Question

我必须在PHP中使用localhost创建一个离线电影网站。不知何故，我无法获得正确的搜索，如果有人能向我解释原因，我会很高兴。搜索应该通过标题，年份，导演类型中的单词进行搜索。不要介意CSS只是希望它现在可以工作：）

这是我到目前为止得到的以及它给出的错误。

<form action="../PHP/moviedirectory.php" method="post">
    <input type="text" name="search" placeholder="zoeken.."/>
</form>

<?php
$link = mysqli_connect("localhost", "*User*", "*Password*", "WEBSITE");
$searchq = null;
if (isset($_POST['search'])) {
    global $searchq;
    $searchq = $_POST['search'];
}

$query = mysqli_query($link, "SELECT * FROM WEBSITE.dbo.movie WHERE title 
LIKE '%$searchq%' ORDER BY publication_year AND title DESC") or die("could 
not search!");
$query = mysqli_query($link, "SELECT * FROM WEBSITE.dbo.movie WHERE 
publication_year LIKE '%$searchq%' ORDER BY publication_year AND title 
DESC") or die("could not search!");
$query = mysqli_query($link, "SELECT * FROM WEBSITE.dbo.movie WHERE genre 
LIKE '%$searchq%' ORDER BY publication_year AND title DESC") or die("could 
not search!");
$query = mysqli_query($link, "SELECT * FROM WEBSITE.dbo.movie WHERE director 
LIKE '%$searchq%' ORDER BY publication_year AND title DESC") or die("could 
not search!");
$count = mysqli_num_rows($query);
$bla = mysqli_fetch_array($query);
$result = mysqli_fetch_all($query);
$title = $bla['title'];
$director = $bla['director'];
$description = $bla['description'];
$cast = $bla['bla'];
$duration = $bla['duration'];
$year = $bla['year'];
$img = $bla['img'];
$url = $bla['url'];
$genre = $bla['genre'];

错误：

警告：mysqli_connect（）:( HY000 / 2002）：因为无法连接   目标计算机已经积极拒绝连接。在   第6行的C：\ xampp \ htdocs \ php \ test \ zoeken.php

警告：mysqli_query（）期望参数1为mysqli，boolean   在第13行的C：\ xampp \ htdocs \ php \ test \ zoeken.php中给出的不能   搜索！

使用数据库（sqlserver）在PhpStorm中搜索电影

0 个答案: