Saya mencoba mengambil data dari database menggunakan Xamp Server tetapi saya mendapatkan kesalahan ini.
Kesalahan fatal: Kesalahan yang tidak tertangkap: Panggilan ke fungsi yang tidak terdefinisi mysql_select_db() in E:\xamp\htdocs\PoliceApp\News\fetch.php:10 Stack trace: #0 {main} thrown in E:\xamp\htdocs\PoliceApp\News\fetch.php on baris 10
Di bawah ini adalah skrip php saya, saya masih baru di php tolong bantu saya dalam hal ini. Tapi saya membaca semua Posting lain di sini tetapi tampaknya itu membingungkan saya, bagaimana saya bisa memperbaikinya.
<?php
$username="root";
$password="namungoona";
$hostname = "localhost";
//connection string with database
$dbhandle = mysqli_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
echo "";
// connect with database
$selected = mysql_select_db("police",$dbhandle)
or die("Could not select examples");
//query fire
$result = mysql_query("select * from News;");
$json_response = array();
// fetch data in array format
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
// Fetch data of Fname Column and store in array of row_array
$row_array['Headlines'] = $row['Headlines'];
$row_array['Details'] = $row['Details'];
$row_array['NewsPhoto'] = $row['NewsPhoto'];
//push the values in the array
array_push($json_response,$row_array);
}
//
echo json_encode($json_response);
?>
Sesuai permintaan Anda, saya telah memodifikasi kode.
<?php
$username="root";
$password="namungoona";
$hostname = "localhost";
//connection string with database
$dbhandle = mysqli_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
echo "";
// connect with database
$selected = mysqli_select_db($dbhandle, "police")
or die("Could not select examples");
//query fire
$result = mysqli_query($dbhandle,"select * from News;");
$json_response = array();
// fetch data in array format
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
// Fetch data of Fname Column and store in array of row_array
$row_array['Headlines'] = $row['Headlines'];
$row_array['Details'] = $row['Details'];
$row_array['NewsPhoto'] = $row['NewsPhoto'];
//push the values in the array
array_push($json_response,$row_array);
}
//
echo json_encode($json_response);
mysqli_free_result($result);
?>
Harap diperhatikan: Anda perlu menambahkan pengecekan kesalahan. Catatan juga baru saja diketik di sini (belum diuji), jadi bersabarlah jika ada beberapa kesalahan.