PHP - One To Many Join To Select Only The Max Value.

I'm trying to figure out the best way to do this if I'm doing it right with my for loop. With how many numAnswers there are for the selected poll its going to put the div with the label and text box for each of those but its giong to go and tie in each of those pollAnswers with what ID of the pollAnswer to the pollAnswer table.


Code: [Select]
$pollsQuery = "
    SELECT
        polls.question,
        polls.statusID,
        polls.numAnswers,
        DATE_FORMAT(polls.dateExpires, '%m/%d/%Y') AS dateExpires
    FROM
        polls
    WHERE
        polls.ID = '" . $pollID . "'";
$pollsResult = mysqli_query ( $dbc, $pollsQuery ); // Run The Query
$row = mysqli_fetch_array ( $pollsResult, MYSQL_ASSOC );

$pollAnswers = "
    SELECT
        pollAnswers.ID,
        pollAnswers.answer
    FROM
        pollAnswers
    WHERE
        pollAnswers.pollID = '" . $pollID . "'";
$pollAnswersResult = mysqli_query ( $dbc, $pollAnswers ); // Run The Query

Code: [Select]
<fieldset class="answerLeg">
        <legend>Edit Poll Answers</legend>
        <?php for ( $j = 0; $j <= $numAnswers; $j++) { ?>
            <div class="field required answers">
                <label for="answer<?php $j?>">Answer <?php $j?></label><input type="text" class="text" name="answer<?php $j?>" id="answer<?php $j?>" title="Answer <?php $j?>"/>
                <span class="required-icon tooltip" title="Required field - This field is required, it cannot be blank, and must contain something that is different from emptyness in order to be filled in. ">Required</span>     
            </div>
        <?php } ?>
    </fieldset>

First off Please bear with me. I am newbie.

Here is a table name 'Employes' and has following Columns.
ID(PK) | FirstName | LastName | SecurityLicence | CrowdLicence | DriversLicence | Password |

And through form all the values are assigned to these columns and user gets registered.
I have done it using this code and its working fine.
Code: [Select]
<?php
session_name('YourVisitID');
session_start();
if(!isset($_SESSION['FirstName']))
        {
                header("Location: http://" . $_SERVER['HTTP_HOST'] . dirname($_SERVER['PHP_SELF']) . "/index.php");
                exit();
        }
else
{
$page_title = 'Register';
include('templates/header.inc');
if(isset($_POST['submit']))
{
        require_once('mysql_connect.php');
        // connect to the db
        
        //Create a function for escaping the data.
        function escape_data($data)
        {
                global $con; // need connection
                if(ini_get('magic_quotes_gpc'))
                {
                        $data = stripslashes($data);
                }
                return mysql_real_escape_string($data, $con);
        }
        $message = NULL;
        if(empty($_POST['firstName']))
        {
                $fn = FALSE;
                $message .= '<p>you forgot to enter your first name!</p>';
        }
        else
        {
                $fn = escape_data($_POST['firstName']);
        }
        
        if(empty($_POST['lastName']))
        {
                $ln = FALSE;
                $message .= '<p>You forgot to enter your last name!</p>';
        }
        else
        {
                $ln = escape_data($_POST['lastName']);
        }
        if(empty($_POST['licenceId']))
        {
                $li = FALSE;
                $message .='<p>You Forgot enter your Security Officer Licence Number!</p>';
        }
        else
        {
                $li = escape_data($_POST['licenceId']);
        }
        if(empty($_POST['crowdLicenceNo']))
        {
                $cln = FALSE;
                $message .='<p>You Forgot to enter your Crowd Controller Licence Number!</p>';
        }
        else
        {
                $cln = escape_data($_POST['crowdLicenceNo']);
        }
        if(empty($_POST['driverLicenceNo']))
        {
                $dln = FALSE;
                $message .='<p>You forgot to enter your Driving Licence Number!</p>';
        }
        else
        {
                $dln = escape_data($_POST['driverLicenceNo']);
        }
        if(empty($_POST['password']))
        {
                $p = FALSE;
                $message .='<p>You forgot to enter your password!</p>';
        }
        else
        {
                        if($_POST['password'] == $_POST['password2'])
                                {
                                        $p = escape_data($_POST['password']);
                                }
                        else
                                {
                                        $p = FALSE;
                                        $message .='<p>Your password did not match the confirmed password</p>';
                                }
        }
        if($fn && $ln && $li && $cln && $dln && $p)
        {
                                
