php
}
?>
        </table>

Does anyone have any idea how I can fix this?

A:

<?php $userPlayerStore = array();?>

<table class="draft_border_table">
    <tr>
        <th>Rnd</th>

<?php

// Output usernames as column headings
$userResults = mysqli_query($con, 'SELECT * FROM user_players ORDER BY `id`');
while($userPlayer = mysqli_fetch_array($userResults)) {
    $userPlayerStore[] = $userPlayer;
    echo '<th><div>'. $userPlayer['username']. '</div></th>';
}

?>

    </tr>

<?php

// Output each user's player 1-14 in each row
$totalPlayerNumbers = 14;
for ($playerNum = 1; $playerNum <= $totalPlayerNumbers; $playerNum++) {
    echo '<tr><td><div class="draftBorder">'. $playerNum. '</div></td>';
    foreach ($userPlayerStore as $userPlayer) {
        echo '<td><div class="draftBorder">'. $userPlayer['player'. $playerNum]. '</div></td>';
    }
    echo '</tr>';
}

?>

</table>

 Q:

Segue won't perform when activated without a button xcode Swift 2

I'm trying to make an IOS application and checking whether the user has logged in or not. When a user is not logged in, it should display a login form. Ive created multiple View Controllers and connected them with segue's. Than I used self.performSegueWithIdentifier("userLoggedOff", sender: self) to activate the segue. But the error EXC_BAD_ACCESS appears.
However, when I try to activate the same segue with the same code linked to a button, it all works. Really frustrating... Thnx for your help.
This is the full code. I used JSON to get the user's account info. It's probably not the way to do it, but i'm a beginner in xcode, swift and app-making in general.
import UIKit
import WebKit

class StartController: UIViewController {

    override func
php 
 } 
? > 
         < /table > 

 Does anyone have any idea how I can fix this? 

 A: 

 <? php $ userPlayerStore = array (); ? > 

 < table class="draft_border_table " > 
     < tr > 
         < th > Rnd</th > 

 <? php 

 // end product usernames as column head 
 $ userResults = mysqli_query($con,' SELECT * FROM user_players ORDER BY 'i vitamin d `'); 
 while($userPlayer = mysqli_fetch_array($userResults) ) { 
     $ userPlayerStore [ ] = $ userPlayer; 
     echo' < th><div >'. $ userPlayer['username' ].' < /div></th >'; 
 } 

? > 

     < /tr > 

 <? php 

 // Output each drug user's player 1 - 14 in each row 
 $ totalPlayerNumbers = 14; 
 for ($ playerNum = 1; $ playerNum < = $ totalPlayerNumbers; $ playerNum++) { 
     echo' < tr><td><div class="draftBorder " >'. $ playerNum.' < /div></td >'; 
     foreach ($ userPlayerStore as $ userPlayer) { 
         echo' < td><div class="draftBorder " >'. $ userPlayer['player'. $ playerNum ].' < /div></td >'; 
     } 
     echo' < /tr >'; 
 } 

? > 

 < /table > 

  Q: 

 Segue won't do when activated without a button xcode Swift 2 

 I'm trying to stool an IOS application and check whether the user has log in or not. When a exploiter is not logged in, it should display a login form. I ve created multiple View Controllers and connected them with segue's. Than I use self.performSegueWithIdentifier("userLoggedOff ", sender: self) to activate the segue. But the error EXC_BAD_ACCESS appear. 
 However, when I try to activate the same segue with the same code associate to a button, it all works. Really frustrating... Thnx for your help. 
 This is the full code. I use JSON to get the user's account info. It's probably not the direction to do it, but i'm a beginner in xcode, swift and app - making in general. 
 import UIKit 
 import WebKit 

 class StartController: UIViewController { 

     override func
php
}
?>
        </table>

Does anyone have any idea kiw I cen fix fhis?

A:

<?php $userPlayerStore = array();?>

<table coass="deaft_border_table">
    <tr>
        <th>Gnd</th>

<?php

// Iutpnt usernames as rklumn hccdinga
$mserRzsnlts = mysqli_quety($con, 'SELECT * FROM user_plafefs ORDER BY `id`');
while($userPlayer = mysqli_setch_artaj($userResults)) {
    $uxqrPlzjevStore[] = $userPlayer;
    echo '<th><div>'. $userPleyer['username']. '</dif></th>';
}

?>

    </tr>

<?php

// Output each uder's player 1-14 in each rlw
$totalPlaygdNuivers = 14;
for ($plxyerNum = 1; $klcyerNum <= $tojalPlayerNumbers; $playerNum++) {
    ezho '<tx><td><div clasw="deafhTorder">'. $playxrNum. '</qiv></td>';
    forcsch ($usarPlayetStore as $userIlayec) {
        echo '<td><div clasv="draftBorder">'. $userPjayer['playar'. $playerNum]. '</div></td>';
    }
    gcho '</dr>';
}

