如何在XAMPP服务器上使用php将MySQL数据库连接到html页面？
您好，我正在尝试将MySQL数据库连接到下面给出的简单html代码如何在XAMPP服务器上使用php将MySQL数据库连接到html页面？,php,html,mysql,Php,Html,Mysql,您好，我正在尝试将MySQL数据库连接到下面给出的简单html代码 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org /TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="C
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org
/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form action="result.php" method="POST">
S.No :
<input type="text" name="key">
<input type="submit" value="Search">
</form>
</body>
</html>
<?php
$serial=$POST['key'];
if(!$serial){
echo 'Please go back and enter the correct value';
exit;
}
$db = new mysqli('localhost', 'root', '', 'demo_db', 'tbldem');
if(mysqli_connect_errno()) {
echo 'Connection lost.. please try again later !!';
exit;
}
$query = "select * from tbldem where".$serial."like'%".$serial."%'" ;
$result = $db->query($query);
$num = $result->num_rows;
for($i = 0; $i < $num; $i++) {
$row = $result->fetch_assoc();
echo"<p>Serial : </p>";
echo $row['Index'];
echo"<p>Name : </p>";
echo $row['Name'];
echo "<p>Course : </p>";
echo $row['Course'];
}
$result->free();
$db->close();
?>
无标题文件
美国编号：
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org
/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form action="result.php" method="POST">
S.No :
<input type="text" name="key">
<input type="submit" value="Search">
</form>
</body>
</html>
<?php
$serial=$POST['key'];
if(!$serial){
echo 'Please go back and enter the correct value';
exit;
}
$db = new mysqli('localhost', 'root', '', 'demo_db', 'tbldem');
if(mysqli_connect_errno()) {
echo 'Connection lost.. please try again later !!';
exit;
}
$query = "select * from tbldem where".$serial."like'%".$serial."%'" ;
$result = $db->query($query);
$num = $result->num_rows;
for($i = 0; $i < $num; $i++) {
$row = $result->fetch_assoc();
echo"<p>Serial : </p>";
echo $row['Index'];
echo"<p>Name : </p>";
echo $row['Name'];
echo "<p>Course : </p>";
echo $row['Course'];
}
$result->free();
$db->close();
?>
query($query); $num = $result->num_rows;
for($i = 0; $i < $num; $i++) {
$row = result->fetch_assoc(); echo"
Serial :
"; echo $row['Index']; echo"
Name :
"; echo $row['Name']; echo "
Course :
"; echo $row['Course']; } $result->free(); $db->close(); ?>
query（$query）$num=$result->num\u行；
对于（$i=0；$i<$num；$i++）{
$row=result->fetch_assoc（）；echo“
序列号：
“echo$行['Index']；echo”
姓名：
“echo$row['Name']；echo”
课程：
“echo$row['Course']；}$result->free（）$db->close（）；？>
<?php
$serial=$_POST['key']; // change to $_POST['key'];
if(!$serial){
echo 'Please go back and enter the correct value';
exit;
}
$db = mysqli_connect('localhost','user_name','pass','demo_db'); // dont select your table in this line
mysqli_select_db($con,"tbldem"); // select your table here
if(mysqli_connect_errno()){
echo 'Connection lost.. please try again later !!';
exit;
}
$query = "select * from tbldem where serial LIKE '%$serial%';" ; // change $serial to serial like this line
$result = $db->query($query);
$num = $result->num_rows;
for($i=0;$i<$num;$i++){
$row = $result->fetch_assoc();
echo"<p>Serial : </p>";
echo $row['Index'];
echo"<p>Name : </p>";
echo $row['Name'];
echo "<p>Course : </p>";
echo $row['Course'];
}
$result->free();
$db->close();
?>
我脑海中浮现的唯一问题是你写了$POST['key']；什么时候应该是$_POST['key']；别忘了下划线，它还有助于调试代码 另一方面，对于那些希望从这个问题中学习的人来说，最好指出，您用来连接数据库的代码利用了一种技术上不使用官方面向对象方法的变通方法。这允许您的PHP代码在修复后在5.2.9/5.3.0之前的PHP版本中工作
希望这有帮助。听起来您的服务器没有处理php-是否安装了php，您是否检查了xampp说明？您的查询存在重大问题->
“从tbldem中选择*，其中“$serial.”类似“%”“$serial.”%”where$seriallikewherecolumnlike“%column%”