php - Add class to img if database value equals 1 -
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- php parse/syntax errors; , how solve them? 11 answers
i know simple, don't find solution.
i trying add class 'inactivelicence' object () when database row 'glider' in table 'users' , 'username = $u' equals '1'.
this code have (i post code matters):
<?php include_once("php_includes/check_login_status.php"); $u = ""; $glider = ""; if(isset($_get["u"])){ $u = preg_replace('#[^a-z0-9]#i', '', $_get['u']); } else { header("location: index.html"); exit(); } $sql = "select * users username='$u' , activated='1' limit 1"; $user_query = mysqli_query($db_conx, $sql); while ($row = mysqli_fetch_array($user_query, mysqli_assoc)) { $glider= $row["glider"]; } if($glider == "1"){ echo "<script>$('#glider').addclass('inactivelicence')</script>"; } ?> <div class="licences"> <img id="glider" src="img/licences/icon_hc.png" > </div>
update:
i have code :
echo "<script>$(document).ready(function(){$('#glider').addclass('inactivelicence')});</script>";
but still error on line.. error: parse error: syntax error, unexpected '(', expecting t_variable or '$' in /home/a3318252/public_html/user.php on line 64
if($glider == "1"){ echo "<script>$(document).ready(function(){$('#glider').addclass('inactivelicence')});</script>"; }
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