PHP - Moved: Simple Database Query, I Think!

Hello All,

Trying to write a log-in script using PDO.  But have one question.  With the below when I query the database and even run a print_r on $RES nothing is outputted?

Any ideas?

Thanks



if (isset($_POST['Submit']))
{
 $email  = $_POST['email'];
 $password  = $_POST['password'];
 
 $QUERY = $dbc->prepare("SELECT email, password FROM tbl WHERE email = :email");
 $QUERY->bindParam(':email', $email);
 $QUERY->execute();
 $RES = $QUERY->fetchColumn();

 echo "Email is:".$RES['email'];
 echo "PW is: ".$RES['password'];
 print_r($RES);
}

I just did a massive hardware/software upgrade - first new computer (a Mac) in 6 1/2 years, first MAMP upgrade in several years, etc. I'm now using these software versions: Apache 2.2.21, PHP 5.3.6, MySQL 5.5.9

Just about everything seems to be working fine except my database connections. I appear to be making a simple mistake, but I can't figure out the solution. I've checked out all sorts of online help pages, but I'm only getting more confused.

Rather than ask what the problem is, it might be better to go back to square one and ask how you would write this query from scratch. I pasted my current, somewhat convoluted code below, to give you an idea of how I was doing it.

But here's what I'm trying to do:

Imagine a website with a static page at World: MySite/World. It displays dynamic pages, like these: MySite/World/France, MySite/World/Spain...just as long as "France" or "Spain" match values stored in a database table (gw_geog). In my query, I represent France, Spain and other place names with the variable $MyURL.

So I want to query that database table. If $MyURL = 'Japan' (e.g. MySite/World/Japan), then $result should = 1, fetching a web page. If $result = 0 (e.g. a misspelled URL, like MySite/World/Japax), then it fetches a 404 error page.

If $result = 2 or more (e.g. MySite/World/Georgia - the name of a country and a U.S. state), then it fetches a duplicate note.

I wrote this script years ago, when I didn't know much about PHP/MySQL (I still don't), and I've upgraded my software, so I need to start fresh. Can anyone tell me how YOU would a script to accomplish this task?

Thanks.

* * * * *

Code: [Select]
<?php
// DATABASE CONNECTION
 $conn = mysql_connect("localhost", "Username", "Password");
 
 if (!$conn) {
    echo "Unable to connect to DB: " . mysql_error();
    exit;
}
  
if (!mysql_select_db("db_general")) {
    echo "Unable to select db_general: " . mysql_error();
    exit;
}

// DATABASE QUERY
 $sql = "
  SELECT COUNT(URL)
  FROM gw_geog
  WHERE URL = '$MyURL'
";
$sql_result = mysql_query($sql,$conn);
$result = $sql_result; 

// DO SOMETHING WITH THE RESULTS
switch ($result)
{
 case 1:
 echo "\n";
 include_once($BaseINC."/$MyPHP/inc/B/DB2/Child/World.php");
 include_once($BaseINC."/$MyPHP/inc/B/Child2.php");
 include_once($BaseINC."/$MyPHP/inc/D/Child.php");
 echo "\n";
 break;

 case 0:
 include_once($BaseINC."/404.php");
 break;  

 default:
 // More than one results, as in Georgia (state and republic)
 include_once($_SERVER['DOCUMENT_ROOT']."/Dupe.php");
 echo "\n";
 break;
}
?>

Hello All,

This is my first post.  I have tried getting some help on the PHP channel in IRC but people there always point you to a reference page.  I need actual help and example with my code.  Possible someone to show/highlight what I did wrong and then show me the correction.

First, my code is supposed to create a form.  The form (server_select.php) is a pre-populated form that has a drop down box that queries a table from the database.  The form then sends the info to get_disk.php page to post the results of the users selected query options.  Here is how it's supposed to work:

User goes to webpage.tld/index.php.
They click on "Select Server Details" page (server_select.php).
A form gets loaded
User then has to select from the drop down menu's: server, date range from and to and then hit submit
User gets sent to get_disk.php

The get_disk.php is supposed to show them all of the following info from their selected server and date range:
hostname, device name, device size, current usage, remaining size, percentage used, mount point, time stamp.

All of that is supposed to show when they user hits submit.

Issue:

My submit page seems to work but my "get_disk.php" page doesn't seem to know how to query the database correctly.  Attached are both of my files.  Can someone tell me why my query on the get_disk.php page is not working and how I get it to display my requested info, please?

Is it possible to query my database and display results from a varible that changes each time from a users search?

I know how to get results from a variable like a field name and display them but i was wondering if for example; the user searches for criteria and gets the result there looking for (postcode); they then click that result in the browser and are redirected to another page with lots more information.(company_name and location).

There result from the original search would be different each time but i would like to know if i could take that result and display all other fields in my table associated with that one result.

