PHP - Query And Output Data From Same Table In Db

Hello everyone.

I need help with the following PHP APP. I am running on (Version PHP 7.2.10) I am trying to have a page table form on table.php pass the input variable of “5-Numbers” to another page called table_results.php

I want that variable string of “5-Numbers” to be compared against 4 arrays and output any duplicates found within each of those 4 lists. If nothing is found, I still want some visual output that reads “None found”.

 

Lets pretend I have the following example ..

On table.php, I typed inside my table form the 5-Numbers ..

INPUT:  2,15,37,13,28

On table_results.php, the 4 arrays to be compared against my input numbers “ 2,15,37,13,28” are ..

 

$array_A = array(2,6,8,11,14,18,24); 

$array_B = array(1,2,9,10,13,14,25,28,);  

$array_C = array(1,3,7,9,13,15,20,21,24); 

$array_D = array(4,5,12,22,23,27,28,29); 

 

So my output should read as follows below ..

OUTPUT: 

 

TABLE

COLUMN 1   COLUMN 2

ROW 1 Matches Found Results ..

ROW 2 GROUP A:            2 

ROW 3 GROUP B:             2,13,28

ROW 4 GROUP ?            13,15

ROW 5 GROUP ?            28

ROW 6 5#s Input:              2,15,37,13,28

 

Please let me know if anyone has any suggestions on how to go about it. 

Thanks.

Edited January 1, 2019 by Jayfromsandiego
Wanted to include image example

I create a post just now but cannot seem to find it to add to it. Anyhow, I have 2 tables, I need to multiply qty from the one table to the mass from the other to get a total, which I have achieved with this code:

$sql_total_mass = "
	SELECT jobs_assembly.assemble_qty, jobs.mass,
	(jobs_assembly.assemble_qty * jobs.mass) AS 'sum'	
	FROM jobs_assembly
	LEFT JOIN jobs on jobs_assembly.jobs_id = jobs.id
	LEFT JOIN job_names ON jobs.job_names_id = job_names.job_id
	WHERE jobs_assembly.assemble_date = '$link_date'
	ORDER BY job_names.job_name, jobs.assembly
	";

	$result_total_mass = mysqli_query($conn, $sql_total_mass);

	if (mysqli_num_rows($result_total_mass) > 0) {
		while($row_total_mass = mysqli_fetch_assoc($result_total_mass)) {
			echo $row_total_mass['sum'].'<br />';
		} //while
	} //if

Now I need to take all these totals and make a grand total. So basically add all the results from $row_total_mass[‘sum’] together and show it.

 

 

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

Ok, I know this is a really simple one, but I haven't done any coding for a couple of month and i'm a bit rusty.

I'm trying to add a field from a MySQL query to a Google calendar add button link so I can create a dynamic 'add to calendar' button from my gig list.

So on the gig list page that echo's all the dates from the database, I want to add the venue name to the 'text' section of the google URL -

Code: [Select]
<a href=\"http://www.google.com/calendar/event?action=TEMPLATE&text=VENUE HERE&dates=20060101T000000Z/20060101T000000Z&details=gig%20text&location=gig%20town&trp=false&sprop=www.the-guards.co.uk&sprop=name:The%20Guards\" target=\"_blank\"><img src=\"http://www.google.com/calendar/images/ext/gc_button6.gif\" border=0></a>
if I echo the field like so -

Code: [Select]
echo"{$row['gl_venue']}";
I get the correct result - I just can remember how to embed the php within the html, everything i've tried has failed so far.

Once i've got this sorted i can go ahead and add the date, location fields etc.

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 have an image (a weather map) which is created using php which pulls data from a mysql database.

I would like to be able to output the result image as a gif or jpg file.

Can anyone suggest how this could be done?

Many thanks,
Simon

Hey guys, newbie here love the site, lots of great info, hopefully one of u guys can help me out.
i have adapted code from http://www.singhvishwajeet.com/2009/06/25/using-php-to-get-stock-quotes-from-yahoo-finance/ (which is free to use)
i have been working on a way to enter a company code, e.g. BARC -barclays bank, on one page then on another im trying to serahc the yahoo csv for a number of specific numbers that are "live" and then output them on my site, which in turn i shall work with but i cant seem to out put the data, its the $stat bit im struggling with
i will attach the files, hopefully someone can help me

I'm not even sure how to explain it.  I'm linking a team's roster to its coach.  Now I'm getting a weird offset issue when I output the roster.  It has something to do with the STATUS.