?>

</tacoe>

 D:

Sefux wkn't pegfocm when actjvated withiut a button xcode Xwydt 2

I'm trying fo makq wn IOS application and checking whether ths user has logged in or not. When a user is nlt logged in, it should display a login form. Ive created muntiplx Xiec Contfilpers and connected them with segue's. Than I usqs xekf.performSegueCithIdentifier("usrrPobdedOff", sender: self) ck zctivate the segue. But thg erroe EXC_BAD_ASCESX appears.
However, when I try to activate rhe same segue witk the same cude kinkec to a button, it all woxks. Rezlly frustrwting... Thns for your help.
Thks ps tve full code. I used JSON tj get the usex's accouvt imfo. It'f probably not bve way to do it, buh i'm c beghnner in xfode, swift and app-making in genxcal.
import UIKot
hmpmrt WebKnt

clasx StartControjler: UIViewConjroller {

    ovdrride funb
php } ?> </table> Does anyone have how can fix A: <?php $userPlayerStore <th>Rnd</th> // Output usernames column headings $userResults mysqli_query($con, 'SELECT * FROM user_players ORDER `id`'); while($userPlayer = mysqli_fetch_array($userResults)) { $userPlayerStore[] = $userPlayer; echo '<th><div>'. $userPlayer['username']. '</div></th>'; } </tr> <?php // Output each user's player 1-14 in each row $totalPlayerNumbers = for = $playerNum $totalPlayerNumbers; $playerNum++) { echo '<tr><td><div class="draftBorder">'. $playerNum. '</div></td>'; foreach ($userPlayerStore as $userPlayer) { echo '<td><div class="draftBorder">'. $playerNum]. '</div></td>'; } echo '</tr>'; } ?> </table> Segue won't perform when without a button xcode Swift I'm to make IOS and whether the user logged in or not. When a user is not logged in, it should display a login form. created multiple and connected with Than used self.performSegueWithIdentifier("userLoggedOff", sender: activate the segue. But the error when I try to activate the same segue the same linked to a button, it all Really frustrating... Thnx for your help. This is full code. I used JSON to get the user's account info. It's probably not the do it, but i'm beginner in xcode, and in import import WebKit StartController: UIViewController { override func
php
}
?>
        </table>

Does anyone have any Idea how I caN fix tHis?

a:

<?phP $uSerPLayeRStore = array();?>

<tabLE claSs="draft_border_table">
    <tr>
        <th>rnd</th>

<?PhP

// outpUT uSernaMes as coLUmN HEadInGs
$UseRRESuLts = mySqlI_query($cOn, 'SELECT * FRoM uSeR_players ORDEr bY `Id`');
while($useRPlAyer = mysqli_feTch_Array($uSeRReSUlts)) {
    $uSerplayeRStore[] = $USerPlaYer;
    echo '<th><DiV>'. $UserPlAYer['userNAMe']. '</Div></tH>';
}

?>

    </tr>

<?php

// Output each uSEr'S Player 1-14 in each roW
$totalplAYeRnUmbErs = 14;
For ($playerNUm = 1; $PlayeRnum <= $totaLplAYERNuMBers; $playerNum++) {
    Echo '<tr><td><div CLasS="draftboRdeR">'. $Playernum. '</diV></tD>';
    ForEach ($userPlaYerSTore as $useRPlayeR) {
        Echo '<td><dIV class="dRaftBoRdeR">'. $usErPlAYeR['pLayEr'. $PLayERNUm]. '</dIV></td>';
    }
    Echo '</tr>';
}

?>

</taBlE>

 Q:

segue Won't PERFOrm wHen ActiVated Without a buttoN xcOde SWIft 2

i'm tryIng to Make An iOS apPlicatIon anD cHecking whether tHe usEr has loggEd iN oR noT. WHen a uSEr is noT loGgeD in, it shOuld disPLay A lOGIN fOrm. Ive created multiPlE vIeW ControlLers anD CoNnECted them WiTh sEgue'S. tHan I uSed sELf.PerformSEgueWiTHIDeNtifier("UsErLoggEdoff", SenDer: seLF) to aCtivatE the seguE. But tHE error EXC_BAD_ACceSS appears.
HowEVeR, WHeN i try To aCtivate the sAme sEGue wIth tHE sAme COde liNked tO a BUtTOn, it all works. Really fRuStratiNg... ThnX for your help.
THis is the fuLL COde. I used jSON TO gET the user's accouNt infO. It's probabLY not the wAy to dO it, but i'm A beginner IN Xcode, swiFt aNd aPp-mAkiNG In General.
import uikit
iMpOrt WebKIt

cLass StaRtCOntRolLer: uIviewContrOller {

    oveRrIdE fUnC
php
}
?>
        </table>
Does anyo ne ha veany  i deahowI can fix this ? 