                $query = "SELECT ID FROM Employes WHERE SecurityLicence='$li'";
                $result = @mysql_query($query);
                if(mysql_num_rows($result) == 0)
                { 
                
                
                $query = "INSERT INTO Employes (FirstName, LastName, SecurityLicence, CrowdLicence, Driverslicence, Password)
                VALUES ('$fn', '$ln', '$li', '$cln', '$dln', PASSWORD('$p'))";
                $result = @mysql_query($query);
                        if($result)
                                {
                                        echo '<p>You have been registered</p>';
                
                                }
                        else
                        {
                                $message = '<p>We apologise there is a system error.</p><p>' . mysql_error(). '</p>';
                        }
                } 
                else 
                {
                        $message = '<p>That Security Licence is already registered</p>';
                }
                        mysql_close();
        }
        else
        {
                $message .='<p>Please try again</p>';
        }
}

//print the message if there is one
if (isset($message))
{
        echo '<font color="red">', $message, '</font>';
}
}
?>
<script type="text/javascript">
function validate_form()
{
        var f = document.forms["regForm"]["firstName"].value;
                if(f==null || f=="")
                        {
                                alert("First Name must be filled out");
                                return false;
                        }
        var l = document.forms["regForm"]["lastName"].value;
                if(l==null || l=="")
                        {
                                alert("Last Name must be filled out");
                                return false;
                        }
        var sl = document.forms["regForm"]["licenceId"].value;
        var s = /^\d{5,}$/g.test(sl);
        var sll = sl.length;
                if(s==false)
                        {
                                alert("Security Licence No must be filled out in digits");
                                return false;
                        }
                else if(sll>=7)
                        {
                                alert("Invalid Security Licence No");
                                return false;
                        }
        var csl = document.forms["regForm"]["crowdLicenceNo"].value;
        var k = /^\d{5,}$/g.test(csl);
        var csll = csl.length;
                if(k==false)
                        {
                                alert("Crowd Controller Licence No must be filled out in digits");
                                return false;
                        }
                else if(csll>=7)
                        {
                                alert("Invalid Crowd Controller Licence No");
                                return false;
                        }
        var d = document.forms["regForm"]["driverLicenceNo"].value;
        var v = /^\d{6,}$/g.test(d);
        var dl = d.length;
                if(v==false)
                        {
                                alert("Driver's Licence No must be filled out in digits");
                                return false;
                        }
                else if(dl>=11)
                        {
                                alert("Invalid Driver's Licence No");
                                return false;
                        }
       
}
</script>


<h3>Employment Registration Form</h3>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" name="regForm" method="post" onsubmit="return validate_form()">
<fieldset>
<legend>Enter informations in the form below</legend>
<table>
<tr><th>First Name:</th><td><input type="text" name="firstName" /></td></tr>
<tr><th>Last Name:</th><td><input type="text" name="lastName" /></td></tr>
<tr><th>SO Licence No:</th><td><input type="text" name="licenceId" size="5" /></td></tr>
<tr><th>CC Licence No:</th><td><input type="text" name="crowdLicenceNo" size="5" /></td></tr>
<tr><th>Driver's Licence No:</th><td><input type="text" name="driverLicenceNo" size="10" /></td></tr>
<tr><th>Create Password:</th><td><input type="password" name="password" size="10" /></td></tr>
<tr><th>Confirm Password:</th><td><input type="password" name="password2" size="10" /></td></tr>
</table>
<input type="submit" name="submit" value="Register" />
<input type="reset" value="Reset" />
</fieldset>   
</form>

<?php
include('templates/footer.inc'); 
?>
 

Here is another Table "Jobs", with the following columns.
JobID(PK) | ID(FK) | JobDate | JobStart | JobFinish | JobLocation | RequestedBy |

For Example I am the owner of the company and want to assign jobs to my registerd workers using a form.
I assign JobDate=12/12/12 JobStart=1900 JobFinish=2300 JobLocation=Perth Requestedby=John. And I do it using form. JobID will be incremented automatically.
Now what i dont understand is how do I assign ID(FK) to it?? I am newbie this question may sound stupid to many of you but hey please help.
Or should I replace ID(FK) with the SecurityLicence of the worker, Like this
JobID(PK) |ToSecurityLicence | JobDate | JobStart | JobFinish | JobLocation | RequestedBy |

What I want is when the workers signs in He should see the job assigned to him. Can I do it using inner join??? if so how to right PHP code for comparing these two tables???
I am way too much confused.

