Выводит ошибку Notice: Undefined variable: number in Z:\home\localhost\www\php\indexGalery.php on line 45 как решить вопрос с этой переменной, это фото галерея, Спасибо! <?php include ("blocks/bd.php"); error_reporting(E_ALL); ini_set('display_errors',5); ?> <!DOCTYPE html> <html lang="en"> <head> <meta charset="utf-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css"> <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script> <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script> <script type="text/javascript" src="js/js/jquery.js"></script> <script type="text/javascript" src="js/js/myscript.js"></script> </head> <body> <div class="container"> <div class="panel panel-default"> <?php include ("blocks/Header.php");?> <div class="panel-body"> <?php include ("blocks/button.php");?> <div class="container"> <div class="row"> <?php $result=mysql_query("SELECT image, number FROM imagetable ",$db); $myrow=mysql_fetch_array($result); do{ printf (" <a href='#? number=%s' data-toggle='modal' data-target='.pop-up-1'> <img src='%s' width='150' class='img-responsive img-rounded center-block' alt=''></a> ",$myrow ['number'],$myrow ['image']);} while( $myrow=mysql_fetch_array($result)); ?> <?php $result=mysql_query("SELECT id,image , name FROM imagetable WHERE number='$number'",$db); $myrow=mysql_fetch_array($result); ?> <div class='modal fade pop-up-1' tabindex='-1' role='dialog' aria-labelledby='myLargeModalLabel-1' aria-hidden='true'> <div class='modal-dialog modal-lg'> <div class='modal-content'> <div class='modal-header'> <button type='button' class='close' data-dismiss='modal' aria-hidden='true'>×</button> <h4 class='modal-title' id='myLargeModalLabel-1'><?php echo $myrow ['name']; ?></h4> </div> <div class='modal-body'> <img src="<?php echo $myrow ['image']; ?>" class='img-responsive img-rounded center-block' alt=''> </div> </div> </div> </div> </div> </div> </div> </div> <?php include ("blocks/Footer.php");?> </div> </body> </html>
$number замени на $myrow['number'] а вообще PHP: $result=mysql_query("SELECT image, number FROM imagetable ",$db); $myrow=mysql_fetch_array($result); do{ printf(" <a href='#? number=%s' data-toggle='modal' data-target='.pop-up-1'><img src='%s' width='150' class='img-responsive img-rounded center-block' alt=''></a>", $myrow ['number'],$myrow ['image']);} while( $myrow=mysql_fetch_array($result)); $result=mysql_query("SELECT id,image , name FROM imagetable WHERE number='$number'",$db); $myrow=mysql_fetch_array($result); похлопать можно
поменял, ошибку не выводит, но и не работает, так как бы вместо WHERE number='1' подставить --- Добавлено --- спасибо!
поменял, ошибку не выводит, но и не работает, так как работает если вместо $number подставить number='1' спасибо!