Birch Posted July 26, 2009 Share Posted July 26, 2009 Hello PHP Gurus! () I've made a db that stores information about issues of a comics magazine. After having a hard time writing and fixing the select and view script *phew* I started writing the search script. Each issue has 5 parameters, ID (number of the issue, auto-increasing and primary key), cover story, pages, date and price. So, I wanted 2 ways of searching, one by using the ID, and one by using cover story or date. However, I get php errors everytime I change something to fix the previous error, so I cannot remember which error started it all . Please tell me what's wrong. I am using WAMP for the testing. <html> <head> <meta content="text/html; charset=utf-8" http-equiv="Content-Type" /> <meta content="el-gr" http-equiv="Content-Language" /> <title>KomixDB</title> </head> <body> <h1>KomixDB Search Results</h1> <?php // create short variable names $searchtype=$_POST['searchtype']; $searchterm=$_POST['searchterm']; $searchterm= trim($searchterm); if (!$searchtype || !$searchterm) { echo 'You have not entered search details. Please go back and try again.'; exit; } if (!get_magic_quotes_gpc()) { $searchtype = addslashes($searchtype); $searchterm = addslashes($searchterm); } @ $db = new mysqli("localhost", "root", "password", "komixdb") or die(mysql_error());; if (mysqli_connect_errno()) { echo 'Error: Could not connect to database. Please try again later.'; exit; } if ($searchtype == 'id') { mysqli_query($db, "set names utf8"); $query = "select * from issues where id == '%".$searchterm."%'"; $result = $db->query($query); $row = mysqli_query($result, MYSQLI_NUM) or die("Error: ".mysqli_error($result)); echo '<p><strong> Ιστορία Eξώφυλλου: '; echo htmlspecialchars(stripslashes($row['coverstory'])); echo '</strong><br />Νούμερο: '; echo stripslashes($row['id']); echo '<br />Ημερομηνία: '; echo stripslashes($row['issuedate']); echo '<br />Σελίδες: '; echo stripslashes($row['pages']); echo '<br />Tιμή: '; echo stripslashes($row['price']); echo '</p>'; $result->free(); $db->close(); } else { mysqli_query($db, "set names utf8"); $query = "select * from issues where ".$searchtype." like '%".$searchterm."%'"; $searchResult = $db->query($query); $num_results = $searchResult->num_rows; while ($row = mysqli_fetch_assoc($searchResult)) { echo '<p>Number of issues found: '.$num_results.'</p>'; for ($i=0; $i <$num_results; $i++) { $row = mysqli_fetch_assoc($searchResult); echo '<p><strong>'.($i+1).'. Ιστορία Eξώφυλλου: '; echo htmlspecialchars(stripslashes($row['coverstory'])); echo '</strong><br />Νούμερο: '; echo stripslashes($row['id']); echo '<br />Ημερομηνία: '; echo stripslashes($row['issuedate']); echo '<br />Σελίδες: '; echo stripslashes($row['pages']); echo '<br />Tιμή: '; echo stripslashes($row['price']); echo '</p>'; } } $searchResult->free(); $db->close(); } ?> </body> </html> Quote Link to comment Share on other sites More sharing options...
ignace Posted July 26, 2009 Share Posted July 26, 2009 modify from: select * from issues where id == '%".$searchterm."%' to: select * from issues where id = '".$searchterm."' Quote Link to comment Share on other sites More sharing options...
Birch Posted July 26, 2009 Author Share Posted July 26, 2009 ok, that reduced the error levels when searching for ID but there are still 2 msgs Warning: mysqli_query() expects parameter 1 to be mysqli, object given in C:\wamp\www\results.php on line 47 Warning: mysqli_error() expects parameter 1 to be mysqli, object given in C:\wamp\www\results.php on line 48 Error: And forgot to post the problem for cover story and date searching: KomixDB Search Results Number of issues found: 1 1. Ιστορία Eξώφυλλου: (coverstory was supposed to be there) Νούμερο: ID Ημερομηνία: Date Σελίδες: Pages Tιμή: Price That's all I get, when the row that matches is only one. When there are more, only the last result is empty Quote Link to comment Share on other sites More sharing options...
vineld Posted July 26, 2009 Share Posted July 26, 2009 I don't read Greek and I have never used mysqli myself but the problem is that you send the $result object and not the mysqli to the mysqli_query and mysqli_error functions. They only accept an instance of mysqli. Since you have already executed the query you do not wish to execute another one but operate on the result class instead as you do in the other places of your code. There are plenty of ways you can do this. Look here. Quote Link to comment Share on other sites More sharing options...
Birch Posted July 26, 2009 Author Share Posted July 26, 2009 What do you mean by 'They only accept an instance of mysqli.'? How should I write it instead? Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.