Question

我的页面有一个搜索表单，我从 index.php 运行，通过调用 function.php 中的搜索功能并与setup.php建立连接。

搜索字段显示正常，并且没有数据库连接错误。然而，弹出的错误是：index.php（）中的Notice: Undefined variable: search_results。确切的行是它在表单下方显示的位置：<?php if ( $search_results ) : ?>。我认为这是因为 $ search_results 未在第一个if语句中声明为全局。但在将其全局化后，错误仍然存在。

非常感谢任何有关错误原因的见解。

以下是错误的快照：

index.php的内容：

<?php require_once  ('setup.php');
require_once  ('function.php');    
?>

<!DOCTYPE html>
<html>
<body>


<?php
if ( isset( $_GET['s'] ) ) {
  global $search_results;
  $search_term = htmlspecialchars($_GET['s'], ENT_QUOTES);
  $search_results = search($conn, $search_term);     
  }
?>


<form role="search" method="get" class="search-form" action="">
                <label>
                <input type="search" class="search-field" placeholder="Search" value="searchitem" name="s" 
                title="Search after:">
                </label>
                <input type="submit" class="search-submit" value="Search">              
</form>

  <?php if ( $search_results ) : ?>
    <div class="results-count">
      <p><?php echo $search_results['count']; ?> results found</p>
    </div>
    <div class="results-table">
      <?php foreach ( $search_results['results'] as $search_result ) : ?>
      <div class="result">
        <p><?php echo $search_result->title; ?></p>
      </div>
      <?php endforeach; ?>
    </div>
    <div class="search-raw">
      <pre><?php print_r($search_results); ?></pre>
    </div>
    <?php endif; ?>

<body>
</html>

Function.php ：

<?php

function search($conn, $search_term) {
    $sanitized = $conn->mysqli->real_escape_string($search_term);

    $query = $conn->mysqli->query("
      SELECT *
      FROM news
      WHERE title LIKE '%{$sanitized}%'
      OR body LIKE '%{$sanitized}%'
      OR sources LIKE '%{$sanitized}%'
      OR date LIKE '%{$sanitized}%'
    ");


    if ( ! $query->num_rows ) {
      return false;
    }


    while( $row = $query->fetch_object() ) {
      $rows[] = $row;
    }

    $search_results = array(
      'count' => $query->num_rows,
      'results' => $rows,
    );

    return $search_results;
  }
 ?>

Setup.php：

<?php
$hn = 'localhost';
$db = 'boh';
$un = 'water';
$pw = 'water'; 

#DB Connection:
 $conn = new mysqli($hn, $un, $pw, $db);
 if ($conn->connect_error) die($conn->connect_error);
?>

正如回复中所建议的，我尝试了if ( isset($search_results ))，这样就摆脱了错误，但是当我运行搜索（点击按钮）时，出现了以下错误：

Notice: Undefined property: mysqli::$mysqli in C:\xampp\htdocs\form\function.php on line 4

Fatal error: Call to a member function real_escape_string() on null in C:\xampp\htdocs\form\function.php on line 4

第4行：$ sanitized = $ conn-＆gt; mysqli-＆gt; real_escape_string（$ search_term）;

PHP：运行搜索功能时未定义的变量错误

0 个答案: