PHP - Cant Get Table To Show Up In Html Area

Hi, okee so I'm new to everything that has to do with scripting, especially when it comes to PHP and stuff,
but what is the best way to display information from a table on a page?
I'm creating this site where users can post their events, but I want the Location to be based on the Location they inserted during the registration process, so that the location will be displayed directly.
How am I able to do that? I've already got an login script, it looks like this:
Code: [Select]
<?php
ob_start();
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="whats_happening_db"; // Database name
$tbl_name="members"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

// Define $myusername and $mypassword
$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];

// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);

$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row

if($count==1){
// Register $myusername, $mypassword and redirect to file "login_success.php"
session_register("myusername");
session_register("mypassword");
header("location:membersarea.php?user=$myusername");
}
else {
echo "Wrong Username or Password";
}

ob_end_flush();
?>
Sorry if my English is terrible haha.

Dante.

Hello Guys.   I am having a trouble viewing a specific column content both in phpMyAdmin and console.It only shows a partial info about the contents inside that column.   I have attached a screenshot of this to give you a demonstration inside that link:   http://i59.tinypic.com/2n6ymqf.jpg   as you can see , the user picked hamburger ,steak and ground beef (and some more), but it seems that it stops showing it right in the middle of the word. Is there any limit for characters in this case?   What would be possible to figure that out please?   Thanks in advance..
Edited by osherdo, 17 December 2014 - 09:11 PM.

Hello, i am coding a backend portal, this portal will have staff members lets call them 'M1' and each staff member will have a client 'C1'

What is the best way to put this into a database?

do i have a table for Staff? then a seperate table for Members? or one table consisting of staff, with the members details written into the staff's row?

I need to copy a table from one code to another. But I couldn't figure out how to do it properly. Just tried to copy code from one file to another but it doesn't work the same. I need to find where to change the code to show right columns.

https://easyupload.io/maoa9h - zipped cart.php (original code with right table) and review.php (the one that doesn't show right table)
 

We need to copy the table from cart.php:
https://ibb.co/MZ7K7S5 - table look

But it doesn't work. The quantity column doesn't show in review.php - as you can see here https://ibb.co/brP0yy1

I´ve reached to

    <?php require_once('Connections/conexxion.php'); ?>
    <?php
    mysql_select_db($database_conexxion, $conexxion);
    $query_consulta = "SELECT venta,compra,taller,regula_mas,regula_menos,movimiento.id_item,sum(compra+regula_mas-venta-taller-regula_menos)  as stock, cilindro, esfera FROM movimiento  join item on item.id_item=movimiento.id_item join rx on rx.id_rx=item.id_rx join cilindro  on cilindro.id_cil=rx.id_cil  join esfera  on esfera.id_esf=rx.id_esf GROUP BY movimiento.id_item ORDER BY esfera desc, cilindro desc";
    $consulta = mysql_query($query_consulta, $conexxion) or die(mysql_error());
    $row_consulta = mysql_fetch_assoc($consulta);
    $totalRows_consulta = mysql_num_rows($consulta);
    ?>
    <!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>test</title>
    </head>
     
    <body>
     
       <?php
    echo "<table align=center>";
     
    $columnes = 10; # Número de columnas (variable)
     
    if (($rows=mysql_num_rows($consulta))==0) {
      echo "<tr><td colspan=$columnes>No hay resultados en la BD.</td></tr> ";
    } else {
      echo "<tr><td colspan=$columnes>$rows Resultados </td></tr>";
    }
     
    for ($i=1; $row = mysql_fetch_row ($consulta); $i++) {
    $resto = ($i % $columnes); # Número de celda del <tr> en que nos encontramos
    if ($resto == 1) {echo "<tr>";} # Si es la primera celda, abrimos <tr>
       echo "<td>$row[1]</td>";
    if ($resto == 0) {echo "</tr>";} # Si es la última celda, cerramos </tr>
    }
    if ($resto <> 0) { # Si el resultado no es múltiple de $columnes acabamos de rellenar los huecos
    $ajust = $columnes - $resto; # Número de huecos necesarios
    for ($j = 0; $j < $ajust; $j++) {echo "<td>&nbsp;</td>";}
    echo "</tr>"; # Cerramos la última línea </tr>
    }
    mysql_close($connexion);
    echo "</table>";
    ?>
     
     
     
    </body>
    </html>
    <?php
    mysql_free_result($consulta);
    ?>

it "works", but results doesn´t start at number 1 but number 2   anyways, The intent of this table is to show the stock of a product, a lens, that has a range of Rx, we use to see this range this way, and show the stock this way would be very productive, but I can´t figure out how to achive it, any sugestion??  
Edited by gralfitox, 16 December 2014 - 06:45 AM.

