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搜索时PHP不会加载MySQL数据_Php_Mysql - Fatal编程技术网

搜索时PHP不会加载MySQL数据

搜索时PHP不会加载MySQL数据,php,mysql,Php,Mysql,我正试图在我的网站上制作一个搜索表单,打印出数据库中所有可能的内容。然而,当我寻找某样东西时,它总是告诉我它什么也找不到。我该如何解决这个问题 PHP代码: $host = 'localhost'; $user = '111042'; $password = 'jcbvrjd8'; $db_name = '111042'; $search = $_POST["search"]; if(isset($search)) { $db = mysqli_connect($host, $user,

我正试图在我的网站上制作一个搜索表单,打印出数据库中所有可能的内容。然而,当我寻找某样东西时,它总是告诉我它什么也找不到。我该如何解决这个问题

PHP代码:

$host = 'localhost';
$user = '111042';
$password = 'jcbvrjd8';
$db_name = '111042';
$search = $_POST["search"];
if(isset($search)) {
    $db = mysqli_connect($host, $user, $password, $db_name);
    $wild_search = "%".$search."%";
    $findname = "SELECT `name`,`surname`
                FROM `Account` 
                WHERE `name` LIKE '".$wild_search."'
                    OR `surname` LIKE '".$wild_search."'
                    OR CONCAT(`name`, `surname`) LIKE '".$wild_search."'
                    OR CONCAT(`name`, `surnameprefix`, `surname`) LIKE '".$wild_search."';";
    $query = mysqli_query($db,$findname);
    $results = mysqli_fetch_all($query);

    if($result) {
        echo "<div id='searchresult'>\n";
        echo "<h1>People Found:</h1>\n";
        echo "<table id='searchresult'>\n";
        foreach($result as $rowno => $row) {
            echo "<tr class='searchtablerow'>\n";
            echo "<td>".$row['name'].", ".$row['surname']."</td>\n";
            echo "</tr>\n";
        }
        echo "</table>\n";
    } else {
        echo "<div id='searchresult'><h1>People Found:</h1>\n";
        echo "<p>No one was found...</p>\n";
        echo "</div>\n";
    }
} else {
}
HTML格式:

<div id="searchform">
    <h1>Search friends:</h1>
    <form name="searchform" method="post" action ="searchlink.php">
        <input type="text" name="search" id="search" autofocus placeholder="e.g. John Smith..."></input> <br>
        <input type="submit" name="submitsearch" value="Search" id="searchbutton"></input>
    </form>
</div>
谢谢你们帮助我

请不要烤我,我只是没有你先进

$query = mysqli_query($db,$findname);
$results = mysqli_fetch_all($query);

if ($results) {
...

在结果的最后,他不见了。或者删除$results=mysqli\u fetch\u all$query中的s

您的网站是否可以从外部访问,然后删除您在此处的登录数据!!最有可能$query返回false,用mysqli_error$db检查MySQL向您抛出的错误;错误在于$result的使用,而$results首先修复了这个问题。您的脚本有可能会查看在使用时发生了什么,这通常意味着查询失败