Too many of you I have made a fool of myself asking this question. But I believe a person who doesnt ask the question is fool foreva.

If somebody could Help I would really really aprreciate it.

Thanks.

I'm trying to pull from my database every spell or attack that is equal to `All`, but it doesn't seem to be working. All it is doing is pulling all the spells, regardless of class = `All`, and also 0 of the attacks. It should be producing a single attack that is = to `All` as no spells are = `All`.

Any help would be greatly appreciated! Thanks! (This is my first attempt at a JOIN statement...)


$query = "SELECT attacks.id, attacks.name, attacks.price, attacks.class, attacks.descript, spells.id, spells.name, spells.price, spells.class, spells.descript ".
"FROM attacks, spells ".
"WHERE attacks.class = 'All' || spells.class = 'All' order by attacks.name, spells.name asc";

I'm an amateur PHP/mySQL person, trying to learn some more things, and I've hit a snag trying to do something I haven't done before.

Okay, here's what I'm trying to do. I have 3 tables:

Items (which includes `id` as the primary key) and a TON of information about each item.
Shop (which includes `id` as the primary key) and shop name and location.
Sale which has shop_id and item_id.

So, what I'm trying to do is have a form where someone can input in an item name and a shop name and have that item then added to the shop. This is where the Sale table comes into play too, as that's going to house each of the items and what shop it's in.

I currently have the following queries:


       $query5 = $db->execute("SELECT id FROM items LEFT JOIN sale ON items.id=sale.item_id");
       $query6 = $db->execute("SELECT id FROM shop LEFT JOIN sale ON shop.id=sale.shop_id");


I have the following form:


        if ($name == "") { $errors++; $errorlist .= "Name is required.<br />"; }
        if ($location == "" ) { $errors++; $errorlist .= "Location is required.<br />"; }
if (!isitem($itemname)) die("That item does not exist."); }

        if ($errors == 0) { 
            $query = doquery("INSERT INTO `sale` SET `item_id`=`$item`");
            admindisplay("New Item Added.","Add Items");

$page = <<<END

<b><u>Add Item</u></b><br /><br />
<form action="create_shop.php?do=additem" method="post">
<table width="90%">
<tr><td width="20%">Name:</td><td><input type="text" name="itemname" size="30" maxlength="255" value="" />**Be sure the item is already in the database and matches it IDENTICALLY.</td></tr>
<tr><td width="20%">Location:</td><td><input type="text" name="location" size="30" maxlength="55" value="" /><br /><span class="small">Where is the item obtained (unique shop name, please!)</span></td></tr>
</table>
<input type="submit" name="submit" value="Submit" /> <input type="reset" name="reset" value="Reset" />
</form>


What next steps do I need to take? Am I even on the right path?

Thanks for any and all help anyone provides!! I really appreciate it!

Hi

I want to know I can select which ID I get from an inner join?

Not sure If I should post here or in the MYSQL forum, but as it's using OOP I posted here.


Code: [Select]
$sql = "SELECT * FROM area_county INNER JOIN area_country ";
$sql .= "ON area_county.country_id = area_country.id ";
$sql .= "ORDER BY area_country.country, area_county.county ASC ";
$sql .= "LIMIT {$per_page} ";
$sql .= "OFFSET {$pagination->offset()}";
$list = Area_county::find_by_sql($sql);

foreach($list as $lists){
$ID = $lists->id;
$country_id = $lists->country_id;
$county = ucwords($lists->county);
        $country = ucwords($lists->country);

echo $ID;
}


I want it to show the ID of the county, NOT the id of the country.
both tables in the database have the ID column called id.

I know this can be done, but not to sure how.


Thanks

Not sure if it is because it is too early in the morning but I am having a problem.

Code: [Select]
$mpid = #

$query = "SELECT table1.title, table2.* ".
 "FROM table1, table2 ".
"WHERE table1.mpid = table2.mpid";

How do I only show the info from the 2 tables where mpid=#

Hi guys.

I have a SQL query running within my php. But if i add and inner join to link another table, it slows the performance(refresh, loading time) of the page by at least 2-3 seconds... I am working from localhost wampp serever. Should this happen when using inner Join?

