PHP - Using Php And Mysql To Navigate A Database

I'm trying to use php code that is stored in the sql database, but It doesn't seem to be executing the code.  when I see the page source, its there but the server is not executing the command how do I accomplish this.

Here is a simple code snippet to show what I am trying to do.

$result = mysql_query("select * from data");
$row = mysql_fetch_array($result);
echo $row['code'];

In the code field in data table this is whats there.
<?php echo "testing."; ?>

Ok, I got someone to help me fix this but he had no idea what the error was...

I have 2 tables, one called points and the other called members.

In members i have got:
id
name

In points i have got:
id
memberid
promo

I have the following code:
Code: [Select]
<?php
$con = mysql_connect("localhost","slay2day_User","slay2day");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("slay2day_database",$con);
$sqlquery="SELECT Sum(points.promo) AS score, members.name, members.id = points.memberid Order By members.name ASC";
$result=mysql_query($sqlquery,$con);
while ($row = mysql_fetch_array($result))
{

//get data
$id = $row['id'];
$name = $row['name'];
$score = $row['score'];

echo "<b>Name:</b> $name<br />";
echo "<b>Points: </b> $score<br />" ;
echo "<b>Rank: </b>";
if  ($name == 'Kcroto1'):
  echo 'The Awesome Leader';
else:
if ($points >= '50'):
  echo 'General';
elseif ($points >= '20'):
  echo 'Captain!';
elseif ($points >= '10'):
  echo 'lieutenant';
elseif ($points >= '5'):
  echo 'Sergeant';
elseif ($points >= '2'):
  echo 'Corporal';
else:
  echo 'Recruit';
endif;
endif;
echo '<br /><br />';

}

?>
I am getting the following error when i do the query in mysql:
Code: [Select]
#1109 - Unknown table 'points' in field list
And when i open the webpage i get the following error:
Code: [Select]
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/slay2day/public_html/points/members.php on line 18
Please Help me?

Hi,

I cant connect to my Mysql database.
I get this problem:
Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'esolarch_databas'@'localhost' (using password: YES) in /home7/esolarch/public_html/new/storescripts/connect_to_mysql.php on line 21
could not connect to mysql

Code: [Select]
<?php  
/*  
1: "die()" will exit the script and show an error statement if something goes wrong with the "connect" or "select" functions. 
2: A "mysql_connect()" error usually means your username/password are wrong  
3: A "mysql_select_db()" error usually means the database does not exist. 
*/ 
// Place db host name. Sometimes "localhost" but  
// sometimes looks like this: >>      ???mysql??.someserver.net 
$db_host = "localhost"; 
// Place the username for the MySQL database here 
$db_username = "esolarch_database";  
// Place the password for the MySQL database here 
$db_pass = "Password";  
// Place the name for the MySQL database here 
$db_name = "esolarch_admin2"; 

// Run the actual connection here  
mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
mysql_select_db("$db_name") or die ("no database");              
?>

Hi,

I wonder if you could help me try to find what i'm looking for, i have a problem with bogus users on my site.

I created a register page where the details are sent to the database, this is fine but someone is registering with the username: 1 and password: 1 multiple times.
I have about 50 of these now and i would like to know what to actually search for (how to word it) to find out how to stop this?

What would be the name of the script? i've looked for fake username script, multiple username/password prevention script, i'm just not getting it, sorry.

If any of you have any ideas i'd like to hear from you, many thanks in advance for that.

If you need anymore to go on please ask, once again thank you.

Im trying to connect to a database from php. Heres the code:
<?php
  $dbc = mysqli_connect('192.168.0.122', 'boyyo', 'KiaNNa11', 'aliendatabase')
    or die('Error connecting to MySQL server.');
   
   
  $query = "INSERT INTO aliens_abduction (first_name, last_name, " .
    "when_it_happened, how_long, how_many, alien_description, " .
   "what_they_did, fang_spotted, other, email) " .
   "VALUES ('Sally', 'Jones', '3 days ago', '1 day', 'four', " .
   "green with six tentacles', 'We just talked and played with a dog', " .
   "'yes', 'I may have seen your dog. Contact me.', " .
   "'sally@gregs-list.net')";
   
   $result = mysqli_query($dbc, $query)
   or die('Error querying database.');
   ?>

I think i have a theory that my MySQL server location is wrong but i dont know. I use HostGator to do this and im using Phpmyadmin. But everytime i type in a form that i created it says Error querying database. Can someone tell me whats wrong with this code. Oh by the way im using head first into PHP and MySQL