I'm have trouble getting my script to show images in a table across the screen. I want to show the picture and have the description shown below it. Right now it shows it in a column with the name next to it. Any help would be greatly appreciated.
Thanks in advance. Below is the script.

<?php
include 'config1.php';

// Connect to server and select database.
mysql_connect($dbhost, $dbuser, $dbpass)or die("cannot connect");
mysql_select_db("vetman")or die("cannot select DB");


$result = mysql_query("SELECT * FROM $dbname WHERE year = '1954'") or die(mysql_error());
// store the record of the "" table into $row
//$current = '';

// keeps getting the next row until there are no more to get
if($result && mysql_num_rows($result) > 0)

{
    $i = 0;
    $max_columns = 3;
    echo "<table>";
echo "<br>";
    while($row = mysql_fetch_array($result))
   {
       // make the variables easy to deal with
       extract($row);

       // open row if counter is zero
       if($i == 0)

          echo "<tr>";

          echo "<td align=center>";
          ?>
          <div>
               <a><img src="<?php echo $tn; ?>"</a><a><?php echo $title; ?></a></div>



          <?php
          echo "</td>";

       // increment counter - if counter = max columns, reset counter and close row
       if(++$i == $max_columns)
       {
           echo "</tr>";
           $i=0;
       }  // end if
   } // end while
} // end if results

// clean up table - makes your code valid!
if($i > 0)
{
    for($j=$i; $j<$max_columns;$j++)
        echo "<td>&nbsp;</td>";
   echo '</tr>';
}
mysql_close();
 ?>
</table>

Hi,

How can I get others informations on another page by clicking on one row's value

I have this code, but it doesn't work:


while($row = mysql_fetch_array($result))
  {
 echo "<tr><p>";
 
  echo "<th><p>" . $row['fname'] . "&nbsp;&nbsp;&nbsp;" . $row['lname'] . "</p></th>";
  echo"<th><p><a href='student.php?id='".$row['topic']."'\'>".$row['topic']."</p></a></th>";

  echo "</tr>";
  }
echo "</table>";


Student.php


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

mysql_select_db("college", $con);

$result = mysql_query("SELECT * FROM students");
$topic = $_Get['topic'];
while($row = mysql_fetch_array($result))
  {
  
 echo "<tr><p>";
  echo "<th><p>" . $row['month']."&nbsp;" .$row['datetime']."</p></th>";
  echo"<th><p>". $row['topic'] ."</p></th>"; 
  echo "<th><p>" . $row['gender'] . "</p></th>";
  echo "<th><p>" . $row['fname'] . "&nbsp;&nbsp;&nbsp;" . $row['lname'] . "</p></th>";
  echo "<th ><p>" . $row['id'] . "</p></th>";
  echo "</tr>";
  
echo "</table>";

mysql_close($con);
  }
?> 

i am making online jobs portal where people send data after finish work and then we check that update thier earning stat
for that i am confused
i want that if user earn 50$ per month we add those but when i next month add earning of that user that show show also 50$ and if i include 50 more there i can show last month earning one table and in one this month earn and in one table total earning till now and when USer earning total earning reach on 100$ or grater 100$ automatically show him 3 buttons where he can choose how he witdhraw the money and how much minimum $100 after once when he send us request for withdraw money those buttons again disapear and total balance - how much he requested us for withdraw


so tell me how i make table strcute and which code i'll use

Hello,

I need some help. Say that I have a list in my MySQL database that contains elements "A", "S", "C", "D" etc... Now, I want to generate an html table where these elements should be distributed in a random and unique way while leaving some entries of the table empty, see the picture below. But, I have no clue how to do this... Any hints?



