如何在同一页面上显示div的php结果

Question

我无法弄清楚如何将我的搜索结果显示在具有搜索栏的div中的同一页面上... 这是我的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>";

?>

// 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>"; ?>

这是html表单输入：

Answer 1

嗯......相当低效。您执行两次查询。只需获取匹配行的数量，然后再使用LIMIT子句获取结果的一个“页面”。

使用MySQL，您可以将其组合到一个查询中：

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

有了这个，mysql仍然会计算出没有limit子句会匹配多少行，但仍然只返回LIMIT中指定的行。然后，您可以使用更简单（并且更轻量级）检索总行数：

SELECT found_rows();

Answer 2

您可以通过AJAX执行此操作。但是，您必须使代码不那么程序化。

Answer 3

您不能仅使用PHP / HTML执行此操作，您需要使用AJAX / JavaScript或将PHP表单发布到嵌套在搜索页面中的IFRAME。