststartdate);
  if (mktime(0,0,0,$m,$d,$y) > strtotime('+ 30 days')) {
    echo "Beyond 30 days";
  } else {
    echo "OK";
  }
?>

     

    Barand,

     

    Thank you!! I'm still learning a lot of PHP and haven't dabbled in mktime yet. I'd looked at it, but wasn't sure how to use it, but that worked perfectly. I can sleep tonight! :)

    Date > A future date

    in PHP Coding Help

    Posted

    Hi,

     

    This one has been driving me up the wall, so hopefully some kind person can help me.

     

    I'm trying to make a validation script. $creststartdate comes in from a form as a UK formatted date (d/m/Y), and if $creststartdate is more than a month ahead of today, then it gets rejected. here's the code I have now...

     

    
            $today = date("m/d/y");
            $onemonth = strtotime ('+1 month', strtotime($today));
            $nextmonth = date ('d/m/Y', $onemonth);
            
            $csd = date("m/d/y", $creststartdate);
            $strcsd = strtotime($csd);
            $newdate = date ('d/m/Y', $strcsd);

            if ($newdate > $nextmonth)
                {
                    $creststartdatefailed = "The Crest Start Date cannot be more than 1 month in the future.";
                    $creststartdatevalid = "NO";
                }

     

    As it stands, this version of the code means that nothing is getting rejected.

     

    Please help??  :shrug:

     

    Cheers

    Dealing with stupid column names

    in Microsoft SQL - MSSQL

    Posted

    Hi,

     

    I am trying to connect to our company software database via PHP. I've got the connection working and can retrieve records etc. However one of my tables has a ridiculous column name that I can't get working.

     

    Here's the code...

     

    $mssql_server = "sql-primary";
$connectionInfo = array( "Database"=>"FOUR_I_CORE", "UID"=>"...", "PWD"=>"...");

$connnection = sqlsrv_connect( $mssql_server, $connectionInfo);

$sql = "SELECT [REF (SEDOL) No.] FROM [sTO_CORE] WHERE [REF (SEDOL) No.] = $sedol";

$stmt = sqlsrv_query( $connnection, $sql );

				if( $stmt === false) 
					{
    						die( print_r( sqlsrv_errors(), true) );
					}


				while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) )

					{
						//echo '<b>Client Ref:</b> ' . $row['REF (SEDOL) No.'] . '<br>';
					}

     

    I keep getting the following error message:

     

    Array ( [0] => Array ( [0] => 42S22 [sqlSTATE] => 42S22 [1] => 207 [code] => 207 [2] => [Microsoft][sql Server Native Client 10.0][sql Server]Invalid column name 'B7N4L15'. [message] => [Microsoft][sql Server Native Client 10.0][sql Server]Invalid column name 'B7N4L15'. ) )

     

    FYI, B7N4L15 is the value being entered as $sedol.

     

    I'm pretty sure the problem is with the column name REF (SEDOL) No., but I've no idea how to contain the dodgy name.

     

    Unfortunately, I don't have the luxury of renaming the column, so I have to make do. Is there anything I can do to make this work?

     

    Cheers

    Automatic table problem

    in PHP Coding Help

    Posted

    Hi CPD,

     

    It is adding 4 weeks every time. Every column should display 11 dates (so column 1 works perfectly), and column 2 should carry on where column 1 left off. It's this that is the problem, it isn't carrying on, it's starting again.

     

    Thanks

    Automatic table problem

    in PHP Coding Help

    Posted

    I'm trying to create a table that looks like this:

     

    25/11/201128/09/201202/08/2013

    23/12/201126/10/201230/08/2013

    20/01/201223/11/201227/09/2013

    17/02/201221/12/201225/10/2013

    16/03/201218/01/201322/11/2013

    13/04/201215/02/201320/12/2013

    11/05/201215/03/201317/01/2014

    08/06/201212/04/201314/02/2014

    06/07/201210/05/201314/03/2014

    03/08/201207/06/201311/04/2014

    31/08/201205/07/201309/05/2014

     

    Basically it formats the dates automatically every 4 weeks into 3 columns. So far, I have the following code:

     

    <?php
$startDate = '2011-11-25';
$columns = 3;
$number_of_dates = 33;

// calculate number of rows per column
$rows_per_column = ceil($number_of_dates / $columns);
// make your table
echo "<table>\n";

//here we changed the condition to $i < $rows
for($i = 0; $i < $rows_per_column; $i++) {

echo "<tr>\n";

//here will run another loop for the amount of columns
for($j = 0; $j < $columns; $j++) 
{
	$weekOffset = $i * 4;
    	$nextDate = strtotime("{$startDate} +{$weekOffset} weeks");
    	echo "<td width = '100'>" . date('d/m/y', $nextDate) . "</td>\n";
    }
echo "</tr>\n";
}
echo "</table>\n";
?>

     

    Which creates the first column fine, but then repeats the same column two more times...