Thanks in advance,
Vero

everything seems to be working fine except the form wont show up.
the site is here http://davisgutierrez.com/testing_area/form.php

<?php
if($_POST['formSubmit'] == "Submit") 
    {
if(empty($_POST['formName'])) 
        {
$errorMessage .= "<li>You forgot to enter your Name.</li>";
}

if(empty($_POST['address'])) 
        {
$errorMessage .= "<li>You forgot to enter your address.</li>";
        }
        
if(empty($_POST['city'])) 
        {
$errorMessage .= "<li>You forgot to enter your city.</li>";
}
        
if(empty($_POST['state'])) 
        {
$errorMessage .= "<li>You forgot to enter your state.</li>";
}

if(empty($_POST['zip'])) 
{
$errorMessage .= "<li>You forgot to enter your zip code.</li>";
}

if(empty($_POST['phone'])) 
        {
$errorMessage .= "<li>You forgot to enter your phone number.</li>";
}

    if(empty($_POST['email'])) 
        {
$errorMessage .= "<li>You forgot to enter your email.</li>";
}
        
        }


$varformName = $_POST['formName'];
$varaddress = $_POST['address'];
$varcity = $_POST['city'];
$varstate = $_POST['state'];
$varzip = $_POST['zip'];
$varphone = $_POST['phone'];
$varemail = $_POST['email'];



 
        
        
//generated by fatcow


$link = mysql_connect(''); 
if (!$link) { 
    die('Could not connect: ' . mysql_error()); 
   }  
echo 'Connected successfully'; 
$db_selected = mysql_select_db("learnphp"); 

   if (!$db_selected) {
    die(' blast!' . mysql_error()); 
   }
       if(!empty($errorMessage))  {
   echo("<p>There was an error with your form:</p>\n");
   echo("<ul>" . $errorMessage . "</ul>\n");
 }
     
 

  $fs = fopen("memberdata.csv","a");
 
fwrite($fs,$varformName . ", " .
$varAddress . ", " .
$varCity . "," .
$varState . "," .
$varZip . "," .
$varPhone . "," .
$varEmail . "," .
"/n");
fclose($fs);


//header("Location: thankyou.html");
exit;


 
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<meta http-equiv="Content-type" content="text/html;charset=UTF-8">

<head>
    <title>membership test</title>
</head>

<body>
<?php
    if(!empty($errorMessage)) 
    {
    echo("<p>There was an error with your form:</p>\n");
    echo("<ul>" . $errorMessage . "</ul>\n");
            }
        ?>
       
<form  action="form.php" method="post" name="ohta_membership_application" >
Name: <input type="text" name="formName" maxlenghth="45" value="<?=$varformName;?>" />
Address: <input type="text" name="address" maxlenghth="4" value="<?=$varAddress;?>" />
City: <input type="text" name="city" maxlenghth="4" value="<?=$varCity;?>" />
State: <input type="state" name="state" maxkength="2" value="<?=$varState;?>">
Zip: <input type="text" name="zip" maxlenghth="4" value="<?=$varZip;?>" />
Phone: <input type="text" name="phone" maxlenghth="4" value="<?=$varPhone;?>" />
Email: <input type="text" name="email" maxlenghth="4" value="<?=$varEmail;?>" />
<input type="submit" name="formSubmit"  value="Submit" />
 <input type="reset" name="reset" value="Reset" >

</form>

</body>
</html>

Hi all,

The below example is a workable code, taken from tutor_profile.sql table
Code: [Select]
<?php
    $query = "SELECT tutor_id, religion_id FROM tutor_profile WHERE tutor_id = '" . $_GET['tutor_id'] . "'";
    
$data = mysqli_query($dbc, $query)
or die(mysqli_error($dbc));

// The user row was found so display the user data
if (mysqli_num_rows($data) == 1) {
$row = mysqli_fetch_array($data);

print_r($row);
    if ($row != NULL) {
  $religion_id = $row['religion_id'];
  $tutor_id = $row['tutor_id'];

    }
    else {
      echo '<p class="error">There was a problem accessing your profile.</p>';
    }
  }

<!--Religion-->   
  <tr>
      <td class="label">Religion:</td>
<td>
<select id="religion_id" name="religion_id">
<option value="1" <?php if (!empty($religion_id) && $religion_id == '1') echo 'selected = "selected"'; ?>>Buddhism</option>
<option value="2" <?php if (!empty($religion_id) && $religion_id == '2') echo 'selected = "selected"'; ?>>Christianity</option>
<option value="3" <?php if (!empty($religion_id) && $religion_id == '3') echo 'selected = "selected"'; ?>>Hinduism</option>
<option value="4" <?php if (!empty($religion_id) && $religion_id == '4') echo 'selected = "selected"'; ?>>Islam</option>
<option value="5" <?php if (!empty($religion_id) && $religion_id == '5') echo 'selected = "selected"'; ?>>Taoism</option>
<option value="6" <?php if (!empty($religion_id) && $religion_id == '6') echo 'selected = "selected"'; ?>>Others</option>
</select>
</td>
  </tr>
?>

As you can see I have hard coded the names of the religion in html code
example - <option value="3" <?php if (!empty($religion_id) && $religion_id == '3') echoselected = "selected"'; ?>>Hinduism</option>

And if our record shows that the tutor has previously selected '3', it will reflect as 'hinduism' in his profile. View profile.jpg for example
example - <option value="3" <?php if (!empty($religion_id) && $religion_id == '3') echo 'selected = "selected"'; ?>>Hinduism</option>

In fact, these names can be found in another table called religion.sql, but I hard coded it anyway, without using loop (while function), since there are only 8 names
Code: [Select]
<!--Religion-->  
  <tr>
      <td class="label">Religion:</td>
<td>
<select id="religion_id" name="religion_id">
<option value="1" <?php if (!empty($religion_id) && $religion_id == '1') echo 'selected = "selected"'; ?>>Buddhism</option>
<option value="2" <?php if (!empty($religion_id) && $religion_id == '2') echo 'selected = "selected"'; ?>>Christianity</option>
<option value="3" <?php if (!empty($religion_id) && $religion_id == '3') echo 'selected = "selected"'; ?>>Hinduism</option>
<option value="4" <?php if (!empty($religion_id) && $religion_id == '4') echo 'selected = "selected"'; ?>>Islam</option>
<option value="5" <?php if (!empty($religion_id) && $religion_id == '5') echo 'selected = "selected"'; ?>>Taoism</option>
<option value="6" <?php if (!empty($religion_id) && $religion_id == '6') echo 'selected = "selected"'; ?>>Others</option>
</select>
</td>
  </tr>

Currently I am facing an issue, I guess I will need to use looping, as there are 22 names in another table which I will need to call forth, tutor_educational_level.sql, and the number of names get more and more in other tables.

My question is, how do I pull out the entire list of names into a drop down box and yet showing the selected name which the user has chosen, more elaboration can be seen in profile.jpg. In profile.jpg - as you can see the list of names are shown in the drop down box and the system is able to decipher the chosen name.

Another
Example
1) N level
2) O level
3) A level
4) University

User selected '3', which is A level, and system would still show the list of educational_names in a drop down box,, but selecting A level as the one to appear.

Example
1) N level
2) O level
3) A level (selected)
4) University

It should have the same overall result as the religion which I have stated above, however this time round, it is using looping function (while) to retrieve the entire list of names, select and show the name which the user has chosen


Below is my code, and I know it is wrong, but generally would like to relate my idea across.
Code: [Select]
<?php
<!--Teaching Credentials-->   
  <tr>
      <td class="label">Teaching Credentials:</td>
<td>
<?php
echo '<select name="educational_level" id="educational_level">';

$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
or die(mysqli_error($dbc));
$query = "SELECT tp.educational_id, el.educational_name AS educational_name, el.educational_id AS list_educational_id " . 
  "FROM tutor_profile AS tp " .
  "INNER JOIN * tutor_educational_level AS el USING (educational_id) " .
  "WHERE tp.tutor_id = '" . $_GET['tutor_id'] . "'";
$sql = mysqli_query($dbc, $query)
or die(mysqli_error());



while($data = mysqli_fetch_array($sql)) {
echo'<option value="'.$data['list_educational_id'].'">'.$data['educational_name'].'</option>';
if (!empty ($data['educational_id']) && ($data['educational_id']) == ($data['list_educational_id'])) {
echo 'selected = "selected"';
}
}
echo '</select><br/>';

mysqli_close($dbc);
?>
</td>
  </tr>
?>

I have a page that shows entries in a guestbook I'm making, and below the entries there is supposed to be a form to write an entry. Except, none of the HTML after the script to show the entries shows up on the page. I have no clue what's wrong. 

