如何在同一页面上的div中显示php结果

Question

I cannot figure out how to have my search results displayed on the same page, in a div, that has the search bar... here is my 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","PW"); //(host, username, password) //specify database ** EDIT REQUIRED HERE ** mysql_select_db("DB") or die("Unable to select database"); //select which database we're using // Build SQL Query $query = "select * from TABLE where ID like \"%$trimmed%\" order by ID"; // 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: &quot;" . $trimmed . "&quot; 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: &quot;" . $var . "&quot;</p>"; // begin to show results set echo "<p>Results</p>"; $count = 1 + $s ; // now you can display the results returned while ($row= mysql_fetch_array($result)) { $title = $row["name"]; $picture = $row["thumbnail"]; $code = $row["thumbnail"]; echo "<p>$count.&nbsp;$picture&nbsp;$title</p>" ; $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-$limit); print "&nbsp;<a href=\"$PHP_SELF?s=$prevs&q=$var\">&lt;&lt; Prev 10</a>&nbsp&nbsp;"; } // 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+$limit; echo "&nbsp;<a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 &gt;&gt;</a>"; } $a = $s + ($limit) ; if ($a > $numrows) { $a = $numrows ; } $b = $s + 1 ; echo "<p>Showing results $b to $a of $numrows</p>"; ?> 

here is the html form input:

Answer 1

Hmm... rather inefficient. You execute the query twice. Once just to get the number of matching rows, then again with a LIMIT clause to fetch just one "page" of results.

With MySQL you can combine that into a single query:

$sql = "SELECT SQL_CALC_FOUND_ROWS ... LIMIT $x,$limit"

With this, mysql will still figure out how many rows would have matched without the limit clause, but still only return the rows specified in the LIMIT. You can then retrieve the total row count with a simplier (and much more lightweight):

SELECT found_rows();

Answer 2

You can do this via AJAX . However, you will have to make your code less procedural.

Answer 3

您不能仅使用PHP / HTML做到这一点，您将需要使用AJAX / JavaScript或将PHP表单发布到嵌套在搜索页面中的IFRAME。