搜索时PHP不会加载MySQL数据
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: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,
$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