Here is the script to show the entries. $n is name, $d is date, $s is site (optional), and $m is message.
$file = fopen("posts.txt", 'rb');
flock($file, LOCK_SH);

while(!feof($file)){

$entry = fgetcsv($file, 0, '|');

if(empty($entry)){exit;}

$d = $entry[1]; $n = $entry[2]; $s = $entry[3]; $m = $entry[4];

echo '
<table style="border: #3399AA 1px solid;"><tr style="background: #3399AA; font: bold 10px verdana,sans-serif; color: #FFFFFF;">
<td width="170">'.$n.'</td>
<td align="right" width="170">'.$d.'</td>
</tr>
';

if($s != 'none'){
echo '
<tr><td colspan="2" style="font: 10px verdana,sans-serif; color: #3399AA;">
<b>Site: </b><a href="'.$s.'">'.$s.'</a>
<center><div style="width: 250; height: 1px; background: #3399AA; margin: 10px;"></div></center></td></tr>
';
}

echo '
<tr>
<td colspan="2" style="font: 10px verdana,sans-serif; color: #3399AA;">'.$m.'</td>
</tr></table><br>
';

}

flock($file, LOCK_UN);
fclose($file);

The file it is reading looks like this:
Code: [Select]
|11:51 am, 3rd Nov 2010|Memoria|none|Hello. :)
|11:51 am, 3rd Nov 2010|Memoria|http://sitehere|Hello again.
|11:51 am, 3rd Nov 2010|Memoria|none|How are you doing?
Thanks! 

I know I'm doing it something right, but can someone tell me why only one table is showing up? Can you help me fix the issue?

Heres my code:


function showcoords()
{
echo"J3st3r's CoordVision";
$result=dbquery("SELECT alliance, region, coordx, coordy FROM ".DB_COORDFUSION."");
dbarray($result);
$fields_num = mysql_num_fields($result);

echo "<table border='1'>";
// printing table headers


    echo "<td>Alliance</td>";
    echo "<td>Region</td>";
    echo "<td>Coord</td>";


// printing table rows
while($row = mysql_fetch_array($result))
{
    

    // $row is array... foreach( .. ) puts every element
    // of $row to $cell variable
    foreach($row AS $Cell)
     echo "<tr>";     
     echo "<td>".$row['alliance']."</td>\n";
     echo "<td>".$row['region']."</td>\n";
     echo "<td>".$row['coordx'].",".$row['coordy']."</td>\n";
     echo "</tr>\n";
    
}
echo "</table>";
mysql_free_result($result);

}



I have 2 rows inserted into my coords table. Just frustrated and ignorant to php.

create table mimi (mimiId int(11) not null, mimiBody varchar(255) );
<?php
//connecting to database
include_once ('conn.php');
$sql ="SELECT mimiId, mimiBody FROM mimi";

$result = mysqli_query($conn, $sql );

$mimi = mysqli_fetch_assoc($result);

$mimiId ='<span>No: '.$mimi['mimiId'].'</span>';

$mimiBody ='<p class="leading text-justify">'.$mimi['mimiBody'].'</p>';
?>
//what is next?

i want to download pdf or text document after clicking button or link how to do that

I want to show a cookie of a referral's username on a sign up page. The link is like this, www.mysite.com/signup?ref=johnsmith.

The cookie doesn't show if I go to that url page. But it does show up once I reload the page.  So I'm wondering if it's possible to show the cookie the first time around, instead of reloading the page?

Here is my code.

// This is in the header
$url_ref_name = (!empty($_GET['ref']) ? $_GET['ref'] : null);

if(!empty($url_ref_name)) {

  $number_of_days = 365;
  $date_of_expiry = time() + 60 * 60 * 24 * $number_of_days;
  setcookie( "ref", $url_ref_name, $date_of_expiry,"/");

} else if(empty($url_ref_name)) {
  if(isset($_COOKIE['ref'])) {
    $user_cookie = $_COOKIE['ref'];
  }
} else {}

// This is for the sign up form
if(isset($_COOKIE['ref'])) {
  $user_cookie = $_COOKIE['ref'];
  ?>
  <fieldset>
    <label>Referred By</label>
    <div id="ref-one"><span><?php if(!empty($user_cookie)){echo $user_cookie;} ?></span></div>
    <input type="hidden" name="ref" value="<?php if(!empty($user_cookie)){echo $user_cookie;} ?>" maxlength="20" placeholder="Referrer's username" readonly onfocus="this.removeAttribute('readonly');" />
  </fieldset>
  <?php
}