Okay, I have been following this tutorial: http://www.freewebmasterhelp.com/tutorials/phpmysql/1 to achieve exactly what I wanted to get done. I also followed the way to have it formatted in tables and added extra columns that I needed, included an "Options" column which houses three links, including "edit" and "delete"

Now, I have everything working fine but I am stumped on how to get the "edit" and "delete" links to work for each individual entry that is listed. I have to have these features so the entries can be edited and deleted without having to physically go into the MySQL database to do it. The tutorial explains how to do it in Step 6, but I am confused. I'm not quite sure where to place the code for the links, which are generated automatically every time a new entry is inputted into the database.

Anybody available to help me out? Thanks!

Hi guys,

I need your help. I am checking on a database as I want to see if I have the same value in the url and in the database.


Code: [Select]
<?php
session_start();
    define('DB_HOST', 'localhost');
    define('DB_USER', 'mydbuser');
    define('DB_PASSWORD', 'mydbpass');
    define('DB_DATABASE', 'mydbtable');
       
    $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']);
    $password = clean($_GET['pass']);
    $test = clean($_GET['test']);
    $public = clean($_GET['public']);
   
   
   if (isset($_GET['user']) && (isset($_GET['pass']))) {
    if($username == '' || $password == '') {
  $errmsg_arr[] = 'username or password are missing';
  $errflag = true;
    }
} elseif (isset($_GET['user']) || (isset($_GET['test'])) || (isset($_GET['public']))) {
    if($username == '' || $test == '' || $public == '') {
  $errmsg_arr[] = 'user or others are missing';
  $errflag = true;
    }
  }

    if($errflag) {
  $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
  echo implode('<br />',$errmsg_arr);
   }
   else {
  $qry="SELECT * FROM members WHERE username='$username' AND passwd='$password'";
  $result=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error());


if ($username && $password) {
  if(mysql_num_rows($result) > 0) {
     $qrytable1="SELECT images, id, test, links, Public FROM user_list WHERE username='$username'";
    $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error());

    while ($row = mysql_fetch_array($result1)) {
        echo "<p id='test'>";
        echo $row['test'] . "</p>";
        echo '<p id="images"> <a href="images.php?test=test&id='.$row['id'].'">Images</a></td> | <a href="http://' .  $row["links"] . '">Link</a> </td> | <a href="delete.php?test=test&id='.$row['id'].'">Delete</a> </td> | <span id="test">'.$row['Public'].'</td>';
     }
   } else {
    echo "user not found";
  }
} elseif($username && $test && $public) {
    $qry="SELECT * FROM members WHERE username='$username'";
    $result1=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error());
    
    if(mysql_num_rows($result1) > 0) {     
      $qrytable1="SELECT Public FROM user_list WHERE username='$username' && test='$test'";
      $result2=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error());
      
      if(mysql_num_rows($result2) > 0) {
        $row = mysql_fetch_row($result2);
        mysql_query("UPDATE user_list SET Public=('$_GET[public]') WHERE username='$username' AND test='$test'");
        echo "update!";
      } else {
        echo "already updated!";
      }
    } else {
      echo "user not found";
    }
  }
} 
?>
This is the function I use to check the value in the database:

Code: [Select]
if (mysql_affected_rows($result2) > 0) {
         mysql_query("UPDATE user_list SET Public=('$_GET[public]') WHERE username='$username' AND test='$test'");
         echo "you have update it!";
       } else if (mysql_affected_rows($result2) < 0) {
         echo "it is not on the database";
       } else {
         echo "you have already updated!";
       }

When i input the different value in a url bar while the records are not the same as the value in the url and in the database, i can't get passed and I am keep getting "you have already updated!!" when the value in a database are different than I have input in a url.

Do you know how i can get pass it when I have input the different value in the url while it is not the same in the database?

Any advice would be much appreicated.

