Question

我正在构建一个返回数据的搜索引擎，如果数据库中没有结果，我将无法显示消息。我是MYSQLI的初学者。我该怎么办呢？我累了计算行数，如果它们小于0则显示它但是我在编写时遇到了错误。

代码：

<?php
$db = mysqli_connect('localhost','root', '', 'searchengine');

if(!$db) {
    die('sorry we are having some problbems');
}


$searchTerm = mysqli_real_escape_string($db,$_GET['term']);

$searchTerm = trim($searchTerm);

if ($searchTerm == '') {
    echo("no key words searched please try again");
}

else
{
$sql = mysqli_query(
    $db,
    sprintf(
        "SELECT * FROM searchengine WHERE name LIKE '%s' LIMIT 0,20",
        '%'. $searchTerm .'%'
    )
);

while($ser = mysqli_fetch_array($sql)) {
    echo "<a href='$ser[pageurl]'>$ser[img]</a>";
}
}

mysqli_close($db);
?>

Answer 1

使用mysqli_num_rows ...

$sql = mysqli_query(
    $db,
    sprintf(
        "SELECT * FROM searchengine WHERE name LIKE '%s' LIMIT 0,20",
        '%'. $searchTerm .'%'
    )
);
if(mysqli_num_rows($sql) === 0) {
    echo "No results.";
}

Answer 2

因为mysqli_fetch_array()在没有结果时返回null，所以你也可以放

if ($ser === NULL){
    echo "no results";
}

最后

PHP简单的搜索引擎？

2 个答案: