Venkatesh Posted November 11, 2006 Share Posted November 11, 2006 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '5' at line 1[b]Code: [/b] <?//Connect to DBmysql_connect("localhost","user","pass") or die("Unable to connect to SQL server");mysql_select_db("Web") or die("Unable to SELECT DB");$pagenum = 5; //Number of results per page$SearchString=$_POST["SearchString"]; // Get the search tearmIf($SearchString == "") $SearchString=$_GET["SearchString"]; // Get the search tearmIf($SearchString == "") {Echo"Nothing to Search For";exit();}$page=$_GET["page"]; //Get the page number to showIf($page == "") $page=1; //If no page number is set, the default page is 1//Get the number of results$SearchResult=mysql_query("SELECT * FROM Hyd_Contacts") or die(mysql_error());$NumberOfResults=mysql_num_rows($SearchResult);//Get the number of pages$NumberOfPages=ceil($NumberOfResults/$pagenum);$SearchResult=mysql_query("SELECT * FROM Hyd_Contacts" . ($page-1)*$pagenum . ",$pagenum") or die(mysql_error());While($row = mysql_fetch_object($SearchResult)) {Echo $row->ArticleTitle . "<BR>";}$Nav="";If($page > 1) {$Nav .= "<A HREF=\"Search.php?page=" . ($page-1) . "&SearchString=" .urlencode($SearchString) . "\"><< Prev</A>";}For($i = 1 ; $i <= $NumberOfPages ; $i++) {If($i == $page) {$Nav .= "<B>$i</B>";}Else{$Nav .= "<A HREF=\"Search.php?page=" . $i . "&SearchString=" .urlencode($SearchString) . "\">$i</A>";}}If($page < $NumberOfPages) {$Nav .= "<A HREF=\"Search.php?page=" . ($page+1) . "&SearchString=" .urlencode($SearchString) . "\">Next >></A>";}Echo "<BR><BR>" . $Nav;?> Quote Link to comment https://forums.phpfreaks.com/topic/26914-php-error-message/ Share on other sites More sharing options...
wildteen88 Posted November 11, 2006 Share Posted November 11, 2006 Its not a PHP error by a MySQL error.One of you queries has an error which MySQL has reported.From looking at your code the error is coming from this query:[code]$SearchResult=mysql_query("SELECT * FROM Hyd_Contacts" . ($page-1)*$pagenum . ",$pagenum") or die(mysql_error());[/code]Change the above query to this:[code=php:0]$SearchQuery = "SELECT * FROM Hyd_Contacts" . ($page-1)*$pagenum . ", $pagenum)";$SearchResult = mysql_query($SearchQuery) or die("Unable to perform query<br />\n" . $SearchQuery . "<br /><br />" . mysql_error());[/code]Now re run your script again. Post the full error message it returns here. Quote Link to comment https://forums.phpfreaks.com/topic/26914-php-error-message/#findComment-123093 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.