PHP - Select From 2 Tables

hi im trying to echo out all the possibilities from one column in 2 different tables, both with the same structure. if that makes sense

this is what i have

$query = mysql_query("SELECT * FROM stanjamesprem,centrebetprem GROUP BY eventname ORDER BY eventtime ");
while($row = mysql_fetch_assoc($query))
   {
   echo $row[eventname];
   }

if i take one of the tables out it works fine but if i have 2 table names it doenst work

please help me

cheers matt

any questions please ask away

This topic is now in MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=357554.0

Hi guys,

This is my first time posting here - im just getting into PHP - i got a question;

I have two databases:

profile(id, name, interests, dob, gender, join_date, email)
interests(id, profile_id, interests)
id being the primary key, and profile_id being the foreign key from profile.

I want to script that returns profile information and all the matching interests (one user can have multiple interests).

This is what i have so far, though it does not work, and i knew it wouldnt;

function get_profile($id) {
   $connection = mysql_open();
   $query = "SELECT * ";
   $query .= "FROM profiles, interests ";
   $query .= "WHERE profile.id=" . $id;
   $query .= " AND interests.profile_id=" . $id;
   $result = @ mysql_query($query, $connection);
   // Transform the result set to an array (for Smarty)
   $entries = array();
   while ($row = mysql_fetch_array($result)) {
      $entries[] = $row;
   }
   mysql_close($connection) or show_error();;
   return $entries;   
}


Can someone please advise on how this can be done? or do i need to have two query's one for each table ?

Thank you in advance!!

Hi ho, I am trying to assemble a table so that it shows article in order by id descending. The current one i am using is working but not displaying as i hoped, i need them aligned on the left and right, so id=1 align left, id=2 align right etc. Here's my current code

Code: [Select]
<table style='table-layout:fixed' align="center" cellpadding="2" cellspacing="0" colspan="" class="Main"><br>

<?
$select_paper=mysql_query("SELECT * FROM paper ORDER BY id DESC");
while($the=mysql_fetch_object($select_paper)){
?>
     <tr>
        <td class="subtableheader" width="25%"><?php echo "$the->title"; ?></td>
      </tr>

     <tr>
        <td class="profilerow"  width="100%"><?php echo "$the->news"; ?></td>
        </tr>
     <tr>
        <td class="subtableheader"  width="25%"><?php echo "Article By $the->by - $the->date"; ?></td>
        </tr>

I'm having a hard time grasping how to select data from several tables.

Here's the code

Code: [Select]
<?php 
require_once "config.php";
$gamedate = $_SESSION['date'];
echo "These umpires are not currently scheduled on this date, and have not asked for the day off:<br />";

$umpirelist = Array('umpire 1 name', 'umpire 2 name', 'umpire 3 name');

$dbc = mysql_pconnect($host, $username, $password);  
mysql_select_db($db,$dbc);  

foreach($umpirelist as $data) { 

//now get stuff from a table           
$sql = "SELECT calendar.date, calendar.ump1, calendar.ump2, calendar.ump3, calendar.ump4, calendar.ump5, vacation.date, vacation.umpire FROM umps, calendar, vacation WHERE umps.full_name = $data AND $gamedate = `calendar.date` AND $gamedate = `vacation.date` AND $data NOT IN ('calendar.ump1', 'calendar.ump2', 'calendar.ump3', 'calendar.ump4', 'calendar.ump5') AND $data NOT IN ('vacation.umpire') order by umps.full_name asc"; 

$rs = mysql_query($sql,$dbc);  
$matches = 0; 
while ($row = mysql_fetch_assoc($rs))  {
$matches++; 
echo "$row[$data]<br />";
}
}
?>

The table CALENDAR holds the date of the game, who is playing, and what umpires are schedule for that game.  it does have a unique "row_number" column as well, but didn't think I'd need it in this select.

The table VACATION has the date, and umpire's name that is on vacation.

The table UMPS contains the umpires name (same names as the array in the code) and email addresses and login info.

I'm getting this error:

Quote

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource on line 152



This is line 152

Quote

while ($row = mysql_fetch_assoc($rs))  {



Can someone please point me in the right direction as to where I'm going wrong?

Thanks!

I want to select the user from super_administrators, administrators, teachers and students, and give the user permission based from what table he "came". But is giving the "Query failed" error...

