php and sql ?

[mac_shinobi]

Hi

I have phpmyadmin setup on my web server and have created a database called gecko with one table called feedback.

I am trying to create a php page that will execute a the Select field1, field2, field 3 from feedback but am just getting it returning

Connected Succesfully and Resource id #3

What does that mean and how do I fix it ?

<?php
    $user="gecko";
    $host='localhost:/var/run/mysqld/mysqld.sock';
    $password="password";
    $database="gecko";

$link = mysql_connect($host, $user, $password);
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
?>
<p>
<?php
$sql = 'SELECT Forename , Surname , Comment FROM feedback; ';
$result = mysql_db_query($database, $sql) or die("Failed Query of : " . $sql);  //do the query

print($result);
mysql_close($link);    
    
?>

I did find this :

PHP/MYQSL brings back wrong value (resource ID #3) - Dynamic Drive Forums

which basically says

"You cannot echo your raw query. You must first grab the information it is returning with a function such as mysql_fetch_array()"

Can someone correct my php code so that it displays what I have in my database table feedback.

[Jona]

something like

 print_r(mysql_fetch_array($result)); 
  

Should do the job, to display it in a more formatted way you'll want to assign the array to a variable and then do a for loop such as:

 $rows = mysql_fetch_array($result);
foreach($rows as item){
print $item['field1'];
print $item['field2'];
} 
  

Hope that helps.

Cheers
Jona

Ninja edit: To correct typo.

[Gatt]

if u have PHPMyAdmin running, then tehre should be an option somehwere to run a query, then get it do display the PHP coding for that query..

[mac_shinobi]

something like

 print_r(mysql_fetch_array($result)); 
  

Should do the job, to display it in a more formatted way you'll want to assign the array to a variable and then do a for loop such as:

 $rows = mysql_fetch_array($result);
foreach($rows as item){
print $item['field1'];
print $item['field2];
} 
  

Hope that helps.

Cheers
Jona

hi

ok tried that and got this error :

Parse error: syntax error, unexpected ')', expecting T_PAAMAYIM_NEKUDOTAYIM in /home/gecko/www/feedback.php on line 19

[Gatt]

 $rows = mysql_fetch_array($result);
foreach($rows as item){
print $item['field1'];
print $item['field2];
} 
  

Should read

 $rows = mysql_fetch_array($result);
foreach($rows as item){
print $item['field1'];
print $item['field2'];
} 
  

missing ' after field2

[Jona]

As above, typing error, will correct in the orignal post.

[mac_shinobi]

ok lets try again :

<?php
$link = mysql_connect($host, $user, $password);
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
?>
<p>
<?php
$sql = 'SELECT `Forename` , `Surname` , `Comment` FROM feedback LIMIT 0 , 30';
$result = mysql_db_query($database, $sql) or die("Failed Query of : " . $sql);  //do the query

$rows = mysql_fetch_array($result);
foreach($rows as item){
print $item['Forename'];
print $item['Surname'];
print $item['Comment'];
}  


mysql_close($link);    
    
?>

I get that error using the above code which has the commas etc in the correct place yet if I put a dollar symbol on this part :

$rows = mysql_fetch_array($result);
foreach($rows as $item){
print $item['Forename'];
print $item['Surname'];
print $item['Comment'];
}

As you can see in the foreach part I have put a dollar symbol aka $ before the item and then each time I run it it just returns

mmmmmm

What good is that - I have emptied my database and have not got any fields then I go to my add.php which works fine and it adds a record into the database but I have not once entered mmmmmm

[Gatt]

ignore this...

[mac_shinobi]

As you can see from my PHP above I only want it to return

$sql = 'SELECT `Forename` , `Surname` , `Comment` FROM feedback LIMIT 0 , 30';

The Forename, Surname and Comment that each user has made in the order they were posted.

What PHP do I need to do this please.

[Gatt]

This is an exert from one of my PHP scripts:

            <? 
                $query = "SELECT * FROM dvds ORDER BY dvd_name ASC";
                $result = mysql_query($query);
                $numofrows = mysql_num_rows($result);
 
                while ($i < $numofrows) :
                      $information = mysql_result($result, $i, "dvd_name");
                      $i++;
                endwhile;
            ?>  

SO yours would be:

            <? 
                $query = "SELECT Forename, Surname, Comment FROM Feedback LIMIT 0, 30";
                $result = mysql_query($query);
                $numofrows = mysql_num_rows($result);
 
                while ($i < $numofrows) :
                      $information = mysql_result($result, $i, "forename");
                      $i++;
                endwhile;
            ?>  

[mac_shinobi]

This is an exert from one of my PHP scripts:

            <? 
                $query = "SELECT * FROM dvds ORDER BY dvd_name ASC";
                $result = mysql_query($query);
                $numofrows = mysql_num_rows($result);

                while ($i < $numofrows) :
                      $information = mysql_result($result, $i, "dvd_name");
                      $i++;
                endwhile;
            ?>  

