SQL Server Database Query with PHP - php

I need to get some data from a Microsoft SQL Server database at work. When I have the data I need, I need to make an Excel spreadsheet that can be saved locally on my computer.
I found PHPExcel which seems to do the job on the Excel part, but what about getting the data from the Database?
I can't seem to find anything that's recent. Only old tutorials.

Use this way to Fetch the Records :
<?php
$hostname = "192.168.3.50";
$username = "sa";
$password = "123456";
$dbName = "yourdb";
MSSQL_CONNECT($hostname,$username,$password) or DIE("DATABASE FAILED TO RESPOND.");
mssql_select_db($dbName) or DIE("Database unavailable");
$query = "SELECT * FROM dbo.table";
$result = mssql_query( $query );
for ($i = 0; $i < mssql_num_rows( $result ); ++$i)
{
$line = mssql_fetch_row($result);
print( "$line[0] - $line[1]\n");
}
?>
This will fetch each rows from the Data Retrieve and Print on the Page. Use your Required format into that. I mean, Use html Table to show the data in well format.

Use this code to get an data from Database.
<?php
// Server in the this format: <computer>\<instance name> or
// <server>,<port> when using a non default port number
$server = '192.168.3.50';
// Connect to MSSQL
$link = mssql_connect($server, 'sa', 'sa');
if (!$link) {
die('Something went wrong while connecting to MSSQL');
}
else{
echo "connected ";
mssql_select_db('Matrix') or die("Wrong DATAbase");
//mssql_query("SELECT Seq_no from dbo.Trans_R WHERE Seq_no = 000001",$link) or die("cannot execute the query");
$query = mssql_query("SELECT Tr_Date,Tr_Time,Tr_Data from Matrix.dbo.Trans_R");
$f = mssql_fetch_array($query);
echo $f['Tr_Date'];
}
?>
Can i know why Negative Vote??
He asked me to :
" but what about getting the data from the Database?"

Related

How to connect to database file to extract the data? [duplicate]

This question already has answers here:
How to connect to a SQLite3 db with PHP
(3 answers)
Closed 5 years ago.
I need some help with my PHP code. I want to connect to the database file called myChannel.db to extract the data from the rows, but I have got no idea how to do that by using this code:
Here is the config:
<?php
define('DB_HOST', 'localhost');
define('DB_USER', 'myusername');
define('DB_PASSWORD', 'mypassword');
define('DB_DATABASE', 'mydbname');
$errmsg_arr = array();
$errflag = false;
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$link)
{
die('Failed to connect to server: ' . mysql_error());
}
$db = mysql_select_db(DB_DATABASE);
if(!$db)
{
die("Unable to select database");
}
?>
Here is the get-listing.php script:
<?php
$errmsg_arr = array();
$errflag = true;
$link;
//Connect to the database
require_once('config.php');
$qrytable1="SELECT id, channels, programme_title, programme_description, programme_start_date, programme_end_date FROM tvguide WHERE channels='$channels' && id='$id'";
$result1 = mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error());
while ($row = mysql_fetch_array($result1))
{
//output the data for channels, programme_title, programme_description, programme_start_date, programme_end
}
?>
The code will only allow me to connect to mysql and nothing is else.
Can you please show me an example how I could connect to myChannel.db and extract the data from the columns called channels, programme_title, programme_description, programme_start_date, programme_end where the channels is matched?
You asked for an example... this was my solution for querying geocoordinates from mySQL database. I call all of this in my php file...
//set up a MySQL query, this is the data you are asking for. This asks for all data from tabele 'shelters'. Be sure to call your specific table.
$sql = "SELECT * FROM shelters;";
if(!$results = $link->query($sql)){
die("Query Unsuccessful");
}
//Here I create a value map and pull out the results I care about from my request query.
$valueMap = array();
while ($data = $results->fetch_assoc()){
$valueMap[] = array(
'title' => $data['title'],
'lat' => $data['lat'],
'lng' => $data['lng'],
);
}
//encoding it into JSON is more manageble for me, but you will have to parse it into a javascript object in your index file.
$JSONData = json_encode($valueMap);
//heere you want to echo your data in order to retrieve it in your index file. Has the functionality of connecting to an API
echo $JSONData;
Hit me up in the comments with any more specific questions.

php local connection mysql database

I try to make a simple IOS app that can connect to mysql database and read one table. But my php code does't work and really have no idea why, it's seems correct to me. The database is in a raspberry phpmyadmin server and the server works great.
I will put my code here and please tell me what's wrong.
<?php
$host = "192.168.2.193";
$db = "produtos";
$user = "root";
$pass = "1234";
$connection = mysql_connect($host, $user, $pass);
if(!$connection)
{
die("Database server connection failed.");
}
else
{
//attempt to select the database
$dbconnect = mysql_select_db($db, $connection);
//check to see if we could select the database
if(!dbconnect)
{
die("Unable to connect to the specified database!");
}
else
{
$query = "SELECT * FROM produtos";
$resultset = mysql_query($query, $connection);
$records = array();
//loop throught all our records and add them to our array
while ($r = mysql_fetch_assoc($resultset))
{
$records[] = $r;
}
echo json_ecode($records);
echo $resultset;
}
}
?>
Based on the question:
use mysqli_connect rather than mysql_connect because mysql_connect is deprecated and will not work someday. Also what is the the error you are getting? change your die() statement to something more helpful die(mysqli_error($connection));
Based on your comment:
That error would suggest that you either A) don't have the right IP address or B) there is a network issue between your host server and the SQL server, is this code running on the same server that is hosting the SQL database? if so then you can probably just use localhost for your $host

Passing data between MySql and Objective C

I am working on a small social web application as a final project for my iOS class. I have a profile view controller where all the info about the user from the database is supposed to be displayed on the labels. The problem is that I don't really know the best way to do this. Here is my php script:
<?
// Database credentials
$host = 'localhost';
$db = 'blabla';
$uid = 'blabla';
$pwd = 'blabla';
// Connect to the database server
$link = mysql_connect($host, $uid, $pwd) or die("Could not connect");
//select the json database
mysql_select_db($db) or die("Could not select database");
// Create an array to hold our results
$arr = array();
//Execute the query
$rs = mysql_query("SELECT IdUser, username, fullname, phonenumber, facebook, instagram FROM login");
// Add the rows to the array
while($obj = mysql_fetch_object($rs)) {
$arr[] = $obj;
}
//return the json result.
echo '{"users":'.json_encode($arr).'}';
?>
So here I get the info about all the users in the database. I am sure this is not the right way to go, so I guess I need to change the SQL query to retrieve the data for the current user only. But how can I do this? Should I put the username which I enter on the login page into an extra variable and then pass it with JSON to this php script and add the 'WHERE username = 'blabla' statement to the SQL query then? If so, how can I pass the variable to this script with JSON?
Can you please give me some sample code? Or is there a different way to do this?
Thank you so much!
<?php
// Database credentials
$host = 'localhost';
$db = 'blabla';
$uid = 'blabla';
$pwd = 'blabla';
// Connect to the database server
$link = mysql_connect($host, $uid, $pwd) or die("Could not connect");
//select the json database
mysql_select_db($db) or die("Could not select database");
//Execute the query
$rs = mysql_query("SELECT IdUser, username, fullname, phonenumber, facebook, instagram FROM login");
// Add the rows to the array
$data = mysql_fetch_array($rs);
foreach($data as $rec){
echo "user: $rec<br>";
}
?>

Import csv from Yahoo Finance into MySQL using PHP

Using Johnboy's tutorial on importing .csv files to MySQL, I tried to make a script that would take exchange rate data from Yahoo finance, and write it to the MySQL database.
<?php
//connect to the database
$host = "****"
$user = "****"
$password = "****"
$database = "****"
$connect = mysql_connect($host,$user,$password);
mysql_select_db($database,$connect);
//select the table
if ($_FILES[csv][size] > 0) {
//get the csv file
$symbol = "ZARINR"
$tag = "l1"
$file = "http://finance.yahoo.com/d/quotes.csv?e=.csv&f=$tag&s='$symbol'=x";
$handle = fopen($file,"r");
//loop through the csv file and insert into database
do {
if ($data[0]) {
mysql_query("INSERT INTO ZAR_to_INR(exchange_rate) VALUES
(
'".addslashes($data[0])."',
)
");
echo "done";
}
} while ($data = fgetcsv($handle,1000,",","'"));
//redirect
header('Location: import.php?success=1'); die;
}
else{
echo "nope";
}
?>
I added the echos in the hope that they'd tell me whether or not the script worked. It doesn't work at all. There are no error messages or anything. When I run the script by opening it in my webhost, it simply does not run.
I'd appreciate any advice on how to make this script work (or even an alternate way of solving the problem).
try using mysql debugs :
mysql_select_db($database) or die('Cant connect to database');
$result = mysql_query($query) or die('query fail : ' . mysql_error());
$connect = mysql_connect($host,$user,$password)
or die('Cant connect to server: ' . mysql_error());
to find this outputs you need to check your php error_log : where-does-php-store-the-error-log

How to check if my ODBC Data Source exist in PHP?

I have a simple project and that is to create a function that will check for mysql and odbc connection. I'm already done in creating the function for mysql, here's my sample code:
function check() {
$serverName = 'localhost';
$userName = 'root';
$password = '123';
$db = 'sample';
$conn = mysql_connect($serverName, $userName, $password);
mysql_select_db($db, $conn);
$trans = 'SELECT * FROM Labels';
$trans_result = mysql_query($trans, $conn);
if(!$trans_result) {
die(mysql_error());
} else {
echo "connected";
}
}
Well this one works for me when checking for the mysql connection. Now, my question is, is it possible to create something like this for checking my odbc data source connection? So that would be like
$conn = odbc_connect("spmuse1","" ,""); # Open connection.
$trans = "SELECT French FROM Labels";
$trans_result = odbc_exec($conn, $trans);
if(!$trans_result) {
echo "error?";
} else {
echo "connected";
}
You know what I mean? When I use this code, I always have 2 this error
Warning: odbc_connect() [function.odbc-connect]: SQL error: [Microsoft][ODBC Driver Manager] Data source name not found and no default driver specified, SQL state IM002 in SQLConnect
Warning: odbc_exec(): supplied argument is not a valid ODBC-Link resource
Please help! Thanks.
First you need to decide vendor of odbc driver, I hope below example will works for you
<?php
// Configure connection parameters
$db_host = "server.mynetwork";
$db_server_name = "Dev_Server";
$db_name = "Dev_Data";
$db_file = 'c:\dbstorage\dev.db';
$db_conn_name = "php_script";
$db_user = "dbuser";
$db_pass = "dbpass";
$connect_string = "Driver={Adaptive Server Anywhere 8.0};".
"CommLinks=tcpip(Host=$db_host);".
"ServerName=$db_server_name;".
"DatabaseName=$db_name;".
"DatabaseFile=$db_file;".
"ConnectionName=$db_conn_name;".
"uid=$db_user;pwd=$db_pass";
// Connect to DB
$conn = odbc_connect($connect_string,'','');
// Query
$qry = "SELECT * FROM my_table";
// Get Result
$trans_result= odbc_exec($conn,$qry);
if(!$trans_result) {
echo "error?";
} else {
echo "connected";
}
?>
I spent several days looking for a simple answer, and came up with this, which works for me:
if (#odbc_connect("DBName","un","pw",SQL_CUR_USE_ODBC) == FALSE){
echo "Database does not exist";
} else {
$connection=odbc_connect("DBName","un","pw",SQL_CUR_USE_ODBC);
echo "Database exists";
}
The # suppresses the basic error if the database does not exist, so the connection try will just return false. Of course if the connection is good, then it creates the connection object.

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