Code: [Select]
<?php

//Start session

session_start();



//Include database connection details

require_once('../config/config.php');



//Array to store validation errors

$errmsg_arr = array();



//Validation error flag

$errflag = false;



//Connect to mysql server

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if(!$link) {

die('Failed to connect to server: ' . mysql_error());

}



//Select database

$db = mysql_select_db(DB_DATABASE);

if(!$db) {

die("Unable to select database");

}



//Function to sanitize values received from the form. Prevents SQL injection

function clean($str) {



$str = @trim($str);



if(get_magic_quotes_gpc()) {

$str = stripslashes($str);

}



return mysql_real_escape_string($str);

}



//Sanitize the POST values

$email = clean($_POST['email']);

$password = clean($_POST['password']);



//Input Validations

if($email == '') {

$errmsg_arr[] = 'O campo Email nao foi preenchido.';

$errflag = true;

}



if($password == '') {

$errmsg_arr[] = 'O campo Senha nao foi preenchido.';

$errflag = true;

}



//If there are input validations, redirect back to the login form

if($errflag) {

$_SESSION['ERRMSG_ARR'] = $errmsg_arr;

session_write_close();

header("location: ../index.php");

exit();

}



//Create query

$qry = "SELECT * FROM super_administrators,administrators,teachers,students WHERE email = '$email' AND passwd = '".md5($_POST['password'])."'";

$result = mysql_query($qry);

$member = mysql_fetch_assoc($result);

$table = mysql_field_table('$result', '0');



//Check whether the query was successful or not



if($result) {



if(mysql_num_rows($result) == 1) {



if($table == 'super_administrators')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/sadmin/index.php");

exit();

}



if($table == 'administrators')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_SCHOOL_ID'] = $member['school_id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/admin/index.php");

exit();

}



if($table == 'teachers')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_SCHOOL_ID'] = $member['school_id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/prof/index.php");

exit();

}



if($table == 'students')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_SCHOOL_ID'] = $member['school_id'];

$_SESSION['SESS_CLASS_ID'] = $member['class_id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_REGISTRATION'] = $member['registration'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/aluno/index.php");

exit();

}



} else {



$errmsg_arr[] = 'Suas informacoes de login estao incorreta. Por favor, tente novamente.';

$errflag = true;

$_SESSION['ERRMSG_ARR'] = $errmsg_arr;

session_write_close();

header("location: ../index.php");

exit();

}



} else {

die("Query failed");

}

?>

I have 2 tables I have settup called Users_Messages and Users_Message_Replies inside each of these tables I have a row called DateSent, I'm trying to select from both of these tables and only display the latest Sent item by ID and order them all by date. I can get it to display the items correctly using the below code, but I can't get them to order correctly by the latest date in both of the Tables.
Code: [Select]
$page_query = mysql_query("
                         SELECT  MessageID, DateSent
                         FROM Users_Messages
                         WHERE ToID = '$user_ID'
                           
                           UNION
                         
                         SELECT  MessageID, DateSent
                         FROM Users_Messages
                         WHERE FromID = '$user_ID'

                         
                         ORDER BY DateSent DESC
                         ");
 
                      while ($replycheck = mysql_fetch_assoc($page_query)){

                          $message_idmainnew = $replycheck['MessageID'];
                          $date = $replycheck['DateSent'];

                          $sql2 = "SELECT  MainMessageID FROM Users_Message_Replies WHERE MainMessageID = '$message_idmainnew '";
                          $sql_result2 = mysql_query($sql2);
                          $replycheck2 = mysql_fetch_assoc($sql_result2);
                          $newreplyID = $replycheck2['MainMessageID'];
           
                          $sql4 = "SELECT MessageID FROM Users_Messages WHERE MessageID= '$message_idmainnew'";
                          $sql_result4 = mysql_query($sql4);
                          $messagecheck2 = mysql_fetch_assoc($sql_result4);
                          $newmessageID = $messagecheck2['MessageID'];

                          if ($newreplyID == NULL){
                          $sql2 = "SELECT * FROM Users_Messages WHERE MessageID= '$newmessageID' ORDER BY DateSent ASC";
                          $sql_result2 = mysql_query($sql2);
                          $message_row = mysql_fetch_assoc($sql_result2);
                         
                                $message_01 = $reply_row['Message'];
                      $date = $reply_row['DateSent'];
                      $date1 = strtotime($date);
                      $datemain = date('F j, Y, g:i a', $date1);
                     
                          }else{

                          $sql2 = "SELECT * FROM Users_Message_Replies WHERE MessageID= ' $newreplyID ' ORDER BY DateSent DESC";
                          $sql_result2 = mysql_query($sql2);
                          $reply_row = mysql_fetch_assoc($sql_result2);
 
                      $message_01 = $reply_row['Message'];

                           
                      $date = $reply_row['DateSent'];
                      $date1 = strtotime($date);
                      $datemain = date('F j, Y, g:i a', $date1);
                           }

                      {
Is there an easier way to do this? And how would I get the dates to line up, with a Join? Thanks.

