PHP - How To Format The Data Retreived From An Sql Query With Php

Coming from the world of Excel, I can easily format numbers as $1,500.00, or 27%.  When I uploaded a large chunk of data into SQL to be read back through a table, the values all come out as exactly what they were uploaded as.  For example, I have an SQL column set as Decimal(19,4) that I want formatted like currency, but which shows up as 1500.0000 in my table, or another column with type decimal(5,2) which shows up as 5500.0000, but which I want to show up as 55%.  How do I do this?

Sorry for the beginner question, here I'm trying to retrieve data from the database and display it in the table format. But only table headers are printed, and not the actual values. The count variable is echoing 2 saying that data is present and correctly retrieved. Can anyone help?

<?php
include 'connect.php';
error_reporting(E_ALL ^ E_DEPRECATED);
error_reporting(E_ERROR | E_PARSE);
$sql="SELECT * FROM `resources` as r INNER JOIN `project_resources` as pr
ON r.res_id =pr.res_id WHERE project_id='$_POST[project_id]'";
$result=mysql_query($sql);

$count=mysql_num_rows($result);

if($result === FALSE) {
die(mysql_error()); 
}
echo "$count";
echo '<table>
<tr>
<th>Resource ID</th>
<th>Resource Name</th>
<th>Email</th>
<th>Phone Number</th>
<th>Reporting Manager</th>
<th>Role</th>
<th>Designation</th>
</tr>';

while ($row = mysql_fetch_array($result)) {
echo '
<tr>
<td>'.$row['res_id'].'</td>
<td>'.$row['res_name'].'</td>
<td>'.$row['email'].'</td>
<td>'.$row['phone_number'].'</td>
<td>'.$row['reporting_manager'].'</td>
<td>'.$row['role'].'</td>
<td>'.$row['designation'].'</td>
</tr>';

}

echo '
</table>';

?>

Edited by mac_gyver, 22 September 2014 - 07:25 AM.
code tags please

My website sends me an e-mail when there is an error, and I would like it to look like this...
Code: [Select]
Date: 2012-03-17 12:36:46pm
Results Code: EMAIL_USER_NOT_LOGGED_IN_2127
Error Page: /members/change_email.php
Member ID: 0
IP Address: 127.0.0.1
Host Name: localhost

However my PHP code isn't giving me that...

$body = "A website error has occurred...\n\n";
$body .= "Date: \t\t" . date('Y-m-d g:i:sa', time()) ."\n";
$body .= "Results Code: \t\t" . $resultsCode . "\n";
$body .= "Error Page: \t\t" . $errorPage . "\n";
$body .= "Member ID: \t\t" . $memberID . "\n";
$body .= "IP Address: \t\t" . $ip . "\n";
$body .= "Host Name: \t\t" . $hostName . "\n";


What is wrong?

Thanks,


Debbie

OK, Here is the code

$sql = "SELECT
TrialListing.listingID AS Trial, TrialClass.classID AS Class, place.place_name AS Place,
CONCAT_WS(
' ',
pedigree.pretitle,
pedigree.`Name`) AS Hound,
CONCAT_WS(
' ',
ped2.pretitle,
ped2. NAME
)AS Sire,
CONCAT_WS(
' ',
ped3.pretitle,
ped3. NAME
)AS Dam,
pedigree.Breeder,
pedigree.`Owner`,
CASE
WHEN placement.place_id < 5 THEN
TRUNCATE(TrialClass.number_of_entrants / placement.place_id,2)
WHEN placement.place_id  = 5 THEN
''
ELSE
0
END

AS Score
FROM
TrialListing
Left Join TrialClass ON TrialListing.listingID = TrialClass.listingID

JOIN placement ON placement.event_id = TrialClass.trialClassID
JOIN pedigree ON pedigree.PedigreeId = placement.hound_id
LEFT OUTER JOIN pedigree AS ped2 ON pedigree.SireId = ped2.PedigreeId
LEFT OUTER JOIN pedigree AS ped3 ON pedigree.DamId = ped3.PedigreeId
LEFT JOIN place ON place.place_id = placement.place_id

WHERE
TrialListing.listingID = 11
ORDER BY Class, FIELD(place.place_id, '1', '2', '3', '4', '0')


";      // Database Query  

$result = mysql_query("$sql");  // Database Query result

$num_rows = mysql_num_rows($result);