Code: [Select]
$sql="SELECT * FROM cars INNER JOIN postcodes ON postcodes.Pcode=cars.Location where id>='1'"
Thanks in anticipation

"                                                                                        SELECT  event.*,
epr1.participantId AS team_1,
epr1.participantRoleId AS team_1_role,
epr2.participantId AS team_2,
epr2.participantRoleId AS team_2_role
  FROM  bb_EventParticipantRelation epr1,
   bb_EventParticipantRelation epr2,
bb_Event event
  WHERE 
   event.parentId='".$leage_type."' AND 
epr1.id != epr2.id AND
   epr1.parentParticipantId = '0' AND
epr2.parentParticipantId = '0' AND
epr1.eventId=event.id AND
epr2.eventId=event.id

ORDER BY epr1.participantRoleId ASC, event.startTime, team_1, team_2 ASC

"


This sql gets 2 rows and combines them as one, it does work but the issue is it seems to be taking too long in loading the page also I just need some advice on a better version


Thank you

I have a database with two tables, they are designed like this

Table1 has two fields (log_number, name)
Table2 has two fields (log_number, city)

I would like to do a search for a specific log number and display only one result for that matching log number (that we searched for) and display something like this 

Log Number: 59
Name: Mark
City: Chicago

this is my code so far


my search form

Code: [Select]
<form method="get" action="search.php" />   
      Log Number
     <input type="text" id="log_number" name="log_number" />
     <input type="submit" name="submit" value="Search" />

the search.php file

Code: [Select]
      <?php
 require ('/sqlconnect.php');
  $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
 
        $log_number = $_GET['log_number'];
        $name = $_GET['name'];
        $city = $_GET['city'];
 
  $query = "SELECT * FROM table1, table2 WHERE table1.log_number = table2.log_number";

  $result = mysqli_query ($dbc, $query);
     while ($row = mysqli_fetch_array($result)) {       

    echo '<b>Log Number:</b> '.$row['log_number'].'</td>';
    echo '<br/>';
    echo '<b>Name:</b> '.$row['name'].'</td>';
    echo '<br/>';
    echo '<b>City:</b> '.$row['city'].'</td>';
      
}
  mysqli_close($dbc);

?>


the problem I am having with my current code is that it does not display the results of the log number i searched for, but instead returns information for all the entries (two in my case) that have matching log numbers in both table1 and table2
when i search for log number 59, i get the following result

Log Number: 59
Name: Mark
City: Chicago

Log Number: 77
Transport Company: Doug
Customer Number: San Francisco



the result that I want is when i search for log number 59 ... i want it to only display that one record
Log Number: 59
Name: Mark
City: Chicago


hope someone can help
 

Hi,

Been looking for a little while on this problem, but can't seem to find a solution.

Ok, so I have two tables,

gallery & users

Gallery contains images, image info etc.
Users contains usernames, emails etc.

In users there is a 'photo' column, which references to an ID in the gallery table. This is so each user can have a 'profile' image from a bank of images stored in gallery.

What i'm trying to do is have a query that will output all of the images, and then show the usernames of assigned to each photo if there is one.

Heres what i've got:

Code: [Select]
$result = mysql_query("SELECT *,COUNT(photo) AS usercount FROM gallery LEFT JOIN users ON gallery.img_id=users.photo GROUP BY img_id ORDER BY date_uploaded ASC");
Its all fine, and displays the username, though if there are multiple users using the same photo, it only shows one username.
However, when displaying 'usercount', it will show that there are 2+ or whatever.

Short of nesting queries within other loops, is there an alternative solution??

Thanks!
Edd

When I have more than one table i don't know how to echo out a field if it is the same in both fields.  here's my code and I want to echo out the id from the people table:


$query =  "SELECT * FROM people INNER JOIN ratings ON people.id = ratings.rater_id AND ratings.event_id = '$eventid' AND ratings.complete = 'n' "; 

$result = mysql_query($query);
while ($row=mysql_fetch_assoc($result)){
echo $row['people.id'];
}


I tried  $row['people.id'] but it was blank. 

Hi guys,

Would like to check if I have done my INNER JOIN function correctly, as I have received and an error msg

Warning: mysqli_error() expects exactly 1 parameter, 0 given in D:\inetpub\vhosts\111.com\httpdocs\111.com\registration3.php on line 598.