SO yours would be:

            <? 
                $query = "SELECT Forename, Surname, Comment FROM Feedback LIMIT 0, 30";
                $result = mysql_query($query);
                $numofrows = mysql_num_rows($result);

                while ($i < $numofrows) :
                      $information = mysql_result($result, $i, "forename");
                      $i++;
                endwhile;
            ?>  

ok tried that and am getting this :

Connected successfully
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/gecko/www/feedback.php on line 19


The script I am using is this :

<?php
    $user="gecko";
    $host='localhost:/var/run/mysqld/mysqld.sock';
    $password="mypassword";
    $database="gecko";

$link = mysql_connect($host, $user, $password);
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
?>
<p>
<?php
$sql = 'SELECT `Forename` , `Surname` , `Comment` FROM feedback LIMIT 0 , 30';
//$result = mysql_db_query($database, $sql) or die("Failed Query of : " . $sql);  //do the query

                //$query = "SELECT Forename, Surname, Comment FROM Feedback LIMIT 0, 30";
                $result = mysql_query($sql);
                $numofrows = mysql_num_rows($result);

                while ($i < $numofrows) :
                      $information = mysql_result($result, $i, "forename");
                      $i++;
                endwhile;
            


mysql_close($link);       
?>

[mac_shinobi]

ok have this code now :

<?php
    $user="gecko";
    $host='localhost:/var/run/mysqld/mysqld.sock';
    $password="mypassword";
    $database="gecko";
$link = mysql_connect($host, $user, $password);
mysql_select_db("gecko", $link);
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
?>
<p>
<?php
//$sql = 'SELECT `Forename` , `Surname` , `Comment` FROM feedback LIMIT 0 , 30';
//$result = mysql_db_query($database, $sql) or die("Failed Query of : " . $sql);  //do the query

                $query = "SELECT Forename, Surname, Comment FROM feedback LIMIT 0, 30";
                $result = mysql_query($query, $link);
                $numofrows = mysql_num_rows($result);

                while ($i < $numofrows) :
                      $information = mysql_result($result, $i, "Forename");
                      $i++;
                endwhile;
            


mysql_close($link);       
?>

It just states that its connected successfully but no errors so how do I make it print each result ie Forname, Surname and Comment ??

[Gatt]

Sorry my fault - add a line to decalre what $i is

 $i=0 
  

Just after your DB details:

 <?php
    $user="gecko";
    $host='localhost:/var/run/mysqld/mysqld.sock';
    $password="mypassword";
    $database="gecko";
 
$i=0;  //Declare counter here..
$link = mysql_connect($host, $user, $password);
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
?>
<p>
<?php
$sql = 'SELECT `Forename` , `Surname` , `Comment` FROM feedback LIMIT 0 , 30';
//$result = mysql_db_query($database, $sql) or die("Failed Query of : " . $sql);  //do the query
 
                //$query = "SELECT Forename, Surname, Comment FROM Feedback LIMIT 0, 30";
                $result = mysql_query($sql);
                $numofrows = mysql_num_rows($result);
 
                while ($i < $numofrows) :
                      $information = mysql_result($result, $i, "forename");
                      $information = mysql_result($result, $i, "Surname");
                      $information = mysql_result($result, $i, "Comment");
                      $i++;
                endwhile;
            
 
 
mysql_close($link);       
?>  

Does it work now?

[mac_shinobi]

Nope

getting this error :

Connected successfully
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/gecko/www/feedback.php on line 21

[Gatt]

Hmm.. I can only imagine its those quote marks..

Here is my full php script..

 <?php
    include ("includes/db_connect.php");
?>
<html>
    <head>
        <title> DVD Library - 
            <?php
                echo gmdate("l, dS F, Y");
            ?>
        </title>
    </head>
    <body>
        <h1> DVD Library....</h1>
        <? 
            $counter = 0;
        ?>
        <form method='post' action='vew_disc.php'>
            <select name='dvds' MULTIPLE size=10>
            <? 
                $query2 = "SELECT * FROM dvds ORDER BY dvd_name ASC";
                $result2 = mysql_query($query2);
                $numofrows2 = mysql_num_rows($result2);
 
                while ($i < $numofrows2) :
                      $information = mysql_result($result2, $i, "dvd_name");
                     print "<option value='$information'>$information</option>";
                      $i++;
                endwhile;
            ?>
  
               </select>
            <BR>
            <!-- <input type='submit' name='submit' value='Submit'> -->
        </form> 

Also, this is how PHPMyAdmin generated the PHP Code for another query..

 $sql = 'SELECT Season , Episode_Title , Stardate FROM `tng` LIMIT 0, 30 ';