Thanks,
Mark

There is a "PHP ajax cascading dropdown using MySql" at codestips.com/php-ajax-cascading-dropdown-using-mysql/

I want to use this technique but with a XML or array file instead of mysql database, but my knowledge about mysql is very low. How I can modify this code to catch the categories and products from an array, instead of mysql database?
Code: [Select]
$connect=mysql_connect($server, $db_user, $db_pass)
      or die ("Mysql connecting error");
echo '<table align="center"><tr><td><center><form method="post" action="">Category:<select name="category" onChange="CategoryGrab('."'".'ajaxcalling.php?idCat='."'".'+this.value);">';
$result = mysql_db_query($database, "SELECT * FROM Categories");
$nr=0;
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
 $nr++;
 echo "<option value=".'"'.$row['ID'].'" >'.$row['Name']."</option>";
}
echo '</select>'."\n";
echo '<div id="details">Details:<select name="details" width="100" >';
$result = mysql_db_query($database, "SELECT * FROM CategoriesDetails WHERE CategoryID=1");
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "<option value=".$row['ID'].">".$row['Name']."</option>";
}
echo '</select></div>';
echo '</form></td></tr></table>';
mysql_close($connect);
ajaxcalling.php is
Code: [Select]
include("config.php");
$ID=$_REQUEST['idCat'];
$connect=mysql_connect($server, $db_user, $db_pass);
echo 'Details:<select name="details" width="100">';
$result = mysql_db_query($database, "SELECT * FROM CategoriesDetails WHERE CategoryID=".$ID);
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "<option value=".$row['ID'].">".$row['Name']."</option>";
}
echo '</select>';
mysql_close($connect);

Hi,

I want to make a Shoutbox in Php without using a MySQL-database. This is the code I've got:
Code: [Select]
<?php
$nickname = strip_tags($_POST['nickname']);
$message = strip_tags($_POST['message']);
?>



<html>
<body>
<<<HERE<b><h3><center>Shout Box</center></h3></b>
<iframe src="chat.txt" width="700" height="250"></iframe>
<form method="post" name="form">
Nickname: <input type="text" name="nickname" value="<?php echo $nickname;?>"><br>
Message: <input type="text" name="message" value="<?php echo $message;?>">
<input type="submit" value="Submit">
</form>
<a href="#">Refresh</a>


<?php

if ($_POST['nickname'] && $_POST['message']){
$fp = fopen("chat.txt", "a");
$stuff = $nickname . ": " . $message . "n";
fputs($fp, $stuff);
} else {
echo "Please enter nickname and message.";
}

?>

</body>
</html>
Problem is: it isn't working. :-P
It does read things from chat.txt, but I cant get the input-box working (let people add stuff to chat.txt).
What do I do wrong?

For some reason Im not able to connect to mysql database, when i fill in the form and select search, it just basically refreshes the page but does not come up with no error messages or any results from my database. any help is appreciated.

HTML Code

Code: [Select]

<table id="tb1">
<tr>
<td><p class="LOC">Location:</p></td>
<td><div id="LC">
<form action="insert.php" method="post">
<select multiple="multiple" size="5" style="width: 150px;" >
<option>Armley</option>
<option>Chapel Allerton</option>
<option>Harehills</option>
<option>Headingley</option>
<option>Hyde Park</option>
<option>Moortown</option>
<option>Roundhay</option>
</select>
</form>
</div>
</td>

<td><p class="PT">Property type:</p></td>
<td><div id="PS">
<form action="insert.php" method="post">
<select name="property type" style="width: 170px;">
<option value="none" selected="selected">Any</option>
<option value="Houses">Houses</option>
<option value="Flats / Apartments">Flats / Apartments</option>
</select>
</form>
</div>
</td><td>
<div id="ptype">
<form action="insert.php" method="post">
<input type="radio" class="styled" name="ptype" value="forsale"/> For Sale

<p class="increase">
<input type="radio" class="styled" name="ptype" value="forrent"/> To Rent
</p>
<p class="increase">
<input type="radio" class="styled" name="ptype" value="any"/> Any
</p>
</form>
</div>
</td>
</tr>

</table>

<div id="table2">
<table id="NBtable">
<tr>
<td><p class="NBS">Number of bedrooms:</p></td>
<td><div id="NB">
<form action="insert.php" method="post">
<select name="number of bedrooms">
<option value="none" selected="selected">No Min</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
</select> to

<select name="number of bedrooms">
<option value="none" selected="selected">No Max</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
</select>
</form>
</div>
</td>

