아래의 $servername $username $password $db="dbname"; $conn 이 부분을 같은 php 에 넣지 않고 daba/dbconfig.php 에서 불러오려면 어떻게 해야 하나요?
본문
아래의 $servername $username $password $db="dbname"; $conn 이 부분을
같은 php 에 넣지 않고 daba/dbconfig.php 에서 불러오려면 어떻게 해야 하나요?
======= 이 부분 삭제하고 daba/dbconfig.php 에서 불러오기 ??
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$db="dbname";
$conn = mysqli_connect($servername, $username, $password,$db);
?>
======
---------------------------------------------
index2.php
<?php
$result = mysqli_query($conn,"SELECT * FROM category1");
?>
<!DOCTYPE html>
<html lang="en">
<head>
<title>Category1</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.0/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.0/js/bootstrap.min.js"></script>
</head>
<body>
<div class="container">
<form>
<div class="form-group">
<label for="sel1">Category1</label>
<select class="form-control" id="category1">
<option value="">Select Category</option>
<?php
while($row = mysqli_fetch_array($result)) {
?>
<option value="<?php echo $row["id"];?>"><?php echo $row["category_title"];?></option>
<?php
}
?>
</select>
</div>
<div class="form-group">
<label for="sel1">Sub Category</label>
<select class="form-control" id="sub_category">
</select>
</div>
</form>
</div>
<script>
$(document).ready(function() {
$('#category1').on('change', function() {
var category_id = this.value;
$.ajax({
url: "get_subcat.php",
type: "POST",
data: {
category_id: category_id
},
cache: false,
success: function(dataResult){
$("#sub_category").html(dataResult);
}
});
});
});
</script>
</body>
</html>
답변 2
그누보드의 기능만 쓰고 싶으시면
<?php
include_once("./_common.php");
$result = sql_query("SELECT * FROM category1");
?>
로 해보세요.
블랙캣77님, 답변 감사합니다.