Question

我想显示查询的输出，但我无法检索我的变量$ result主要是因为它在不同的文件但我使用require来获取它但我仍然得到未定义的错误变量，我想知道如何检索在function.php上的$结果并在search.php上使用它来显示。

的search.php

<?php

include "db.php";
include "function.php";

if(isset($_GET['keywords'])){
global $connection;
$keyword = ($_GET['keywords']);
 searchData($keyword);
}

?>

<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<form action="search.php" method="GET">
    <label>
        Search
        <input type="text" name="keywords">
    </label>

    <input type="submit" name="search">

    <div class="form-group">

    <div class="result-count">
        Found <?php echo $query->num_rows; ?>result
    </div>
    <?php
        if($query->num_rows){
            while($r = $query->fetch_rows()){
            }
        }   
    ?>

    </div>

  </form>
</body>
</html>

function.php

<?php require_once "db.php";?>
<?php
  function searchData($keyword){
global $connection;

$query = ("
          SELECT username
          FROM users
          WHERE username LIKE '%{$keyword}%'
          ");
$result = mysqli_query($connection, $query);

echo "$result->num_rows". "found";
}   
?>

db.php中

<?php

$connection = new mysqli('localhost', 'root', '',
    'loginapp');

    if(!isset($connection)){
        die("Database connection failed!");
    }

Answer 1

要扩展Arron W。的评论，在你的function.php文件中，你需要返回$ result，然后在search.php中你将使用load返回变量，然后使用该变量访问数据。

function.php

$result = mysqli_query($connection, $query);

echo "$result->num_rows". "found";
return $result
}   
?>

的search.php

    if(isset($_GET['keywords'])){
    global $connection;
    $keyword = ($_GET['keywords']);
     $result = searchData($keyword);
    }

...

<div class="result-count">
        Found <?php echo $result->num_rows; ?>result
    </div>
    <?php
        if($result->num_rows){
            while($r = $result->fetch_row()){
            }
        }   
    ?>

从另一个php文件中检索变量

1 个答案: