Passing php variables through pages / sql -


i have following information displayed 

<?php $my_query="select * games";  $result= mysqli_query($connection, $my_query); if (mysqli_num_rows($result) > 0) while ($myrow = mysqli_fetch_array($result)) { $description = $myrow["game_description"]; $image = $myrow["gamepic"]; $game_id = $myrow["game_id"]; $gamename = $myrow["game_name"];     echo "<div class='cover'>     <a href='game_details.php?gameid={$game_id}'> <img src=\"games/$image\"       class=\"thumbnail\" width=\"200\" height=\"200\"> </a>     </div>"; }    ?>

as can see have created game_details page display specific game_id when image clicked

im having trouble understanding how pull data out game_id in sql on other page.

here attempt on game_details page

 <?php   if (!isset($_get['$game_id']) || empty($_get['game_id'])) { echo "invalid category id."; exit(); }      $game_id = mysqli_real_escape_string($connection, $_get['game_id']);      $sql1 = "select * games game_id={$game_id}'";       $res4 = mysqli_query($connection, $sql1);  if(!$res4 || mysqli_num_rows($res4) <= 0)     {                  while ($row = mysqli_fetch_assoc($res4))                 {                     $gameid = $row['$game_id'];                     $title = $row['game_name'];                     $descrip = $row['game_description'];                     $genre = $row['genretype'];                       echo "<p> {$title} </p>";                  }                } ?>

this attempt giving me "invalid category id" error

would appreciate

there few issues code.

let's start top.

['$game_id'] need remove dollar sign in $_get['$game_id']

then, $row['$game_id'] same thing; remove dollar sign.

then, game_id={$game_id}' throw syntax error.

in first body of code; should use proper bracing conditional statements.

this 1 has none if (mysqli_num_rows($result) > 0) , cause potential havoc.

rewrites:

<?php $my_query="select * games";  $result= mysqli_query($connection, $my_query); if (mysqli_num_rows($result) > 0){ while ($myrow = mysqli_fetch_array($result)) { $description = $myrow["game_description"]; $image = $myrow["gamepic"]; $game_id = $myrow["game_id"]; $gamename = $myrow["game_name"];     echo "<div class='cover'>     <a href='game_details.php?gameid={$game_id}'> <img src=\"games/$image\" class=\"thumbnail\" width=\"200\" height=\"200\"> </a>     </div>"; }     } ?>

sidenote where game_id='{$game_id}' in below. if doesn't work, remove quotes it.

where game_id={$game_id}

2nd body:

<?php   if (!isset($_get['game_id']) || empty($_get['game_id'])) { echo "invalid category id."; exit(); }      $game_id = mysqli_real_escape_string($connection, $_get['game_id']);      $sql1 = "select * games game_id='{$game_id}'";       $res4 = mysqli_query($connection, $sql1);  if(!$res4 || mysqli_num_rows($res4) <= 0)     {                  while ($row = mysqli_fetch_assoc($res4))                 {                     $gameid = $row['game_id'];                     $title = $row['game_name'];                     $descrip = $row['game_description'];                     $genre = $row['genretype'];                       echo "<p> {$title} </p>";                  }            } ?>

use error checking tools @ disposal during testing:


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