// Starts the table
echo "<table class=\"clubList\">\n
<tr>
<th>trialID</th>
<th>ClassID</th>
<th>Place</th>
<th>Hound</th>
<th>Sire</th>
<th>Dam</th>
<th>Score</th>

</tr>";

// Create the contents of the table.
for( $i = 0; $i < $row = mysql_fetch_array($result); $i++){




echo "<tr>\n"
        ."<td>".$row["Trial"]."</td>\n" 
        ."<td>".$row["Class"]."</td>\n" 
        ."<td>".$row["Place"]."</td>\n"  
        ."<td>".$row["Hound"]."</td>\n"  
        ."<td>".$row["Sire"]."</td>\n"                
        ."<td>".$row["Dam"]."</td>\n"
        ."<td>".$row["Score"]."</td>\n"  
        ."</tr>";}



echo "</TABLE>";


Here is the output, I added the TrialID & ClassID for informational purposes, they do not need to be displayed in the live table.

trialID ClassID Place Hound Sire Dam Score 11 1 1st Eaton Brook Tug Hill Tatonka Eaton Brook Hickety Hawk Eaton Brook Gunner's Beulah 43.00 11 1 2nd FC North Bend Igloo FC DFJ Murphy White IFC Brad-Ju's Bella Donna 21.50 11 1 3rd FC North Bend Igloo FC DFJ Murphy White IFC Brad-Ju's Bella Donna 14.33 11 1 4th Rail Road Spike VI Elwell's Mike Elwell's Hannah 10.75 11 1 NBQ FC Fish Creek Spike Fish Creek Bull II Fish Creek Susie [H849395] 11 2 1st Enman Hill Sweet Poppy FTCH Straight Arrow Lucky of Coos 32.00 11 2 2nd Fishflakes Penny At Harehaven FTCH Jill's Fair-Isle Spud FTCH Millbridge Brownie 16.00 11 2 3rd Line Elm Flakers IFC Flakers Rex IFC Line Elm Ginger 10.66 11 2 4th FTCH Fareast Mookie FTCH Mellowrun Sly FTCH Cape Breton Maude 8.00 11 2 NBQ FTCH Mellowrun Sly Mellowrun Skylighter FTCH Mellowrun Becka 11 3 1st Bojangle V Lee Otworth Half Acre's Cocoa Candy 23.00 11 3 2nd Gay Doll Gay Roll II Gay Idol 11.50 11 3 3rd Bruce's Blue Lady FC Kilsock's Blue Creek Bart Bishopville's Zippy 7.66 11 3 4th FC Pearson Creek Barbin FC Pearson Creek Barbarian FC B-Line Stubby 5.75 11 3 NBQ Sims Creek Cricket Ronnie Joe Sims Creek Tiny 11 4 1st FTCH Fareast Mookie FTCH Mellowrun Sly FTCH Cape Breton Maude 26.00 11 4 2nd FTCH Mellowrun Sly Mellowrun Skylighter FTCH Mellowrun Becka 13.00 11 4 3rd Fishflakes Penny At Harehaven FTCH Jill's Fair-Isle Spud FTCH Millbridge Brownie 8.66 11 4 4th Enman Hill Sweet Poppy FTCH Straight Arrow Lucky of Coos 6.50 11 4 NBQ Line Elm Flakers IFC Flakers Rex IFC Line Elm Ginger
Below is what I would like to generate. How do I word or nest the proper PHP code/loops to accomplish this?

ClassID

Place Hound Sire Dam Score 1st Eaton Brook Tug Hill Tatonka Eaton Brook Hickety Hawk Eaton Brook Gunner's Beulah 43.00 2nd FC North Bend Igloo FC DFJ Murphy White IFC Brad-Ju's Bella Donna 21.50 3rd FC North Bend Igloo FC DFJ Murphy White IFC Brad-Ju's Bella Donna 14.33 4th Rail Road Spike VI Elwell's Mike Elwell's Hannah 10.75 NBQ FC Fish Creek Spike Fish Creek Bull II Fish Creek Susie [H849395]

ClassID

Place Hound Sire Dam Score 1st Enman Hill Sweet Poppy FTCH Straight Arrow Lucky of Coos 32.00 2nd Fishflakes Penny At Harehaven FTCH Jill's Fair-Isle Spud FTCH Millbridge Brownie 16.00 3rd Line Elm Flakers IFC Flakers Rex IFC Line Elm Ginger 10.66 4th FTCH Fareast Mookie FTCH Mellowrun Sly FTCH Cape Breton Maude 8.00 NBQ FTCH Mellowrun Sly Mellowrun Skylighter FTCH Mellowrun Becka