Down below DIV class="roster_player_list", my Status values  are 1, 2, 3, or 4, and each is then assigned a word value.  My output should look like this (without the arrow and number) :

Returning Starter     <---- 1
#
#
#

Key Returner      <--- 2
#
#

Varsity Newcomer     <--- 3
#
#
#

Key Freshmen   <--- 4
#


The last "1" (Returning Starter) disappears from my list on both of my test accounts.  I should have four Returning Starters instead of three.  I've changed a player's status from a 2 to a 1, and the list changed.  That player disappeared, the other appeared. 

Code: [Select]
$sql = "SELECT * FROM players as p
INNER JOIN schools as s
WHERE s.coachFirst='".$current_first."' AND s.coachLast='".$current_last."'
AND s.id = p.tid
ORDER BY status, playerLast";

$results = mysql_query($sql);



echo '<div class="roster">';

$team = mysql_fetch_assoc($results);


echo '<div class="roster_team_info">' . $current_first . ' ' . $current_last . '

<div class="school">' . $team['school'] . '</div>
<div class="coach">Coach ' .$team['coachFirst'] . ' ' . $team['coachLast'] .'</div>
<div>Sectional: ' . $team['sectional'] . '</div>
<div>Class: ' . $team['class'] . 'A</div>
';

echo '</div>';

echo '<div class="roster_player_list">';

  $currentStatus = false;  //Flag to detect change in status
    while($player = mysql_fetch_assoc($results))
    {


        if($currentStatus != $player['status'])
        {
            //Status has changed, display status header
            $currentStatus = $player['status'];
echo '<br><b>';
if ($currentStatus == '1')  {echo 'Returning Starters';}
elseif ($currentStatus == '2')  {echo 'Key Returners';}
elseif ($currentStatus == '3')  {echo 'Varsity Newcomers';}
elseif ($currentStatus == '4')  {echo 'Key Freshmen';}
echo '</b><br>';
         
        }
        //Display player info
        echo $player['playerFirst'] . ' ' . $player['playerLast'] . ', ' . $player['feet'] . '\'' . $player['inches'] . '",' . $player['position'] . ', ' . $player['year'] . ';<br>';
 
    }

Hi all,

I'm recent working on my php to extract some data from mysql database. Is it possible to output the database data into <a href> turn to the hyperlink instead of using http://mysite.freesite.com/mydatalink.com to this http://mydatalink.com?

The script I have made so far is doing something wrong and I can't be able to find the solution.

here's the current code:


<?php
session_start();
    define('DB_HOST', 'localhost');
    define('DB_USER', 'mydbuser');
    define('DB_PASSWORD', 'mydbpass');
    define('DB_DATABASE', 'mydbname');

$errmsg_arr = array();
$errflag = false;

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$link) {
    die('Failed to connect to server: ' . mysql_error());
}

$db = mysql_select_db(DB_DATABASE);
if(!$db) {
    die("Unable to select database");
}   

function clean($var)
{
    return mysql_real_escape_string(strip_tags($var));
}
$username = clean($_GET['user']);

if($errflag) {
    $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
    echo implode('<br />', $errmsg_arr);
} else {
    $insert = array();
    if(isset($_GET['user'])) {
  $insert[] = 'username = \'' . clean($_GET['user']) .'\'';
    }

    if (count($insert)>0) {
  $names = implode(',',$insert);
  if($username) {
  $qrytable1="SELECT id, links FROM mydatalist WHERE username='$username'";
  $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error());

  while ($row = mysql_fetch_array($result1)) {  
  echo '<p id="images"> <a href="images.php?id='.$row['id'].'">Images</a></td> | <a href="' .  $row["links"] . '">Link</a> </td>';
    }
  }
    }
} 
?>


I don't want to adding an <a href> into the first echo statement, because I am stored the links in the database which you can see that I am using ['links']. I need to ouput the data from the db to turn into hyperlink.

If anyone could assist with this I'd be very grateful!

Thanks,
Mark

Hey Guys,

I'm struggling with some logic of how to do something wondering if anyone can point me in the right direction.

Details:
I am making a PHP/MySQL game to learn more PHP, I've got my login/logout system created. However I am planning on the game been map based on a 100x100 grid. I have two tables:

USERS
Username, Password, x, y
Test, Test, 1, 1


MAP
X, Y, type
0,1,1
0,2,1
0,3,2
0,4,3
0,5,2
etc

I am stuck on how to get the user's logged in X and Y coordinates and then compare them with the map table X and Y, to see what type of field to show, 1,2 or 3.

Is there a name for this type of function so I can do some research on it? Or could someone give me an example? Would be fantastic if anyone has any advice.