***Freelistings***
id company_name location      postcode

1 Dave's Storage  London        PO Box1

2 Steve's Trucks   Birmingham WS12 3

3 Sue's Tyres       Manchester  M14 6

4 Paul's               Birmingham WS11 7

***Basicpackge***
id company_name location      postcode

1 Storage Ltd       London       LN12 5TW

2 Fran's Grit        Birmingham WS5 37

3 Raj Walls          Manchester  M14 60

4 Paul's Light's     Birmingham WS12 17

***Premiumuser***
id company_name location      postcode

1 Name1               Location1    PC1

2 Name2               Location2    PC2

3 Name3               Location3    PC3

4 Name4               Location4    PC4

I'm sure this can be done, i'm just getting stuck on how to treat the user's search result. Currently it's called '%$var%'.

Can i use:

$query = "(SELECT company_name, location FROM freelistings
             WHERE company_name, location like '%$var%')         
         UNION           
               (SELECT company_name, location FROM basicpackage             
            WHERE company_name, location like '%$var%')         
         UNION           
        (SELECT company_name, location FROM premiumuser             
            WHERE company_name, location  like '%$var%')         
        ORDER BY postcode"; 

So that would be looking at all three tables, getting the details from each field that is associated with that (result) postcode. I hope i have given enough information to explain my problem i'm having. for me quite a tough one? any help would be appreciated, thanks guys in advance!

I have a database called "postvoting", It's basically to store when somebody votes on a particular posts. I store the post_id that the user votes on, the users id that voted on it, and the date.
What I want to do is find the most popular posts in a given time.

So if 4 people voted on the post with the id of 1, and 2 people voted on the post with the id of 2, I want to count the number of rows with post_id='1'

A non working example what I want would look something like:
$count_votes = mysql_query("SELECT * FROM postvoting WHILE post_id=post_id");
print mysql_num_rows($count_votes);
result: 4

or
$count_votes = mysql_query("SELECT * FROM postvoting GROUP BY post_id");
print mysql_num_rows($count_votes);
result: 4

(counting the number of results in a group)

Hope this isn't too confusing. (I've confused myself with this).

Hi --

This query seems to be problematic because the UPDATE is not being performed.  Could you please take a gander and let me know what is the problem?

BTW, the table "teamy" does contain 630 records.  Thanks in advance!

$sql = "SELECT COUNT( * ) AS records FROM teamy" ;
$result = mysql_query( $sql ) ;
if ( ! $result )
{ die ( __line__ . "_teamy_" . mysql_error() ) ; }

$a_count = mysql_fetch_assoc($result);
if ( $a_count['records'] = 0 )
{
   echo "No records found in teamy." ;
}
else
{ 
   /*** Update r_rost_rma ***/
   $sql = "UPDATE r_rost_rma 
             JOIN teamy
               ON r_rost_rma.student_id = teamy.student_id
              SET r_rost_rma.teachername = teamy.team
            WHERE RTRIM( UPPER( r_rost_rma.localcourse ) ) = 'HOMEROOM'" ;

   $result = mysql_query ( $sql ) ;
   if ( ! $result )
   { die ( __line__ . "_update_r_rost_rma_" . mysql_error() ) ; }
}

Code: [Select]
$ids = implode (",", $ibforums->input['checkbox']);
$time = time();
$ids2 = implode (",", $ibforums->input['pendingusers']);


$DB->query("UPDATE friends_pending SET pending='0',date=$time WHERE id IN ($ids) AND toid IN ($ids2)");
weird thing is, it's not bring up any error's or nothing

$ids2 spits out
2,30

and $ids spits out

9,7

for this particular project

doesn't give me mysql error or nothing, script runs fine.  I have mysql error enabling under $dbquery class so nn to  worry, how to get this to work? 

Can i even use 2 IN's in 1 query or???

The following mysql query is not returning rows like I expect it to. '$update_field' is a variable,  matching an actual field name in table 'users'. 'user_task[1]' is an integer value. What am I missing here?
Code: [Select]
$query_update_user = "UPDATE users SET ".$update_field." = 'Y' WHERE user_no = '".$user_task[1]."'";

I've got to be missing something pretty basic here.. considering the query is pretty basic.

I'm trying to figure out how to pull a query as an array so I can compare it against another array (array_diff)

I'm doing a mysql_fetch_array, and I'm getting an error ( mysql_fetch_array(): supplied argument is not a valid MySQL result resource):
Quote

$checker = "SELECT ID FROM edible_uses";

$result2 = mysql_fetch_array($checker) or die(mysql_error());
//echoing to see if I'm getting what I need.
echo $row['ID'];


I've done a mysql_query and I get results. The table name and all that is correct. I'm stumped. This seems like a pretty simple query? I tried mysql_fetch_assoc as well. Same result? I tried it with an extra set of parenthesis around it. nope.

