Php 站点搜索是否显示整个表而不是搜索结果？
我有一个数据库，员工信息显示在一个表中，我已经成功地创建了搜索框，我已经连接到我的数据库，现在的问题是，当我搜索名称，即“Daniel”时，它只会返回我已经拥有的表，因此有两个表而不是“Daniels”信息 这是我的密码-Php 站点搜索是否显示整个表而不是搜索结果？,php,sql,database,mysqli,Php,Sql,Database,Mysqli,我有一个数据库，员工信息显示在一个表中，我已经成功地创建了搜索框，我已经连接到我的数据库，现在的问题是，当我搜索名称，即“Daniel”时，它只会返回我已经拥有的表，因此有两个表而不是“Daniels”信息 这是我的密码- <!doctype html> <html> <head> <title></title> <link rel="stylesheet" href=".
<!doctype html>
<html>
<head>
<title></title>
<link rel="stylesheet" href="../Css/index_stylesheet.css">
</head>
<body>
<p>
<?php
$form = "<html>
<h1>Search!</h1>
<form method=\"get\">
<label>Name:
<input type=\"text\" name=\"keyname\" />
</label>
<input type=\"submit\" value=\"Search\" />
</form>
</body>
</html>";
//capture search term and remove spaces at its both ends if there is any
if(!empty($_GET['keyname'])){
$keyname = $_GET['keyname'];
$searchTerm = trim($keyname);
//database connection info
$host = "localhost"; //server
$db = "development_suite"; //database name
$user = "root"; //dabases user name
$pwd = ""; //password
//connecting to server and creating link to database
$link = mysqli_connect($host, $user, $pwd, $db);
//MYSQL search statement
$query = "SELECT firstname ,surname,address,mobile,email
FROM development_suite.eeg_staff;";
$results = mysqli_query($link,$query);
/* check whethere there were matching records in the table
by counting the number of results returned */
if(mysqli_num_rows($results) >= 1){
$output = "$row_1";
while($row = mysqli_fetch_array($results))
{
$output .= "First Name: " . $row['firstname'] . "<br />";
$output .= "Surname: " . $row['surname'] . "<br />";
$output .= "Address: " . $row['address'] . "<br />";
$output .= "Mobile: " . $row['mobile'] . "<br />";
$output .= "Email: " . $row['email'] . "<br /><br />";
}
}else{
$output = "There was no matching record for the name " .
strip_tags($searchTerm);
}
} else {
$output = "Enter name you are searching for.";
}
echo "$form\n$output";
?>
</body>
</html>
$query = "SELECT firstname ,surname,address,mobile,email
FROM development_suite.eeg_staff
WHERE firstname LIKE '" . $_REQUEST['keyname'] . "';";
内博伊萨（Nebojsa）不过，我应该在这条线上写些什么？出现错误，“警告：mysqli_num_rows（）期望参数1是mysqli_result，在第49行的C:\wamp\www\manholeform\Application\includes\search_function.php中给出的布尔值”第49行是以if（mysqli_num_rows（$results）try echo$query开头的if语句，用于输出查询并尝试直接在数据库上执行该查询。