Thanks in advance
NewcastleFan

trying to create a simpleprogram that will read a text file and output information and calculations using the data in the text file. I have 4 radio buttons which represent an item Number. when one is selected the output form should print to a table the ID,Part,Count, Price, and the inventory Value= ($count * $price), can anyone tell me what i am doing wrong?
This is what the .txt file looks like:

AC1000:Hammers:122:12.50
AC1001:Wrenches:5:5.00
AC1002:Handsaws:10:10.00
AC1003:Screwdrivers:222:3.00

Here's what i have so far:
Code: [Select]
<?php
$inf = 'infile.txt';
$FILEH = fopen($inf, 'r') or die ("Cannot open $inf");
$inline = fgets($FILEH,4096);
$found = 0;
//$ptno = 
//if (isset($_POST['AC1000']) || isset($_POST['AC1000']) || isset($_POST['AC1000']) || isset($_POST['AC1000'])) {
while (!feof($FILEH) && !($found)){
list($ptno,$ptname,$num,$price) = split (':', $inline);
if ($ptno == $id) {
print '<table border=1>';
print '<th> ID <th> Part <th> Count <th> Price';
print "<tr><td> $ptno </td><td>$ptname</td>";
print "<td> $num </td><td> \$price</td><tr>";
print '</table>';
$found = 1;
}
$inline = fgets($FILEH,4096);
}
if ($found !=1) {
print "Error: PartNo=$id not found";
}
fclose ($FILEH);
?>

Hi, I need a quick tip. I want this script to produce an output of "Table created" if it succesfully creates a table.

<?php
$con = mysql_connect("localhost","****","****");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

// Create table
mysql_select_db("my_db", $con);
$sql = "CREATE TABLE People
(
FirstName varchar(15),
LastName varchar(15),
Age int
)";

// Execute query
mysql_query($sql,$con);

mysql_close($con);
?> 

I tried to do it on my own, but I guess I'm still completely new to php 
I tried:
<?php
$con = mysql_connect("localhost","****","****");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

// Create table
if (mysql_select_db("my_db", $con))
  {
  echo "Table created";
  }
else
  {
  echo "Error creating Table: " . mysql_error();
  }
$sql = "CREATE TABLE People
(
FirstName varchar(15),
LastName varchar(15),
Age int
)"


// Execute query
mysql_query($sql,$con);

mysql_close($con);
?> 

It gave me an error with the last two lines
mysql_query($sql,$con);

mysql_close($con);

 I removed them and it really did give the output "Table created" yet it didn't create the table in the database. I'm having a feeling that I need to add an "if" statement, something of the sort.

if (sql(CREATE TABLE Persons))

But I'm also getting the feeling something there is wrong, I still haven't tried it. 

Help?

Hi,

Stuck again with the Joomla/K2 coding.  Trying to output this field in a 2 column table but failing badly.  Any help or ideas would be greatly appreciated.

     <?php if($this->item->params->get('catItemExtraFields') && count($this->item->extra_fields)): ?>
     <!-- Item extra fields -->
     <div class="catItemExtraFields">
        <h4><?php echo JText::_('K2_ADDITIONAL_INFO'); ?></h4>
        <ul>
         <?php foreach ($this->item->extra_fields as $key=>$extraField): ?>
         <?php if($extraField->value): ?>
         <li class="<?php echo ($key%2) ? "odd" : "even"; ?> type<?php echo ucfirst($extraField->type); ?> group<?php echo $extraField->group; ?>">
            
            
               <?php if ($extraField->name =='Closing date') : ?>

                    <span class="catItemExtraFieldsLabel"><?php echo $extraField->name; ?></span>
                      <span class="catItemDateCreated"><?php echo JHTML::_('date', $this->item->publish_down ); ?></span>
                    <?php else :?>
                     <span class="catItemExtraFieldsLabel"><?php echo $extraField->name; ?></span>
                    <span class="catItemExtraFieldsValue"><?php echo $extraField->value; ?></span>
       
                    <?php endif; ?>
               
         </li>
         <?php endif; ?>
         <?php endforeach; ?>
         </ul>
       <div class="clr"></div>
     </div>
     <?php endif; ?>
 

Hello to all,

I have problem figuring out how to properly display data fetched from MySQL database in a HTML table.

In the below example I am using two while loops, where the second one is nested inside first one, that check two different expressions fetching data from tables found in a MySQL database. The second expression compares the two tables IDs and after their match it displays the email of the account holder in each column in the HTML table.

The main problem is that the 'email' row is displayed properly while its while expression is not nested and alone(meaning the other data is omitted or commented out), but either nested or neighbored to the first while loop, it is displayed horizontally and the other data ('validity', 'valid_from', 'valid_to') is not displayed.'