ClassID

Place Hound Sire Dam Score 1st Bojangle V Lee Otworth Half Acre's Cocoa Candy 23.00 2nd Gay Doll Gay Roll II Gay Idol 11.50 3rd Bruce's Blue Lady FC Kilsock's Blue Creek Bart Bishopville's Zippy 7.66 4th FC Pearson Creek Barbin FC Pearson Creek Barbarian FC B-Line Stubby 5.75 NBQ Sims Creek Cricket Ronnie Joe Sims Creek Tiny

ClassID

Place Hound Sire Dam Score 1st FTCH Fareast Mookie FTCH Mellowrun Sly FTCH Cape Breton Maude 26.00 2nd FTCH Mellowrun Sly Mellowrun Skylighter FTCH Mellowrun Becka 13.00 3rd Fishflakes Penny At Harehaven FTCH Jill's Fair-Isle Spud FTCH Millbridge Brownie 8.66 4th Enman Hill Sweet Poppy FTCH Straight Arrow Lucky of Coos 6.50 NBQ Line Elm Flakers IFC Flakers Rex IFC Line Elm Ginger

I was just wondering if it's possible to run a query on data that has been returned from a previous query?  For example, if I do

Code: [Select]
$sql = 'My query';
$rs = mysql_query($sql, $mysql_conn);

Is it then possible to run a second query on this data such as

Code: [Select]
$sql = 'My query';
$secondrs = mysql_query($sql, $rs, $mysql_conn);

Thanks for any help

This topic has been moved to Other.

http://www.phpfreaks.com/forums/index.php?topic=355929.0

Hello,   I am trying to display the data from two tables with proper format. But Its not happening   Here is my 1st table - orders   2.PNG   10.12KB   0 downloads   and my 2nd table - order_line_items    1.PNG   14.92KB   0 downloads   I want to display like this     3.PNG   4.03KB   0 downloads   Here is my code

$query = $mysqli->query("SELECT orders.order_id, orders.company_id, orders.order_for, order_line_items.order_id, order_line_items.item, order_line_items.unit,SUM(order_line_items.unit_cost * order_line_items.quantity) AS 'Total', order_line_items.tax from orders INNER JOIN order_line_items ON orders.order_id = order_line_items.order_id where orders.order_quote = 'Order' GROUP BY order_line_items.id"); ?>

<table id="dt_hScroll" class="table table-striped">
						<thead><tr>
						<th>Order ID</th>
						<th>Company</th>
						<th>Contact Person</th>
						<th>Products</th>
                        <th>Total</th>
                        
                        
						
</tr>

</thead>
<tbody> <?php

while($row = $query->fetch_array())
{
?>
<tr>
<td><?php echo $row['order_id']; ?></td>
<td><?php echo $row['company_id']; ?></td>
<td><?php echo $row['contact_person'] ?></td>
<td><?php echo $row['item']; ?></td>
<td><?php echo $row['Total']; ?> %</td>

</tr>
<?php

}	

But here order ID, Company ID, Contact Person are also repeating thrice with item in order_line_items table   Please suggest me how to do this

Hi,

I want to pull data from db, where sometimes all rows and sometimes rows matching given "username".

Here is my code:

	//Grab Username of who's Browsing History needs to be searched. 
	if (isset($_GET['followee_username']) && !empty($_GET['followee_username'])) 
{ 
    $followee_username = $_GET['followee_username']; 
    
    if($followee_username != "followee_all" OR "Followee_All") 
    { 
        $query = "SELECT * FROM browsing_histories WHERE username = \"$followee_username\""; 
        $query_type = "followee_username"; 
        $followed_word = "$followee_username"; 
        $follower_username = "$user"; 
        echo "$followee_username"; 
    } 
    else 
    { 
        $query = "SELECT * FROM browsing_histories"; 
        $query_type = "followee_all"; 
        $followed_word = "followee_all"; 
        $follower_username = "$user"; 
        echo "all"; 
    } 
} 
	

 

When I specify a "username" in the query via the url:

browsing_histories_v1.php?followee_username=requinix&page_number=1

I see result as I should. So far so good.

 

Now, when I specify "all" as username then I see no results. Why ? All records from the tbl should be pulled!

browsing_histories_v1.php?followee_username=all&page_number=1

This query shouldv'e worked:

	$query = "SELECT * FROM browsing_histories"; 
	

Hi,   I want to display my data in this format  format.JPG   54.52KB   0 downloads   Expense data i am getting like this

 <td><?php
	$in = "SELECT sales_invoice.invoice_id as INID, DATE_FORMAT(sales_invoice.date_invoiced,'%M') AS month, sales_invoice_line_items.invoice_id, SUM(sales_invoice_line_items.sub_total) AS Total, SUM(sales_invoice_line_items.tax_amount) AS total_tax FROM sales_invoice INNER JOIN sales_invoice_line_items ON sales_invoice.invoice_id=sales_invoice_line_items.invoice_id GROUP BY sales_invoice.invoice_id, DATE_FORMAT(sales_invoice.date_invoiced, '%Y-%m')";
	
	$in1 = mysql_query($in) or die (mysql_error());
	$total=0;
	$tax =0;
	while($income = mysql_fetch_array($in1))
	{
		
		$total +=$income['Total']; 
		$tax +=$income['total_tax'];
		echo $total - $tax; }
    ?>
    
    </td>

And Profit i am getting like this

<td><?php
	$in = "SELECT purchase_invoice.invoice_id as INID, DATE_FORMAT(purchase_invoice.date_invoiced,'%M') AS month, purchase_invoice_line_items.invoice_id, SUM(purchase_invoice_line_items.sub_total) AS Total, SUM(purchase_invoice_line_items.tax_amount) AS total_tax FROM purchase_invoice INNER JOIN purchase_invoice_line_items ON purchase_invoice.invoice_id=purchase_invoice_line_items.invoice_id GROUP BY purchase_invoice.invoice_id, DATE_FORMAT(purchase_invoice.date_invoiced, '%Y-%m')";
	
	$in1 = mysql_query($in) or die (mysql_error());
	$total=0;
	$tax =0;
	while($income = mysql_fetch_array($in1))
	{
		
		$total +=$income['Total']; 
		$tax +=$income['total_tax'];
		echo $total - $tax; }
    ?>
    
    </td>

I dont know how to display data like this. Do i need to change the the table structure in my database? Please suggest

Hi...

I tried to use foreach in displaying my table header, but I encountered problem when I tried to display data on the first row , my query only display the last Sum for the last Comp.

here is my code:

<html>
<head>
<title>Half Shell</title>
<link rel="stylesheet" type="text/css" href="kanban.css" />
<?php
  error_reporting(E_ALL ^ E_NOTICE);
  date_default_timezone_set("Asia/Singapore"); //set the time zone  
$con = mysql_connect('localhost', 'root','');

if (!$con) {
    echo 'failed';
    die();
}

mysql_select_db("mes", $con);


?>

<body>
<form name="param" action="" method="post" onSubmit="return false">

<div id="fieldset_PS">
<?php
   echo "<table>";
   
   $sql = "SELECT DISTINCT s.Comp FROM sales_order s, param_settings p WHERE s.Comp = p.Compounds ORDER BY s.Comp";
   $res_comp = mysql_query($sql, $con); 
   while($row_comp = mysql_fetch_assoc($res_comp)){
        $Comp[] = $row_comp['Comp'];
   }
   echo "<th>&nbsp;</th>";
   foreach($Comp AS $Comp){
    echo "<th>$Comp</th>";
  }   

   echo "<tr>
   <td>Total Kg/Compound</td>";
      $sql_sec = "SELECT SUM(TotalKg) AS TotalKg FROM sales_order  WHERE Comp = '$Comp' ORDER BY Comp";
   $res_sec = mysql_query($sql_sec, $con);
   
   while($row_sec = mysql_fetch_assoc($res_sec)){
       $TotalKg[] = $row_sec['TotalKg'];

   }
   foreach($TotalKg AS $TotalKg){
    echo "<td>$TotalKg</td>
    </tr>";
  }   

   ?>


I also attach the correct output that should be and the result from my code.

Thank you

Hi, I have a connection set up to an API using a PHP script - the API sends back data in JSON format, and if I capture it and echo it, it displays in extremely unfriendly format on the screen.

I've tried to find a way to convert this data into readable format, ideally in a HTML table, but although there are articles on converting manually inputted JSON data into a HTML table using Javascript, I can't find a way to do it from a PHP variable.