     

    25/11/201125/11/201125/11/2011

    23/12/201123/12/201123/12/2011

    20/01/201220/01/201220/01/2012

    17/02/201217/02/201217/02/2012

    16/03/201216/03/201216/03/2012

    13/04/201213/04/201213/04/2012

    11/05/201211/05/201211/05/2012

    08/06/201208/06/201208/06/2012

    06/07/201206/07/201206/07/2012

    03/08/201203/08/201203/08/2012

    31/08/201231/08/201231/08/2012

     

    I know that this means there is a problem with my loop resetting itself rather than continuing, but I'm not sure how to fix it. Please could someone point out my stupid mistake?

     

    Thanks

    Trying to find the last working day of the month

    in PHP Coding Help

    Posted

    Hi,

     

    As the title says, I'm trying to get the last working day of the month (by this I mean excluding Saturday and Sunday). My thinking was that I basically need to this:

     

    If (day = Sat)

      {

          -1 day

      }

    elseif (day = Sun)

      {

        -2 days

      }

    else

      {

        keep it as it is

      }

     

    So far, all I have is the below to echo out the day of date("t/m/y"), however, all I'm getting is Thursday, whereas the last day of March is a Saturday

     

    <?php

$lastdateofthemonth = date("t/m/y");
$lastworkingday = date('l', strtotime($lastdateofthemonth));
echo $lastworkingday; 

?>

     

    Can anyone help, or point me in the way of where this may have been answered in the past? I've searched all over Google but can't find what I'm looking for.

     

    Thanks

    Error with UPDATE syntax

    in MySQL Help

    Posted

    Hi,

     

    I am trying to update two columns on one table from a form. The table is called pages, and there are three columns:

    id (auto-increments)

    name

    content

     

    When I try to update 'name' and 'content', I get the following error:

     

    Error: 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 '1' at line 1.

     

    The form simply calls the following script on submittal :

     

    <?php

$name = $_POST[name];
$content = $_POST[content];

include ("../index_files/mysql_include.php");

$result = mysql_query("UPDATE pages SET name = '$name', content = '$content' WHERE id = $id");

if (!mysql_query($result, $connection))
{
die('Error: ' . mysql_error());
}

mysql_close($connection);

?>

     

    Can anyone see where I'm going wrong? I've tried several different variations of the code with no joy.

     

    I should point out that the above code does update MySQL successfully, but it keeps coming out with this error which I'd rather not have :)

     

    Thanks,

    Dave

    Searching through multiple file types

    in Application Design

    Posted

    The PDF file format is not a plain text markup, so opening the file raw and searching it isn't going to yield you reliable results.  You'll need to interpret the file with something that understands the format.  I needed to do something similar and found how sphider made use of xpdf and catdoc for pdf's and doc's respectively.  xpdf has a couple utility programs, pdfinfo and pdftotext which you use to extract the metadata and text which you can in turn search.

     

    Sorry, I forgot to check this post as I've been working on other projects.

     

    I had previously looked at xpdf, but hadn't gotten the results I was looking for. I'll give it a try with Sphider and let you know how I get on. Thanks for the suggestion :)

     

    Dave

    Problem with "If IP equals wildcard"

    in PHP Coding Help

    Posted

    in this case you could remove everything after the last . in $ip giving 10.44.6.

     

    then this would work

     

    if ($ip == '10.44.6.')

     

    this won't work, because for instance, '10.44.6.' does not equal '10.44.6.1'.  You use something like strpos instead.  Alternatively you can use regex to match it, if you want to match ranges.

