0

我有这个脚本：

<?php
/**
 * Template Name: search
 * This is the template that displays all pages by default.
 * Please note that this is the WordPress construct of pages
 * and that other 'pages' on your WordPress site will use a
 * different template.
 *
 * @package WordPress
 * @subpackage Twenty_Twelve
 * @since Twenty Twelve 1.0
 */

get_header(); ?>

    <div id="primary" class="site-content">
        <div id="content" role="main">

            <?php while ( have_posts() ) : the_post(); ?>
                <?php get_template_part( 'content', 'page' ); ?>
                <?php comments_template( '', true ); ?>
            <?php endwhile; // end of the loop. ?>

        </div><!-- #content -->

<form name="form" action="search" 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","",""); //(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  
$query = "select * tabled where field like \"%$trimmed%\"  
  order by 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: &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 "Results";
$count = 1 + $s ;

// now you can display the results returned
  while ($row= mysql_fetch_array($result)) {
  $title = $row["1st_field"];

  echo "$count.)&nbsp;$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-$limit);
  print "&nbsp;<a href=\"$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>";

?>
    </div><!-- #primary -->

<?php get_sidebar(); ?>
<?php get_footer(); ?>

当我进入页面时出现此错误（在我单击提交之前）

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource ,line 66

这是第 66 行的代码：

 $numrows=mysql_num_rows($numresults);

有谁知道为什么会这样，如果可能的话请帮忙。

谢谢

本

4

3 回答 3

2

看起来像一个格式错误的 MySQL 查询：

$query = "select * from tabled where field like \"%$trimmed%\"  
  order by field";

您基本上忘记了查询字符串中的“来自”。

作为旁注，您应该阅读正确转义 MySQL 查询字符串：

http://php.net/manual/en/function.mysql-real-escape-string.php

http://kunststube.net/escapism/

于 2013-04-12T04:49:38.987 回答
2

SQL 查询中缺少 FROM 子句：

$query = "select * FROM tabled where field like \"%$trimmed%\"  
  order by field";
于 2013-04-12T04:52:49.653 回答
1

尝试像这样更改您的 sql 查询：

$criteria = '%' . mysql_real_escape_string($trimmed) . '%';
$query = "select * from tabled where field like '$criteria' order by field";
于 2013-04-12T04:50:09.900 回答