致命错误：在第43行的C：\ xampp \ htdocs \ projectname \ searchit.php中的非对象上调用成员函数mysqli_fetch_array（）[关闭]

I know this question has been asked several times, however I can still not get this to work!!!

I am trying to make a basic search engine which I manually enter pages etc in a phpMyadmin mysql databse. The error that keeps coming up is:

Fatal error: Call to a member function mysqli_fetch_array() on a non-object in C:\xampp\htdocs\projectname\searchit.php on line 43

So here is the code!!!

<html lang="en">
<head>
<meta charset="utf-8">
<title>
Search The Internet
</title>
<link rel="stylesheet" href="main.css">
<div id="menu_bar1" style="padding:10px; margin: 0px;">
<form action="searchit.php" method="GET"> 
<label>Search
</label>
<input type="text" name="lets_search_for" size="50" placeholder="What do you want to search for???"/>
<input type="submit" name="search" value="Lets Search!!!"/>
</form>
</div>
</head>
<body bgcolor="#8E44AD">    
<?php    
  if (isset($_GET['search']))    
  {    
    $mysqli = mysqli_connect("localhost", "root", "Password");  
    $search_val = $_GET['lets_search_for'];
    echo "What Did We Find??? For <b><i> $search_val </i></b>" ;
    print "<br />";    
    $query = "SELECT * FROM search_websitename WHERE keywords LIKE ' . %$search_val% .'"; 
    $result = $mysqli->query($query);

   if($mysqli === FALSE) {
        die(mysqli_error()); 
   }    
   while ($row = $result->mysqli_fetch_array($query) ){  //This is line 43!!!

      $title = $row['title'];
      $link = $row['url'];
      $des = $row['des'];   

      echo "<a href='$link'>$title</a><br />";
      echo $link;
      echo "<p>" .
      $des . "</p><hr /><br />";    
    }    
}    
?>    
</body>    
</html>

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2条回答默认最新

duanjing9739 2014-10-04 06:28

关注

You need to fix the following:

$mysqli = mysqli_connect("localhost", "root", "Password");
// missing database connection

$search_val = $_GET['lets_search_for'];

$query = "SELECT * FROM search_websitename WHERE keywords LIKE ' . %$search_val% .'";
// you do not need a dot (concatenation)

So finally:

$mysqli = mysqli_connect("localhost", "root", "Password", "DATABASE_NAME");
$search_val = $_GET['lets_search_for'];
$search_val = '%' . $mysqli->real_escape_string($search_val) . '%';
$query = "SELECT * FROM search_websitename WHERE keywords LIKE '$search_val' ";
$result = $mysqli->query($query);
while ($row = $result->fetch_assoc() ) {
    echo $row['whaatever_column_name'];
}

本回答被题主选为最佳回答 , 对您是否有帮助呢?