     

    Hi guys,

     

    I went with the strpos idea which worked perfectly. I put this code in the header:

     

    <?php $ip = $_SERVER['REMOTE_ADDR'];
$london = '10.44.';
$office = strpos($ip, $london);
$blackpool_path = 'path';
$london_path = 'path'
?>

     

    And put this where the link would go:

     

    <a href="<?php

if ($office === false) 

{
echo $blackpool_path;						
} 
else 
{
echo $london_path;
};
?>installer.exe">Link name</a>

     

    Many thanks for your help :)

     

    Dave

    Problem with "If IP equals wildcard"

    in PHP Coding Help

    Posted

    Hi,

     

    I have a page on my Intranet which lists a bunch of install files (for example, Microsoft Office), which are stored in the same directory structure as the site itself. I also have a duplicate installs folder structure in another office, on their local server. I want to make it so that the link changes, depending on which office the user is in. To simplify:

     

    If (user is in London)

    display London link

    else

    display Blackpool link

     

    I've been trying to make it work via IP Address. The IP range in London is 10.44.6.* However, I can't make the script work with a wildcard (it works fine if I put a specific address in). The code I have so far is:

     

    <?php
$ip=$_SERVER['REMOTE_ADDR'];
if ($ip == '10.44.6.*') {
?><a href="link1.php">Click Here</a><?php
} else {
?><a href="link2.php">Click Here</a><?php
};
?>

     

    I think it may be an obvious solution, but if some kind person could perhaps point out my mistakes?

     

    Thanks,

    Dave

    Searching through multiple file types

    in Application Design

    Posted

    Hi,

     

    I currently have a 'Research' page on my intranet site. It works by displaying a list of files based on the date selected.

     

    The files are stored in a folder hierarchy on the same server, for example:

     

    2012 <-Year

    --01 <-Month

    ----01 <-Day

    ----02 <-Day

     

    The folders contain a multitude of files that are dropped in by other members of staff, and 99% of the time consist of .msg, .doc and .pdf files. What I want to do is create a textbox which will allow the user to search through the files (as in the file contents, not just the file name). So far, the best thing I have found for this is the following code:

     

    <?php

/**
*	powered by @cafewebmaster.com
*	free for private use
*	please support us with donations
*/



define("SLASH", stristr($_SERVER[sERVER_SOFTWARE], "win") ? "\\" : "/");

$path	= ($_POST[path]) ? $_POST[path] : dirname(__FILE__) ;
$q		= $_POST[q];



function php_grep($q, $path){

	$fp = opendir($path);
	while($f = readdir($fp)){
		if( preg_match("#^\.+$#", $f) ) continue; // ignore symbolic links
		$file_full_path = $path.SLASH.$f;
		if(is_dir($file_full_path)) {
			$ret .= php_grep($q, $file_full_path);
		} else if( stristr(file_get_contents($file_full_path), $q) ) {
			$ret .= "$file_full_path\n";
		}
	}
	return $ret;
}


if($q){
	$results = php_grep($q, $path);
}



echo <<<HRD

<pre >
<form method=post>
	<input name=path size=100 value="$path" /> Path 
	<input name=q size=100 value="$q" /> Query
	<input type=submit>
</form>

	$results

</pre >

HRD;

?>

     

    This obviously uses GREP, which works well, albeit slow. However, it doesn't search through PDF's.

     

    I have contemplated several solutions, including finding a way to convert all the files in to text files or in to a mysql database, but I haven't found anything useful. I think the answer is going to be no, but I'm asking whether anybody knows, or has successfully implemented a similar system?

     

    Big big thanks in advance if you can help in any way!

    Dave

    Date function help

    in PHP Coding Help

    Posted

    You could test for the day number, and only echo it if it's less than 6.

     

    do {
	$test_date = $start_date + ($day_incrementer * 60 * 60 * 24);
	if( date('N', $test_date) < 6 ) { // ADDED THIS CONDITIONAL //
		echo '<a href="http://markets.ft.com/RESEARCH/markets/DataArchiveFetchReport?Category=CU&Type=WORL&Date=' . date("m/d/Y", $test_date) . '">' . date("d/m/Y", $test_date) . "\n</a><br>";
	}

     

    Thank you very much for that, it's working now :)

     

    However, I now have another question - I want to split the list in to months, so that there is a list for January 2012, another list for December 2011 and so on, rather than one indigestible long list. This normally wouldn't be a problem, as I could just repeat the script and hard code the dates for the start and end of each month. However, if I do it for this month and in the future months, I would get the following output:

     

    16/01/2012

    17/01/2012

    18/01/2012

    19/01/2012

    20/01/2012

     

    I don't want any future dates to be shown, rather I would prefer for it to show up until yesterdays date (16/01/2012). Any ideas?

     

    Cheers,

    Dave

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