paullanesra Posted February 4, 2011 Share Posted February 4, 2011 hi i am having a problem with a php script. i have 3 files index.php, home.html.php, form.html.php. home.html.php is working fine, if has a form that is submitting fine but when it submits i get a problem. maybe someone can help me out please i would be grateful as i am new. the is nothing wrong with anything else mysql database is running fine and there is no problem with connection. index.php (you can ignore what is commented out) <?php include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php'; //$_GET['id'] is categoryid if (isset($_GET['id'])) { include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php'; $id = mysqli_real_escape_string($link, $_GET['id']); $sql = "SELECT businessid FROM businesscategory WHERE categoryid LIKE '$id'"; $result = mysqli_query($link, $sql); if ($result !='') { $businessid = mysqli_fetch_array($result); $sql = "SELECT content FROM business WHERE id LIKE '$businessid'"; $result = mysqli_query($link, $sql); } if ($result !='') { $content = mysqli_fetch_array($result); include 'form.html.php'; exit(); } } /* // The basic SELECT statement $select = 'SELECT content'; $from = ' FROM business'; $where = ' WHERE TRUE'; $id = mysqli_real_escape_string($link, $_GET['id']); if ($category != '') // An owner is selected { $where .= " AND categoryid='$categoryid"; } $categoryid = mysqli_real_escape_string($link, $_GET['category']); if ($categoryid != '') // A category is selected { $from .= ' INNER JOIN businesscategory ON id = business'; $where .= " AND categoryid='$categoryid'"; } $text = mysqli_real_escape_string($link, $_GET['text']); if ($text != '') // Some search text was specified { $where .= " AND content LIKE '%$text%'"; } $result = mysqli_query($link, $select . $from . $where); if (!$result) { $error = 'Error fetching businesses.'; include 'error.html.php'; exit(); } while ($row = mysqli_fetch_array($result)) { $businesses[] = array('id' => $row['id'], 'text' => $row['content']); } include 'form.html.php'; exit(); } */ $result = mysqli_query($link, 'SELECT id, name FROM category ORDER BY name'); if (!$result) { $error = 'Error fetching categories: ' . mysqli_error($link); include 'error.html.php'; exit(); } while ($row = mysqli_fetch_array($result)) { $categories[] = array('id' => $row['id'], 'text' => $row['name']); } include 'home.html.php'; ?> form.html.php <?php include_once $_SERVER['DOCUMENT_ROOT'] . '/includes/helpers.inc.php'; ?> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html> <head> <meta http-equiv="content-type" content="text/html; charset=windows-1250"> <meta name="generator" content="PSPad editor, www.pspad.com"> <title>Business</title> </head> <body> it works <?php echo $content; ?> </body> </html> Link to comment https://forums.phpfreaks.com/topic/226667-php-script-not-working-mysql/ Share on other sites More sharing options...
litebearer Posted February 4, 2011 Share Posted February 4, 2011 1. please use the code tags 2. What exactly is the error/problem you are experiencing? Are you getting error messages? Link to comment https://forums.phpfreaks.com/topic/226667-php-script-not-working-mysql/#findComment-1169780 Share on other sites More sharing options...
paullanesra Posted February 4, 2011 Author Share Posted February 4, 2011 sorry, there is no error message at all, all i get is "it works" from the form.html.php page. the objective is to retieve data from database and echo it out on form.html.php i am guessing the problem is in the top half of the index.php script. index.php (you can ignore what is commented out) <?php include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php'; //$_GET['id'] is categoryid if (isset($_GET['id'])) { include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php'; $id = mysqli_real_escape_string($link, $_GET['id']); $sql = "SELECT businessid FROM businesscategory WHERE categoryid LIKE '$id'"; $result = mysqli_query($link, $sql); if ($result !='') { $businessid = mysqli_fetch_array($result); $sql = "SELECT content FROM business WHERE id LIKE '$businessid'"; $result = mysqli_query($link, $sql); } if ($result !='') { $content = mysqli_fetch_array($result); include 'form.html.php'; exit(); } } /* // The basic SELECT statement $select = 'SELECT content'; $from = ' FROM business'; $where = ' WHERE TRUE'; $id = mysqli_real_escape_string($link, $_GET['id']); if ($category != '') // An owner is selected { $where .= " AND categoryid='$categoryid"; } $categoryid = mysqli_real_escape_string($link, $_GET['category']); if ($categoryid != '') // A category is selected { $from .= ' INNER JOIN businesscategory ON id = business'; $where .= " AND categoryid='$categoryid'"; } $text = mysqli_real_escape_string($link, $_GET['text']); if ($text != '') // Some search text was specified { $where .= " AND content LIKE '%$text%'"; } $result = mysqli_query($link, $select . $from . $where); if (!$result) { $error = 'Error fetching businesses.'; include 'error.html.php'; exit(); } while ($row = mysqli_fetch_array($result)) { $businesses[] = array('id' => $row['id'], 'text' => $row['content']); } include 'form.html.php'; exit(); } */ $result = mysqli_query($link, 'SELECT id, name FROM category ORDER BY name'); if (!$result) { $error = 'Error fetching categories: ' . mysqli_error($link); include 'error.html.php'; exit(); } while ($row = mysqli_fetch_array($result)) { $categories[] = array('id' => $row['id'], 'text' => $row['name']); } include 'home.html.php'; ?> form.html.php <?php include_once $_SERVER['DOCUMENT_ROOT'] . '/includes/helpers.inc.php'; ?> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html> <head> <meta http-equiv="content-type" content="text/html; charset=windows-1250"> <meta name="generator" content="PSPad editor, www.pspad.com"> <title>Business</title> </head> <body> it works <?php echo $content; ?> </body> </html> Link to comment https://forums.phpfreaks.com/topic/226667-php-script-not-working-mysql/#findComment-1169785 Share on other sites More sharing options...
BlueSkyIS Posted February 4, 2011 Share Posted February 4, 2011 i find it helpful to have mysql stop everything and tell me if one of my queries is wrong and display the SQL if so. $result = mysqli_query($link, $sql) or die(mysqli_error($link) . " IN $sql"); you might find this helpful, too. Link to comment https://forums.phpfreaks.com/topic/226667-php-script-not-working-mysql/#findComment-1169841 Share on other sites More sharing options...
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