Hi,   How can I select multiple columns from two tables and run a search through multiple fields?   My tables a   t_persons (holds information about persons) t_incidents (holds foreign keys from other tables including t_persons table)   What I want is to pull some columns from the two tables above and run a search with a LIKE criteria, something like below. The code originally worked well with only one table, but for two tables it generate errors:

$query = "SELECT
p.PersonID
,p.ImagePath
,p.FamilyName
,p.FirstName
,p.OtherNames
,p.Gender
,p.CountryID
,i.IncidentDate
,i.KeywordID
,i.IncidentCountryID
,i.StatusID

FROM t_incidents AS i
    LEFT JOIN t_persons AS p ON i.PersonID = p.PersonID
WHERE
    FamilyName LIKE '%" . $likes . "%' AND
    FirstName LIKE '%" . $likes . "%' AND
    OtherNames LIKE '%" . $likes . "%' AND
    Gender LIKE '%" . $likes . "%' AND
    IncidentDate LIKE '%" . $likes . "%' AND
    KeywordID LIKE '%" . $likes . "%' AND
    IncidentCountryID LIKE '%" . $likes . "%' AND
    StatusID LIKE '%" . $likes . "%'
ORDER BY PersonID DESC $pages->limit";

Errors a

Column 'IncidentDate' in where clause is ambiguous
Column 'KeywordID' in where clause is ambiguous
Column 'IncidentCountryID' in where clause is ambiguous
Column 'StatusID' in where clause is ambiguous

These columns are foreign keys on t_incidents table. I have also attached the table relationship diagram if it helps.   I will appreciate any better way to do this.   Thanx.   Joseph Attached Files  Model - 3.png   76.6KB   0 downloads

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=348013.0

I am working on a quiz app

image 1 shows the index.php page
image 2 shows the first question
image 3 shows the second question
image 4 shows the third question
image 5 shows the result after completing the quiz
image 6 shows the database 'quizzer' and its tables
image 7 shows the 'questions' table
image 8 shows the 'choices' table

THIS LINK CONTAIN ALL THE CODE (and images) I HAVE DONE SO FAR

https://www.mediafir...o7f5q0fe6y/quiz

1.Now my question is how to select the question RANDOMLY from 'questions' table along with 'choices' (by adding code to the existing file or create a new one).

2.If user refresh/reload the page before starting ('Start Quiz') or click 'Take Again' after finishing the quiz, the question should appear randomly.

3.Basically I want to change the order of question appearing in the browser each time I refresh.

4.My work so far is mentioned above.........Please help me with this "RANDOM" problem !!

P.S - Will it be possible, by creating a random function in PHP which will check for repeat questions in a session and check for the 'id' of the question and if it is new display it on the page.

If so what should I do and if no then how to do?

Below is a page which is supposed to output the name, blog contribution and picture of contributing members of a website.
<div id="blog_content" class="" style="height:90%; width:97%; border:5px solid #c0c0c0; background-color: #FFFFFF;"> <!--opens blog content--> 



                             
<?php


//address error handling

ini_set ('display_errors', 1);
error_reporting (E_ALL & ~E_NOTICE);



//include the config file

require_once("config.php");



//Define the query. Select all rows from firstname column in members table, title column in blogs table,and entry column in blogs table, sorting in ascneding order by the title entry, knowing that the id column in mebers table is the same as the id column in blogs table.


