Lutaaya Huzaifah Idris Lutaaya Huzaifah Idris - 15 days ago 5
PHP Question

Uncaught Error: Call to undefined function mysql_select_db()

I'm trying to fetch data from the database using Xamp Server but am getting this error.


Fatal error: Uncaught Error: Call to undefined function
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
line 10


Below is my php script , am still new in php please help me out on this.

<?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);
?>

Answer

As per your request i have modified code.

<?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);
?>

Please note: you need to add error checking. Also note just typed in here (not tested), so bear with me if there are some errors.

Comments