<td><p class="PR">Price range:</p></td>
<td><div id="PR">
<form action="insert.php" method="post">
<select name="price range">
<option value="none" selected="selected">No Min</option>
<option value="50,000">50,000</option>
<option value="60,000">60,000</option>
<option value="70,000">70,000</option>
<option value="80,000">80,000</option>
<option value="90,000">90,000</option>
<option value="100,000">100,000</option>
<option value="110,000">110,000</option>
<option value="120,000">120,000</option>
<option value="130,000">130,000</option>
<option value="140,000">140,000</option>
<option value="150,000">150,000</option>
<option value="160,000">160,000</option>
<option value="170,000">170,000</option>
<option value="180,000">180,000</option>
<option value="190,000">190,000</option>
<option value="200,000">200,000</option>
<option value="210,000">210,000</option>
<option value="220,000">220,000</option>
<option value="230,000">230,000</option>
<option value="240,000">240,000</option>
<option value="250,000">250,000</option>
<option value="260,000">260,000</option>
<option value="270,000">270,000</option>
<option value="280,000">280,000</option>
<option value="290,000">290,000</option>
<option value="300,000">300,000</option>
<option value="310,000">310,000</option>
<option value="320,000">320,000</option>
<option value="330,000">330,000</option>
<option value="340,000">340,000</option>
<option value="350,000">350,000</option>
</select> to

<select name="price range">
<option value="none" selected="selected">No Max</option>
<option value="50,000">50,000</option>
<option value="60,000">60,000</option>
<option value="70,000">70,000</option>
<option value="80,000">80,000</option>
<option value="90,000">90,000</option>
<option value="100,000">100,000</option>
<option value="110,000">110,000</option>
<option value="120,000">120,000</option>
<option value="130,000">130,000</option>
<option value="140,000">140,000</option>
<option value="150,000">150,000</option>
<option value="160,000">160,000</option>
<option value="170,000">170,000</option>
<option value="180,000">180,000</option>
<option value="190,000">190,000</option>
<option value="200,000">200,000</option>
<option value="210,000">210,000</option>
<option value="220,000">220,000</option>
<option value="230,000">230,000</option>
<option value="240,000">240,000</option>
<option value="250,000">250,000</option>
<option value="260,000">260,000</option>
<option value="270,000">270,000</option>
<option value="280,000">280,000</option>
<option value="290,000">290,000</option>
<option value="300,000">300,000</option>
<option value="310,000">310,000</option>
<option value="320,000">320,000</option>
<option value="330,000">330,000</option>
<option value="340,000">340,000</option>
<option value="350,000">350,000</option>

</select>
</form>
</div>
</td>


</tr>

</table>

PHP Code
Code: [Select]

<?php
$server = "";      // Enter your MYSQL server name/address between quotes
$username = "";    // Your MYSQL username between quotes
$password = "";    // Your MYSQL password between quotes
$database = "";    // Your MYSQL database between quotes

$con = mysql_connect($server, $username, $password);       // Connect to the database
if(!$con) { die('Could not connect: ' . mysql_error()); }  // If connection failed, stop and display error

mysql_select_db($database, $con);  // Select database to use

$result = mysql_query("SELECT * FROM tablename");  // Query database

while($row = mysql_fetch_array($result)) {     // Loop through results
  echo "<b>" . $row['id'] . "</b><br>\n";      // Where 'id' is the column/field title in the database
  echo $row['location'] . "<br>\n";            // Where 'location' is the column/field title in the database
  echo $row['property_type'] . "<br>\n";       // as above
  echo $row['number_of_bedrooms'] . "<br>\n";  // ..
  echo $row['purchase_type'] . "<br>\n";       // ..
  echo $row['price_range'] . "<br>\n";         // ..
}

mysql_close($con);  // Close the connection to the database after results, not before.
?>

To connect to my database I an entering mysql database details just at the first top lines server,username,password and database. is thats correct?

Thank You for your help in adavance.

Can you upload videos to a mysql database? If so how would you go about doing it?

Hello,
I've been having trouble connecting to a MySQL database, I can't find the problem in the code, what am I doing wrong?

Getting the database file in the config file : require_once("db_connect.php");

db_connect.php :
<?php
$db = mysql_connect('83.172.155.14:3306', 'username', 'password') or die(mysql_error());
mysql_select_db('databasename', $db) or die(mysql_error());
?>

I need to connect to a PhpMyAdmin database.
I need this fixed asap since I'm doing this for someone and he wants the site done as quickly as possible.
P.S: The database used to work in php4 and now I need it to work on php5
Thanks in advance,

Hello guys. Trying to connect php with mysql database and then display results on the screen. This is my code:

Code: [Select]
<?php

$dbhost = "localhost";
$dbuser = "username1";
$dbpass = "password1";
$db = "username1_myDB";

$connection = mysql_connect($dbhost, $dbuser, $dbpass)
or die ("Could not connect");
mysql_select_db($connection, $db);

$show = "SELECT Name, Description FROM people";
$result = mysql_query($show);


while($show = mysql_fetch_array($result)){
$field01 = $show[Name];
$field02 = $show[Description];
echo "id: $field01<br>";
echo "description: $field02<p>";
}


?>
However im getting this:
Warning: mysql_select_db() expects parameter 1 to be string, resource given in /home/pain33/public_html/index.php on line 20

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/pain33/public_html/index.php on line 26

Any ideas how to fix this? Thank you.