$sql = "SELECT


    blogs.title,blogs.entry,members.firstname,images.image
FROM
    blogs
LEFT JOIN
    members
ON
    blogs.member_id = members.member_id
LEFT JOIN
    images
ON
    blogs.member_id = images.member_id

ORDER BY
     blogs.title ASC

"; 

$query = mysql_query($sql);

                 if($query !== false && mysql_num_rows($query) > 0)

                 {

         while(($row = mysql_fetch_assoc($query)) !== false)

         {

 echo

                
                              '<div id="blog_content1" style="float:left; position:relative;bottom:18px;left:13px; background-color: #FFFFFF; height:16.7%; width:100%; border:0px none none;"                              
                              <!--opens blog_content1  same as main center top 1 and 2 from index page everything scaled down by a factor of 3, heightwise--> 
        
 
                                   <div class="red_bar" style="height:3%; width:100%; border:1px solid #959595;"> <!--a--> 
 
                                       <div class="shade1" style="height:5px; width:100%; border:0px none none;"> </div>                                      
                                       <div class="shade2" style="height:5px; width:100%; border:0px none none"> </div> 
                                       <div class="shade3" style="height:5px%; width:100%; border:0px none none"> </div>

 
                                   </div> <!-- closes red bar--> 
 
 
 
                                   <div class="content" style="height:28.3%; width:100%; border:0px none none;"> <!----> 
 
                                       
                                           <div class="slideshow" id="keylin" style="float:left; width:20%;  border:0px none none;"> <!--a-->
 
                                                  
                                                  
 
                                                 <div><img header("Content-type: image/jpeg"); name="" alt="" id="" height="105" width="105" src="$row[image]" /></div>                              
 
 
                                       
                                           </div> <!-- closes pic--> 
 
 
 
                                           <div class="content_text"  style="float:right;  position:relative;top:7px;left:0px; max-height:150px; width:78.5%; border-width:4.5px; border-bottom-style:solid; border-right-style:solid; border-color:#c0c0c0; "> <!--a-->'; 
 
                                           

                                               
                                                echo "<h3>".$row['title']."</h3>";

echo "<p>" .$row['entry']."<br />".$row['firstname']."</p>";

         
                                             
                                                                                           
                                           echo                                         
                                           '</div> <!-- closes content text--> 
 
  
                                   </div> <!-- closes content--> 
 
                                   
                                                                                                                                                                                                                            
                            </div> <!-- closes blog_content1-->';



                                                 }

                 }

                 else if($query == false)

                 {

               

 echo "<p>Query was not successful because:<strong>".mysql_error()."</strong></p>";

         echo "<p>The query being run was \"".$sql."\"</p>";

                 }

                 else if($query !== false && mysql_num_rows($query) == 0)

                 {

         echo "<p>The query returned 0 results.</p>";

                 }


                                                 mysql_close(); //Close the database connection.





                                                 ?> 

                      
                        </div> <!-- closes blog content-->


The select query is designed to retrieve all the blog contributions(represented by the fields blogs.title and blogs.entry) from the database, alongside the contributing member (member.firstname) and the member's picture(images.image), using the member_id column to join the 3 tables involved, and outputs them on the webpage.   The title, entry and firstname values are successfully displayed on the resulting page.  However, I can't seem to figure out how to get the picture to be displayed.  Note that the picture was successfully stored in the database and I was able to view it on a separate page using a simple select query.  It is now just a question of how to get it to display on this particularly crowded page.  Anyone knows how I can output the picture in the img tag?  I tried placing the header("Content-type: image/jpeg"); statement at the top of the php segment, then just right below the select query and finally just right above the img tag, but in every case, I just got a big white blank page starring at me.  How and where should I place the header statement? And what else am I to do to get this picture displayed?  Any help is appreciated.

Hey there. Not sure how to word this. I've been flying blind as a newbie trying to figure out some pagination for a left joined query. I've got syntax errors trying to set up the SELECT COUNT function that adds up the results of the search on a previous page so it knows how many results matched both tables.

Right now, I've got this mess, and it's giving me a syntax error,
"You have an error in your SQL syntax; .... near 'LEFT JOIN plantae ON (descriptors.plant_id = plantae.pla' at line 12"
Code: [Select]
$data = "Select (SELECT COUNT(*)
                    FROM
                        descriptors
                    ) AS
                        count1,
                (SELECT COUNT(*)
                    FROM
                         plantae
                    ) AS
                        count2

                    LEFT JOIN
                        plantae ON (descriptors.plant_id = plantae.plant_name)
                    WHERE
                        `leaf_shape` LIKE '%$select1%'
                        AND `leaf_venation` LIKE '%$select3%'
                        AND `leaf_margin` LIKE '%$select4%'";
                   
 
$result = mysql_query ($data);
if (!$result) {
    die("Oops, my query failed.  The query is: <br>$data<br>The error is:<br>".mysql_error());
}

I have a photo album style gallery to build
and
i'm finding it dificult to list all the table names (these are names of photo albums) and then enter the data into a seperate query for each album name (these will change often so i cant keep updating the file as normal. this will then post all the data to the xml file and show the set of photos in the individual albums in a flash file.
can anyone help me where im going wrong at all?