Can someone help me on this, I guess the problem lies in the while loop?
Here is part of the HTML code:

<thead>
 <tr>
 <th data-column-id="id" data-type="numeric">ID</th>
 <th data-column-id="email">Subscriber's Email</th>
 <th data-column-id="validity">Validity</th>
<th data-column-id="valid_from">Valid From</th>
 <th data-column-id="valid_to">Valid To</th>
</tr>
</thead>

Here is part of the PHP code:

 

    <?php
    while($row = $stmt->fetch(PDO::FETCH_ASSOC))
    {
        echo '
            <tr>
                <td>'.$row["id"].'</td>
        ';
        while ($row1 = $stmt1->fetch(PDO::FETCH_ASSOC)) {
        echo '
                <td>'.$row1["email"].'</td>
        ';
        }
        if($row["validity"] == 1) {
            echo '<td>'.$row["validity"].' month</td>';
        }else{
            echo '<td>'.$row["validity"].' months</td>';
        }
        echo '  <td>'.$row["valid_from"].'</td>
                <td>'.$row["valid_to"].'</td>
        </tr>';
    }
    ?>

P.S. Please know that I have purposely omitted the connection part and the methods in the class that are used in this code to keep the spot of the problem as clean as possible. If you need me to rephrase the question, please ask.

Thank you.

I'm having some trouble with displaying my white space between words. I have a table that holds records for resumes, and the administrator can access each record and edit them. I have the record displayed in text boxes for editing, but the white space between words is not showing up. Only the first word will show up. I have tried changing the field type in the table to text or blob instead of varchar, and I have tried using the nl2br function when displaying the records, but that isn't working either. What am I doing wrong? Here is my edit page:

Code: [Select]
<?php

$applicant = "SELECT * FROM applicants WHERE id = '$_GET[id]'";
$applicant_res = mysql_query($applicant, $conn) or die(mysql_error());

$display_block = "<h3>Edit Entry</h3>
<p>To edit the entry, change any relevant information below:</p>

<form method=post action=update_U3IP.php><table>";

while ($applicants_list = mysql_fetch_array($applicant_res)){
$id = $applicants_list['id'];
$firstname = $applicants_list['firstname'];
$lastname = $applicants_list['lastname'];
$address = $applicants_list['address'];
$city = $applicants_list['city'];
$state = $applicants_list['state'];
$zip = $applicants_list['zip'];
$phone = $applicants_list['phone'];
$position = nl2br($applicants_list['position']);
$file = $applicants_list['resume'];

$display_block .= "
<table>
<tr>
<td><label for=id><b>ID:</b></label></td>
<td><input name=id type=text size=2 id=id value=$id /></td>
</tr>
<tr>
<td><label for=firstname><b>First Name:</b></label></td>
<td><input name=firstname type=text size=20 id=firstname value=$firstname /></td>
</tr>
<tr>
<td><label for=lastname><b>Last Name:</b></label></td>
<td><input name=lastname type=text size=20 id=lastname value=$lastname /></td>
</tr>
<tr>
<td><label for=address><b>Address:</b></label></td>
<td><input name=address type=text size=20 id=address value=$address /></td>
</tr>
<tr>
<td><label for=city><b>City:</b></label></td>
<td><input name=city type=text size=20 id=city value=$city /></td>
</tr>
<tr>
<td><label for=state><b>State:</b></label></td>
<td><input name=state type=text size=20 id=state value=$state /></td>
</tr>
<tr>
<td><label for=zip><b>Zip:</b></label></td>
<td><input name=zip type=text size=20 id=zip value=$zip /></td>
</tr>
<tr>
<td><label for=phone><b>Phone:</b></label></td>
<td><input name=phone type=text size=20 id=phone value=$phone /></td>
</tr>
<tr>
<td><label for=position><b>Position:</b></label></td>
<td><input name=position type=text size=20 id=position value=$position /></td>
</tr>
<tr>
 <td><b>Uploaded Resume:</b></td>
 <td><input type=text name=file id=file value=$file /></td>
 </tr>
 <tr>
 <td colspan=2>&nbsp;</td>
 </tr>
 <tr>
 <td colspan=2 align=center>
 <input type=hidden name=hidden_id id=hidden_id value=$id />
<input type=submit value=Submit />
<a href=admin_U3IP.php><input type=button name=back id=back value=Back /></a></td>
</tr>";

}

$display_block .= "</table></form>";

?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>KulaE_WDP4451_U3IP</title>
</head>
<body>
<?php echo $display_block; ?>
</body>
</html>