Hello all, first post here so i hope i'm doing this right and am putting this into the right place.
Im in the process of integrating a blog into my website, mostly it's set up however i'm currently working on the code to update the posts (theres only 2 parts a title and a comments[which is in actual fact the post content itself]). The code succsefully completes without any errors but for some reason it does not actually update the mysql database.. the code i'm using is as follows:

Code: [Select]
<?php
require_once('header.php');
include "../blog/blogconfig.php";

if(isset($_POST['submit'])){
$update="UPDATE eq_blogarticle SET title='".$_POST['title']."',comments='".$_POST['comment']."' WHERE artid='$aid'";
if(!mysql_query($update)){
echo mysql_error();
}else{
header("location:blog.php?action=listmsgs");
exit;
}
}
?>
<?php
// get value of aid that sent from address bar by blog.php?action=listmsgs
$aid=$_GET['aid'];

// Retrieve data from database
$sql="SELECT * FROM eq_blogarticle WHERE artid='$aid'";
$result=mysql_query($sql);

$row=mysql_fetch_array($result);
?>

<form name="form2" method="post" action="update.php">
<script type="text/javascript">var SITE_URL="<?php echo SITE_URL;?>";</script>
<script type="text/javascript" src="<?php echo SITE_URL;?>/includes/js/nicEdit.js"></script>
<script type="text/javascript">
//<![CDATA[
bkLib.onDomLoaded(function() { nicEditors.allTextAreas() });
//]]>
</script>
<body>
<table width="100%" border="0" cellspacing="1">
  <tr>
    <td colspan="2" class="temptitle">Equidisc Blog</td>
  </tr>
  <tr>
    <td width="74%" valign="top">
<table>
Edit Blog Post
<br>
<br>
Title:
<br>
<input name="title" type="text" class="input" id="title" value="<? echo $row['title']; ?>">
<br>
<br>
<span class="style1 style2 style3">Blog Post:</span>
<br>
<br>
<textarea name="comments" cols="55" rows="12" class="input" id="comments"><? echo stripslashes($row['comments']); ?></textarea>
<br>
<br>
<input name="aid" type="hidden" id="aid" value="<? echo $row['artid']; ?>">
<input type="submit" name="submit" value="Submit">
</form>
</table>
</td>
    <td width="26%" valign="top"><table width="100%" border="0" cellspacing="1">
      <tr>
        <td colspan="2"><img src="../blog/images/fb.gif" width="16" height="16" /> <strong>Blog Menu</strong></td>
        </tr>
       <tr>
        <td><a href="blog.php">Home</a></td>
       </tr>
      <tr>
        <td><a href="blog.php?action=newblogpost">New Post </a></td>
        </tr>
      <tr>
        <td><a href="blog.php?action=listmsgs">Manage Posts </a></td>
        </tr>
    </table></td>
  </tr>
</table>
</body>
<?php
require_once('footer.php');
?>
Explanation: header/footer.php obvious
../blog/blogconfig.php holds my mysql connection settings and connects to the sql, this is the same config as is used for creating the new posts which i have no issues with so i dont think the issue lays there. If i run the query on phpmyadmin with dummy data it works fine and updates the entry..
Any help would be very much appreciated as i'm at the end of my tether with this!.
Thanks in advance.
Jo

Hi there,
I have been creating a website which shows products of the companys (www.theadventurestartshere.org) and I have been trying to make some filters for the products using URL Parameters and recordsheet filters...

Can someone advise me on the easiest way to do this? As I have created a method to do it with, (check website) but it has to be entered manually and I was hoping there is an easier way?

Please Note I like to use drop down boxes for the filters but if there is a way to do the checkbox style you see on say Amazon then that would be brilliant...

I use Dreamweaver CS5, and the newest versions of PHP and MySql

Many Thanks,
Paul