ilikephp Posted May 9, 2010 Share Posted May 9, 2010 Hi, I am trying to create a php form to search for some records in mySQL, I found this code, I chaged the required info, but I am getting this error code: Parse error: parse error in C:\www\Ste Therese Prog\searchengine\search.php on line 22 Could you help me plz? The code: <html> <head> <title>designplace.org search script</title> <meta name="author" content="Steve R, http://www.designplace.org/"> </head> <!-- © http://www.designplace.org/ --> <body> <form name="form" action="search.php" method="get"> <input type="text" name="q" /> <input type="submit" name="Submit" value="Search" /> </form> <?php // Get the search variable from URL $var = @$_GET['q'] ; $trimmed = trim($var) //trim whitespace from the stored variable // rows to return $limit=10; // check for an empty string and display a message. if ($trimmed == "") { echo "<p>Please enter a search...</p>"; exit; } // check for a search parameter if (!isset($var)) { echo "<p>We dont seem to have a search parameter!</p>"; exit; } //connect to your database ** EDIT REQUIRED HERE ** mysql_connect("localhost","username","password"); //(host, username, password) //specify database ** EDIT REQUIRED HERE ** mysql_select_db("database") or die("Unable to select database"); //select which database we're using // Build SQL Query $query = "select * from the_table where 1st_field like \"%$trimmed%\" order by 1st_field"; // EDIT HERE and specify your table and field names for the SQL query $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); // If we have no results, offer a google search as an alternative if ($numrows == 0) { echo "<h4>Results</h4>"; echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>"; // google echo "<p><a href=\"http://www.google.com/search?q=" . $trimmed . "\" target=\"_blank\" title=\"Look up " . $trimmed . " on Google\">Click here</a> to try the search on google</p>"; } // next determine if s has been passed to script, if not use 0 if (empty($s)) { $s=0; } // get results $query .= " limit $s,$limit"; $result = mysql_query($query) or die("Couldn't execute query"); // display what the person searched for echo "<p>You searched for: "" . $var . ""</p>"; // begin to show results set echo "Results"; $count = 1 + $s ; // now you can display the results returned while ($row= mysql_fetch_array($result)) { $title = $row["1st_field"]; echo "$count.) $title" ; $count++ ; } $currPage = (($s/$limit) + 1); //break before paging echo "<br />"; // next we need to do the links to other results if ($s>=1) { // bypass PREV link if s is 0 $prevs=($s-$limit); print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><< Prev 10</a>  "; } // calculate number of pages needing links $pages=intval($numrows/$limit); // $pages now contains int of pages needed unless there is a remainder from division if ($numrows%$limit) { // has remainder so add one page $pages++; } // check to see if last page if (!((($s+$limit)/$limit)==$pages) && $pages!=1) { // not last page so give NEXT link $news=$s+$limit; echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>"; } $a = $s + ($limit) ; if ($a > $numrows) { $a = $numrows ; } $b = $s + 1 ; echo "<p>Showing results $b to $a of $numrows</p>"; ?> <!-- © http://www.designplace.org/ --> </body> </html> Link to comment https://forums.phpfreaks.com/topic/201165-search-form/ Share on other sites More sharing options...
kenrbnsn Posted May 9, 2010 Share Posted May 9, 2010 You need a terminating semi-colon on this line: <?php $trimmed = trim($var) ?> Ken Link to comment https://forums.phpfreaks.com/topic/201165-search-form/#findComment-1055426 Share on other sites More sharing options...
ilikephp Posted May 9, 2010 Author Share Posted May 9, 2010 Thx for your help! I fixed it, but I received other parse errors, could you please check the whole code, or is there another simple search form? Thx a lot... Link to comment https://forums.phpfreaks.com/topic/201165-search-form/#findComment-1055428 Share on other sites More sharing options...
ilikephp Posted May 9, 2010 Author Share Posted May 9, 2010 I found the error, while trying to search, I got no result! :S I checked that there is a result, what could be the prob plz? Link to comment https://forums.phpfreaks.com/topic/201165-search-form/#findComment-1055429 Share on other sites More sharing options...
kenrbnsn Posted May 9, 2010 Share Posted May 9, 2010 Add an "or die" clause on the mysql_query() statement: <?php $numresults=mysql_query($query) or die("Problem with the query: $query<br>" . mysql_error()); ?> This might show the problem. Ken Link to comment https://forums.phpfreaks.com/topic/201165-search-form/#findComment-1055431 Share on other sites More sharing options...
ilikephp Posted May 9, 2010 Author Share Posted May 9, 2010 Could you please send me the whole code, because I got lost :S Thank you, Link to comment https://forums.phpfreaks.com/topic/201165-search-form/#findComment-1055443 Share on other sites More sharing options...
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