I'm having a problem with adding icon/avatar to my database..this is a submit form where I just submit the avatar url. I don't know how to change the submit link from avatar_add.php?mode=submit to index.php?mode=submit&x=avatar_add.
http://www.graphics.allmerk-noplay.com/index.php?x=avatar_add


Code: [Select]
<?php
$host =""; // host name
$user =""; // user
$pass =""; // password
$dbase =""; // database name

// connect to the server
mysql_connect("$host","$user","$pass");
// and select the database
mysql_select_db($dbase);

$mode=$_GET["mode"]; // get the mode
$QUERY_STRING = 'x=avatar_add';

if(!isset($mode) || empty($mode)) // if mode is empty, switch to case index
{
$mode='index';
}

switch($mode)
{

case 'index'; // displays the form
?>


<form method="post" action="avatar_add.php?mode=submit">
Avatar URL:<br>
<input type="text" name="avatar_url" size="30"><br><br>
<input type="submit" value="Submit">


<?php
break;

case 'submit'; // form processing mode

$avatar_url=$_POST["avatar_url"];

// check for empty fields
if(empty($avatar_url))

{
echo "Error! Please go back and fill in all the required fields!";
// if you have a footer include, insert it before the die statement
die; // stops the script from executing the rest of the code
}

// let's get the date in YYYY-MM-DD format
$date = date("Y-m-d");

$query="INSERT into icons_db (id,avatar_url) VALUES ('','$avatar_url')";
$result=mysql_query($query) or die ("Couldnot execute query: $query." . mysql_error());

if($result) // if mysql query successful
{ echo "Thank you! The avatar has been added to or database."; }

break;

}

?>

Hi guys, i am currently working on a project that queries an inventory database through the use of radio buttons, for example; If the first button FOR THE Item(HAMMERS)is clicked, the output should be the name of the item, the number sold and the total profit(calculations) into a table. the way i have it now, when io click on the radio button my output comes back as the results for all the items in the table, i want it to display only the information about hammers, then only about each individual item when the corresponding radio button is clicked. any help would be greatly appreciated!!!

Hi
I'm using a couple of queries to search a database.
This works well but is slow as I'm cascading/nesting the queries..

Below is an example of what I mean... (LDAP queries)
Code:

Code: [Select]
<?
   $dn = "cn=user";
     $filter="(objectclass=user)";
     $justthese = array('cn','telephone','guid','ID','email');
     $sr=ldap_search($ds, $dn, $filter, $justthese);
     $info = ldap_get_entries($ds, $sr);
        for ($n=0; $n<$info['count']; $n++) {
$cn = $info[$n]['cn'][0];
$email = $info[$n]['email'][0];
$id= $info[$n]['id'][0];

   $dn = "cn=Device";
     $filter="(objectclass=device)(userid=$id)";
     $justthese = array('cn','system','guid','ID','userid');
     $sr=ldap_search($ds, $dn, $filter, $justthese);
     $info = ldap_get_entries($ds, $sr);
        for ($n=0; $n<$info['count']; $n++) {
$system = $info[$n]['system'][0];
$userid = $info[$n]['userid'][0];

    }
      }
?>
I'm sure there's a better way of writing this... that will make it faster.

Thanks

in database, I have faced error where query does not responded properly it was showing the error message that "too much query, user cannot execute more query".... when I increased the limit, it works for sometime and after sometime, it again generates the error , my question here is: "why the error is occurring ?" "what is the correct way to improve the execution of multiple queries at single instance ??? "

I have a test database set up that I am working on. Right now, just some names like so

firstName
lastName
companyName

I have used phpmyadmin to insert names in the database. Right now just three names are inserted. One first and last name, one first name and one company name

I am getting the results back in an associative array. How can I echo out the information with the break tag, or one field at a time? if I use a break tag, then the loop adds in extra blank lines when it gets to the first name and it's null. Can I test for null values and do it that way? I am a beginner so I have no idea what I am doing.

Code: [Select]
$result = $mysql->query("select * from names") or die($mysql->error);
if($result){
while($row = $result->fetch_assoc()){
$name = $row['firstName'];
$lastName = $row['lastName'];
$company = $row['companyName'];

echo $name . " ";
echo $lastName . " ";
echo $company . " ";
}

}

So I'm creating a website for bar deals... I've tried for hours trying to get the url to take in more than one parameter with no luck, I don't know if my database isn't setup correctly or something but the url would look like this

http://www.site.com/?id=1&day=Monday

and when those parameters are passed the deals for that id and day are then posted in the content area

I'm pretty much stumped. I feel like it's much easier than I am making it.

My database looks like:

ID   Name  Monday           Tuesday        Wednesday -> and so on

1    Bar1   MondayDeal TuesdayDeal WednesdayDeal
2    Bar2   MondayDeal -> and so on