This is what I have so far:

if ($_POST['getcompany']) {
$companyname = $_POST['_Name'];
	
$ch = curl_init();

$data_array2 =  array(
      'token' => $token
);
$make_call2 = json_encode($data_array2);
	//echo 'Token is '.$token;
	$token2 = substr($token, 14);
	$token3 = substr($token2, 0, -3);
	
	//echo 'Token 2 is '.$token3;
	//echo 'Company Name is '.$companyname;

//curl_setopt($ch, CURLOPT_GET, 1);
//curl_setopt($ch, CURLOPT_POSTFIELDS, $make_call2);
curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-Type: application/json','Authorization: '.$token3.'')); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_URL, 'https://connectionurl?countries=GB&name='.$companyname);

//Execute the request
$result = curl_exec($ch);

echo 'Companies: '.$result; //This displays the data provided by the search, but is in JSON format

 

So I need a way to get the data held in $result, into a HTML table. Any idea how I can do this?

Hello All, 
  Need an advise in finishing this task please. 
 

function convertTimeFormat($time12Hour) {
// Initialized required variable to an empty string.
		$time24Hour = ""; 

// Used explode() function to break the string into an array and stored its value in $Split variable. 
		$Split = explode(":",$time12Hour); // print_r(explode (":", $time12Hour)); => Array ( [0] => 09 [1] => 50 [2] => 08AM )

// Retrieved only "hour" from the array and stored in $Hour variable. 
		$Hour = $Split[0];   

    $Split[2] = substr($Split[2],0,2);

// Used stripos() function to find the position of the first occurrence of a string inside another string.	
		if($Hour == '12' && strpos($time12Hour,"AM")!== FALSE) {
			// Code here 
		} elseif(strpos($time12Hour,"PM")!== FALSE && $Hour != "12") {
			// code here
		}
		return  $time24Hour;
	}
	$time12Hour = "09:50:08AM";
	$result = convertTimeFormat($time12Hour);
	print_r($result); 

/*
Input : "09:50:08AM";
Output : "21:50:08PM";
*/

 

Hello,

I hope all of you are safe with your families.

Currently I am starting with PHP Coding and I am trying to do a simple query to MySQL DB using PHP but even when is able to bring the number of rows is not displaying the values in the DB.

This is my code:

<?php
$servername = "localhost";
$database = "mydbtest";
$username = "root";
$password = "root";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $database);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
//echo "Connected successfully";
$myquery = "SELECT * FROM Country";
$result = $conn->query($myquery);
$numf = $result->num_rows;
echo "Number of rows " . $numf . "<br>";
if($numf >0){
	while($row = $result->fetch_object()){
		echo "Code" . $row->countrycode . "<br>";
		echo "Country" . $row->countryname . "<br>";
	}
 
	}else{
		echo '0 results';
	}

mysql_free_result($myout);
mysqli_close($conn);
?>

What is failing?

 

Thanks in advance for the assistance.

Alright, I've spent over a week trying to fix this now - And Im getting frustrated! I asked at other forums, I asked co-workers and I asked friends-of-friends, and nobody can explain what happens. Let's take a look at this first:

$name = mysql_real_escape_string($_POST['name']);
mysql_query(sprintf("UPDATE em_users SET name='%s' WHERE id='" . $in_user['id'] . "'", $name));

This will insert NO data on the Name field in the database. Obviously, I thought the $_POST variable wasn't passed correctly, but echo'ing it just before the query WILL show data. And as I said, I tried everything possible for the last week. Switching variables, adding static text on the $name variable instead of using the $_POST content (this does work). I used very very simple test data on the form, such as my name "Mark" or "test" and "hey". The query is correctly executed everytime.

The truely WEIRD thing is, if I ensure there is content in $name before executing the query it will work as expected everytime. Like this:

$name = mysql_real_escape_string($_POST['name']);
$name && mysql_query(sprintf("UPDATE em_users SET name='%s' WHERE id='" . $in_user['id'] . "'", $name));

Of course I could do this, but I want to know why my code does or doesn't work + it's a lot of work to do for something that worked fine a week ago. It has spread to a lot of forms on my website that $_POST variables aren't processed correctly - and it happened out of nowhere. Even on codes that havnt changed in months.

I really need help on fixing this! This project has been in development for nearly two years, and without a fix it's pretty much lost

I know that I can fairly easily pull my data from the database and view it in a browser.
I can also 'polish' it with some HTML or put it into a table.
Can I get an item from a given VARIABLE to appear inside of an INPUT box, so that it looks the same as when it was initially submitted?
Can it be done with a multiple choice SELECT dropdown, so that the item chosen is viewable again?