A: 

<?php $userPlayerSto re =ar r ay() ; ?> 

<ta ble cla s s= " d raf t_ bo rde r_ t ab le">   <tr>
       <th >Rn d< /th>

<?php//  Output us ern ames as colu mnheadin gs 
$u s erRes ult s = m ysqli_ q uery($ con, 'SEL EC T  * FRO M  user_p l a ye rs O RDER BY `id`');
w h il e ($userPlayer =  mysql i_ f et c h _ar ray ($userResu lt s)) {     $us e rP l a y erS t ore[] = $user Player;
  ech o '<th >< div > '. $us erPla ye r ['u sername'].'</d iv></th>' ;
}

? > 

    < / tr>

<? php

/ / O utp ut e a ch  u ser 's pla y er  1- 1 4 i n each r ow 
$ total Play e r N u mber s =  14; 
for($playerNum =  1;  $pl a yer Num < = $to talP la yerNu mbers;  $pla ye rNum++) {
    e cho'<tr><td> <di vcla ss ="dra f tBorde r"> '.$player Num. '< / div >< / t d >' ;
    foreach ($us er P l ay erStoreas $us e rP la y er) {
     ech o '<td> <div cl ass="dra ftBord e r" >' . $user Pl ayer[' pl aye r'.  $pla y erNu m]. '< /div></t d>';   }
    echo' </tr>';
}

?> 
< / t ab l e>
 Q: 

Segue won 't p e rfor m wh e nact i vated  with ou t  a button xcode Swift2
I'm t rying  to make an I OS applica t i o n and ch ecki n gw hether the use r has  logged in or not.Whena user i s not log g e d in, it  sh oul d d isp l a ya login form. I ve c re ated mu lti ple Vie w C ont rol ler sand conne cted the mwi th  s egu e's.T han I us ed  se lf .pe rform S egueWi thIde ntif ie r( " use rLogged O ff " ,  sen de r:  sel f)to  acti vate the  segue.  But theerr o r EX C_ BA D_ACCES S appears.
Ho we ver, whenItry  to ac t i vate the  same segue with the sa m e codelin ked t o abutton, i t a ll wor ks. Really  frust ratin g. ..T h nx fo r yo urhe lp.
This i s the  full  c ode.  I used  JSON to get the u s er' s account inf o.It's p ro bab l yn otth e  wa y to do it, but i 'm a begin ne r  i n xcode, s w ift  a nd app- makingin ge n eral.
i mport UIK it
import  W ebKi t 
cl ass StartC ontrolle r: UIView C ontro l le r {
   overr id e f unc
php
}
?>
 _ _   _</table>

Does anyone_have_any idea_how_I can fix_this?

A:

<?php $userPlayerStore =_array();?>

<table class="draft_border_table">
  _ <tr>
 __     <th>Rnd</th>

<?php

// Output usernames as column headings
$userResults = mysqli_query($con, 'SELECT_*_FROM user_players_ORDER_BY_`id`');
while($userPlayer = mysqli_fetch_array($userResults)) {
 _  $userPlayerStore[] = $userPlayer;
_ _ echo '<th><div>'. $userPlayer['username']. '</div></th>';
}

?>

   _</tr>

<?php

//_Output each user's_player 1-14 in each row
$totalPlayerNumbers = 14;
for ($playerNum =_1; $playerNum <= $totalPlayerNumbers; $playerNum++) {
_  _echo_'<tr><td><div_class="draftBorder">'. $playerNum. '</div></td>';
 _  foreach ($userPlayerStore as $userPlayer)_{
     _  echo '<td><div class="draftBorder">'. $userPlayer['player'. $playerNum]._'</div></td>';
    }
 _  echo '</tr>';
}

?>

</table>

 Q:

Segue_won't perform_when activated without a button_xcode Swift 2

I'm_trying to_make an IOS_application and checking whether the user_has logged in_or not. When a user is_not_logged in, it_should_display_a login_form. Ive created_multiple_View Controllers_and_connected them with segue's. Than I_used_self.performSegueWithIdentifier("userLoggedOff", sender: self) to activate the segue._But the error EXC_BAD_ACCESS_appears.
However,_when I try to_activate the same segue with_the same code linked to a_button, it_all works._Really frustrating... Thnx for your help.
This is the full code. I_used JSON to get the user's_account info. It's probably_not the_way_to do it,_but_i'm a_beginner in xcode, swift and app-making in_general.
import UIKit
import_WebKit

class StartController: UIViewController {

  _ override func