<?php
$dbname = 'cablard';

if (!mysql_connect('localhost', 'cablard', '')) {
    echo 'Could not connect to mysql';
    exit;
}

$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);

if (!$result) {
    echo "DB Error, could not list tables\n";
    echo 'MySQL Error: ' . mysql_error();
    exit;
}

while ($row = mysql_fetch_row($result)) {
    echo "Table: {$row[0]}\n";
}

mysql_free_result($result);



$query = "SELECT * FROM photo ORDER BY id DESC"; 
$result2 = mysql_query ($query)  or die ("Error in query: $query. ".mysql_error());

   
   while ($row = mysql_fetch_array($result2))  
   {                                       
         echo  "
<image>
<date>".$row['date']."</date>
<title>".$row['title']."</title>
<desc>".$row['description']."</desc>
<thumb>".$row['thumb']."</thumb>
<img>".$row['image']."</img>
</image>

";


   }                      
?>


Thanks
James

I am working on a project where I want a select form to display information from a MySQL table. The select values will be different sports (basketball,baseball,hockey,football) and the display will be various players from those sports. I have set up so far two tables in MySQL. One is called 'sports' and contains two columns. Once called 'category_id' and that is the primary key and auto increments. The other column is 'sports' and contains the various sports I mentioned. For my select menu I created the following code.


<?php
#connect to MySQL
$conn = @mysql_connect( "localhost","uname","pw")
or die( "You did not successfully connect to the DB!" );

#select the specified database
$rs = @mysql_SELECT_DB ("test", $conn )
or die ( "Error connecting to the database test!");
?>

<html>
<head>Display MySQL</head>
<body>
<form name="form2" id="form2"action="" >
<select name="categoryID">
<?php
$sql = "SELECT category_id, sport FROM sports ".
"ORDER BY sport";

$rs = mysql_query($sql);

while($row = mysql_fetch_array($rs))
{
  echo "<option value=\"".$row['category_id']."\">".$row['sport']."</option>\n  ";
}
?>
</select>

</form>
</body>
</html>




this works great. I also created another table called 'players' which contains the fields 'player_id' which is the primary key and auto increments, category_id' which is the foreign key for the sports table, sport, first_name, last_name.

The code I am using the query and display the desired result is as follows


<html>
<head>
<title>Get MySQL Data</title>
</head>
<body>
<?php
#connect to MySQL
$conn = @mysql_connect( "localhost","uname","pw")
or die( "Err:Db" );

#select the specified database
$rs = @mysql_SELECT_DB ("test", $conn )
or die ( "Err:Db");

#create the query
$sql ="SELECT *
FROM sports
INNER JOIN players ON sports.category_id = players.category_id
WHERE players.sport = 'Basketball'";

#execute the query
$rs = mysql_query($sql,$conn);

#write the data
while( $row = mysql_fetch_array( $rs) )
{
echo ("<table border='1'><tr><td>");
echo ("Caetegory ID:  " . $row["category_id"] );
echo ("</td>");
echo ("<td>");
echo ( "Sport: " .$row["sport"]);
echo ("</td>");
echo ("<td>");
echo ( "first_name: " .$row["first_name"]);
echo ("</td>");
echo ("<td>");
echo ( "last_name: " .$row["last_name"]);
echo ("</td>");
echo ("</tr></table>");
}
?>
</body>
</html>





this also works fine. All I need to do is tie the two together so that when a particular sport is selected, the query will display below in a table. I know I need to change my WHERE clause to a variable. This is what I need help with.

thanks

This portion is kind of stumping me. Basically, I have a two tables in this DB: users and users_access_level (Separated for DB normalization)

users: id / username / password / realname / access_level
users_access_level: access_level / access_name

What I'm trying to do, is echo the data onto an HTML table that displays users.username in one table data and then uses the users.access_level to find users_access_level.access_name and echo into the following table data, I would prefer not to use multiple queries if possible or nested queries.

Example row for users: 1234 / tmac / password / tmac / 99
Example row for users_access_level: 99 / Admin

Using the examples above, I would want the output to appear as such:

Username: Access Name: Tmac Admin
I am not 100% sure where to start with this, but I pick up quickly, I just need a nudge in the right direction.

The code I attempted to create just shows my lack of knowledge of joining tables, but I'll post it if you want to see that I did at least make an effort to code this myself.

Thanks for reading!