Thanks


<?php
   $dbc = mysqli_connect('localhost', '111', '111', '111')
   or die(mysqli_error());
   $query = "SELECT sl.subject_level_id, sl.level_id, sl.subject_id, tl.name AS level_name, ts.name AS subject_name " .
    "FROM tutor_subject_level AS sl " .
    "INNER JOIN tutor_level AS tl USING (level_id) " .
    "INNER JOIN tutor_subject AS ts USING (subject_id) ";
   $sql = mysqli_query($dbc, $query);
   
   echo'<table><tr>'; // Start your table outside the loop... and your first row
   $count = 0; // Start your counter
      while($data = mysqli_fetch_array($sql)) {
         /*    Check to see whether or not this is a *new* row
            If it is, then end the previous and start the next and restart the counter.
         */
         if ($count % 5 == 0) {
            echo "</tr><tr>"; $count = 0;
         }
         echo '<td><input name="subject_level[]" type="checkbox" id="'.$data['subject_level_id'].'" value="'.$data['subject_level_id'].'"/>';
         echo '<label for="'.$data['subject_name'].'">'.$data['subject_name'].'</label></td>';
         $count++; //Increment the count
      }
   echo '</tr></table><br/><br/>'; //Close your last row and your table, outside the loop
   
?>


Wilson

How do I join these queries and is it best practice?

$e_query = mysql_query("SELECT `employeeid`, `userid`, `companyid`, `function`, `timestamp` FROM `employees` WHERE `companyid` = '${c_row['companyid']}'");

$e_row = mysql_fetch_assoc($e_query);
         
$u_query = mysql_query("SELECT `id`, `firstname`, `lastname`, `username`, `accounttype` FROM `users` WHERE `id` = '${e_row['userid']}'");

$u_row = mysql_fetch_assoc($u_query);
         
$eQ = "SELECT `capitalrequested` FROM `eQuestions` WHERE `userid` = '${u_row['id']}'";
         
$eQRes = mysql_query($eQ);
         
$reQ = mysql_fetch_assoc($eQRes);
         
$iQ = "SELECT `capitalavailable` FROM `iQuestions` WHERE `userid` = '${u_row['id']}'";
         
$iQRes = mysql_query($iQ);
         
$riQ = mysql_fetch_assoc($iQRes);

I am somewhat of a newbie and think this is a simple fix, but I have been having problems and can't seem to figure it out. I am trying to JOIN two tables, but am not getting the results I would like. I am getting the correct question number displayed whenever I try different instances, but it always defaults to the "DTCAP" question (q_code).   I have included the snippet of code and how the table structure looks. Any help would be greatly appreciated!   -------------------------------------------------------------------------------------   //Testing to ensure that the $_SESSION['code'] is DTCAP, CNROW, or MONFW after user chooses option echo $_SESSION['code'];   $query_question=mysql_query("SELECT * FROM adventure_questions q
            INNER JOIN users_adventures u ON q.q_number = u.ua_question
            WHERE u.ua_username = '$_SESSION[username]' AND u.ua_code='$_SESSION[code=auto:0]'");   $question_info=mysql_fetch_array($query_question);   echo $question_info['q_number'].". ".$question_info['question'];   -------------------------------------------------------------------------------------   adventure_questions q_id     q_code       q_number     question 1          DTCAP              1         DTCAP question 1 2          DTCAP              2         DTCAP question 2 26       CNROW              1        CNROW question 1 27       CNROW              2        CNROW question 2 51       MONFW              1        MONFW question 1 52       MONFW              2        MONFW question 2   users_adventures ua_id    ua_username       ua_code      ua_question    ua_score 1                kjb31ca             DTCAP              1                   0 2                kjb31ca            CNROW              2                 10 3                kjb31ca            MONFW              1                  0
Edited by KjB31ca, 20 May 2014 - 06:39 AM.

I am trying to join these 2 tables below, but its not working, any ideas where i am going wrong??

<?php
require("../include/mysqldb.php");
$con = mysql_connect("$dbhost","$dbuser","$dbpass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("$dbame", $con);



$result = "SELECT *
Reg_Profile_public.pref, Search_profiles_up.search_small_image
FROM
Search_profiles_up
INNER JOIN
Reg_Profile_public_Sex
ON
Search_profiles_up.UIN=Reg_Profile_public.UIN
WHERE
UIN='803272125132009'";

while ($row = mysql_fetch_array($result)) {
    echo $row['search_small_image'];
    echo $row['pref'];
}



?>