The code below is an attempt to connect more than one database using PHP PDO and SQLite. No matter what I tried, it does not accept the select test1.table1. If I remove the database name the select works; so how do I reference more than one database in the select?
<?php
// connect to SQLite3 database
$query = "test1.sqlite3";
$db = new PDO("sqlite:$query");
// connect to second db
$query = "attach test2.sqlite3";
$db->query($query);
$query = "Select * FROM test1.table1 ";
$result = $db->query($query);
$rows = $result->fetchall(PDO::FETCH_ASSOC);
foreach ($rows as $row) {
echo "<pre>".print_r($row)."</pre>";
}
?>
The first database you connect to (test1.sqlite3 in your case) is always called main, regardless of the actual filename. Try SELECT * FROM main.table1.
Related
i want to connect two mysql databases one is located in localhost and other one is located in a server
here is what i have done so far and i am not getting either error or data
<?php
$con=mysql_connect('120.247.201.8:3306','root','root');
$con1=mysql_connect('localhost','root','');
//mysql_connect('localhost','root','');
if(!$con){
die('Try Again');
}
if(!$con1){
die('Try Again');
}
mysql_select_db("iot",$con1);
mysql_select_db("lora_gateway",$con);
$result =mysql_query("SELECT lora_gateway.`server_log`.`created_at`, lora_gateway.`server_log`.`temperature`FROM iot.`device` inner JOIN `lora_gateway`.`server_log` on `lora_gateway`.`server_log`.`gateway_Id` = `iot`.`device`.`gatewayId` where iot.`device`.`deviceId`='23' ORDER BY lora_gateway.`server_log`.`created_at` desc");
$num= mysql_num_rows($result);
print_r($num);
?>
This is a solution for multiple servers, connections and DBs.
Two or more MySQL 5 db servers
Two or more locations (local and/or anything else)
Two or more different databases
Two or more tables and fields
Tested and Works fine :-)
//Define your database connections and select your database want to use. In this example I use two connections and two DBs. But you can use more than two.
<?php
//MySQL Server 1
$dbhost1 = "127.0.0.1";
$dbuser1 = "dbuser1";
$dbpassword1 = "dbpass1";
$db1 = "database1";
$connection1 = mysql_connect($dbhost1,$dbuser1,$dbpassword1) or die (mysql_error());
mysql_select_db($db1,$connection1);
//MySQL Server 2
$dbhost2 = "xxx.xxx.xxx.xxx";
$dbuser2 = "dbuser2";
$dbpassword2 = "dbpass2";
$db2 = "database2";
$connection2 = mysql_connect($dbhost2,$dbuser2,$dbpassword2) or die (mysql_error());
mysql_select_db($db2,$connection2);
//The SQL statement
$sql =" SELECT database1.tablename1.fieldname1 AS field1, database2.tablename2.fieldname2 AS field2 FROM database1.tablename1,database2.tablename2";
//Execute query and collect results in $results
$results = mysql_query($sql);
//Print result until end of records
while($rows = mysql_fetch_array($results)){
print $rows["field1"]." | ".$rows["field2"]."<br>";
}
?>
First of all, try to accustom yourself to using PDO or at least the mysqli_ functions. It's the future. :)
mysql_query's second parameter, the connection link, is optional. If omitted, it uses the last connection opened with mysql_connect. (See php.net Documentation)
Ergo, always use $con or $con1 as 2nd parameter in order to use the correct connection.
Then, provided that your queries are correct, it should work as expected.
what about to add connection to mysql_query() ?
$result = mysql_query("SELECT lora_gateway.`server_log`.... desc", $con);
I am so sorry mybe it is a silly question but as I am new in web language and php I dont know how to solve this problem.
I have a code which is getting ID from user and then connecting to MySQL and get data of that ID number from database table and then show on webpage.
But I would like to what should I add to this code if user enter an ID which is not in table of database shows a message that no data found.
Here is my code:
<?php
//connect to the server
$connect = mysql_connect ("localhost","Test","Test") ;
//connection to the database
mysql_select_db ("Test") ;
//query the database
$ID = $_GET['Textbox'];
$query = mysql_query (" SELECT * FROM track WHERE Code = ('$ID') ");
//fetch the results / convert results into an array
$ID = $_GET['Textbox'];
WHILE($rows = mysql_fetch_array($query)) :
$ID = 'ID';
echo "<p style=\"font-color: #ff0000;\"> $ID </p>";
endwhile;
?>
Thank You.
Sorry if it is so silly question.
You should use PDO (great tutorial here: http://wiki.hashphp.org/PDO_Tutorial_for_MySQL_Developers ). This way, you can develop safer applications easier. You need to prepare the ID before inserting it to the query string, to avoid any user manipulation of the mysql query (it is called sql injection, guide: http://www.w3schools.com/sql/sql_injection.asp ).
The main answer to your question, after getting the results, you check if there is any row in the result, if you got no result, then there is no such an ID in the database. If you use PDO statements $stmt->rowCount();.
$db = new PDO('mysql:host=localhost;dbname=testdb;charset=utf8', 'username', 'password');
$stmt = $db->prepare("SELECT * FROM table WHERE Code=?");
$stmt->bindValue(1, $id, PDO::PARAM_INT); // or PDO::PARAM_STR
$stmt->execute();
$row_count = $stmt->rowCount();
if ($row_count > 0) {
$results = $stmt->fetchAll(PDO::FETCH_ASSOC);
//results are in $results
} else {
// no result, no such an ID, return the error to the user here.
}
Another reason to not use mysql_* functions: http://php.net/manual/en/migration55.deprecated.php
i am having a bit of issue with a mysql query. for some reason i can echo all all associated rows inside of the mysql query but outside of the query it only return the last row. here is my code. any suggestions?
//Get all associated
$q=mysql_query("SELECT * FROM `ACCOUNT` WHERE ACCOUNT_ID='$act_id'");
while ($row = mysql_fetch_assoc($q)){
$act_name=$row['ACT_NAME'];
echo "$act_name<br>"; // This returns all rows fine
}
echo "$act_name<br>"; // This only return the last row. i would like to get all rows.
The only way that you can fetch all of the records is by using PDO or MySQLi. Here is an example:
$conn = new mysqli($hostname, $username, $password, $database);
$query = "SELECT * FROM `ACCOUNT` WHERE ACCOUNT_ID='$act_id'";
$results = $conn->query($query);
$resultArray = $results->fetch_all(MYSQLI_ASSOC);
As #esqew said, you need to stop using the mysql_* functions.
So I am trying to download JSON into an iOS application and I don't really know a lot about PHP.
Currently I'm using a generic php scrip as a delegate that connects the the MySQL database and returns JSON results.
<?php
$host = "localhost"; //database host server
$db = "***"; //database name
$user = "root"; //database user
$pass = "root"; //password
$connection = mysql_connect($host, $user, $pass);
//Check to see if we can connect to the server
if(!$connection)
{
die("Database server connection failed.");
}
else
{
//Attempt to select the database
$dbconnect = mysql_select_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 table";
$resultset = mysql_query($query, $connection);
$records = array();
//Loop through all our records and add them to our array
while($r = mysql_fetch_assoc($resultset))
{
$records[] = $r;
}
//Output the data as JSON
echo json_encode($records);
}
}
?>
The issue with this is that I don't want to encode the entire table in JSON I just want to return a few select data fields from the particular table so I don't download unnecessary data.
I know that you do this in the SELECT query but something is wrong with my syntax for this.
I had been trying with:
$query = "SELECT *[datafield],[otherdatafield],... FROM table";
but that doesnt seem to be working.
Also, In that table there are two separate data fields that are used to build a URL for an image, and Im not sure how to combine the two here so that they are returned in the JSON as one field.
For example: I Have a base string
"http://wwww.mysite.com/"
and i would like the add the two data Fields to that string so that when it returns the JSON objects they have those fields already concatenated so that the app can just use that URL:
"http://wwww.mysite.com/[data field1]/[datafield2]"
The query you should be trying looks like
$query = "SELECT col1,col2 FROM table";// no need of asterisk if you mention col names
Now if you need to combine two columns there itself in the query, try something like
$query = "SELECT CONCAT('mysite.com/',col1,'/',col2) FROM table";
"/" is the separator between two columns.
The query to fetch individual cols is
select col1,col2,col3 from table_name
No need to provide * like you did
$query = "SELECT *[datafield],[otherdatafield],... FROM table";
^........here * is causing the problem.
I have two databases, one online (mysql) and one in my office (SQL Server) which I would like to compare and update where a value is different.
I am using php to connect to the SQL Server database and run a query to retrieve the information, then connecting to the Mysql database running a query. Then I need to compare the two queries and update where necessary.
Is there somewhere I can look for tips on how to do this, I am sketchy on PHP and struggling really.
This is as far as I have got-:
<?php
$Server = "**server**";
$User = "**user**";
$Pass = "**password**";
$DB = "**DB**";
//connection to the database
$dbhandle = mssql_connect($Server, $User, $Pass)
or die("Couldn't connect to SQL Server on $Server");
//select a database to work with
$selected = mssql_select_db($DB, $dbhandle)
or die("Couldn't open database $DB");
//declare the SQL statement that will query the database
$query = "SELECT p.id, p.code, ps.onhand";
$query .= "FROM products p with(nolock)";
$query .= "INNER JOIN productstockonhanditems ps with(nolock)";
$query .= "ON ps.ProductID = p.ID";
$query .= "WHERE ps.StockLocationID = 1";
//execute the SQL query and return records
$get_offlineproduct2 = mssql_query($query);
mysql_connect("**Host**", "**username**", "**password**") or die(mysql_error());
mysql_select_db("Database_Name") or die(mysql_error());
$get_onlineproducts = mysql_query(SELECT w.ob_sku, w.quantity
FROM product_option_value AS w
ORDER BY ob_sku)
or die(mysql_error());
//close the connection
mssql_close($dbhandle);
?>
I am looking to compare the value p.code to w.ob_sku and whenever they match copy the value of ps.onhand to w.quantity so the online database has the correct quantities from the office database.
My question I guess is how close am I to getting this right? Also am I doing this the right way, I don't want to get so far and realise that i am just wasting my time...
Thanks!
You do not need to fetch any record from MySQL, since you actually want to update it.
I would do something like this:
$query = 'SELECT p.code, ps.onhand FROM (...)';
// execute the SQL query and return a result set
// mssql_query() actually returns a resource
// that you must iterate with (e.g.) mssql_fetch_array()
$mssqlResult = mssql_query($query);
// connect to the MySQL database
mysql_connect("**Host**", "**username**", "**password**") or die(mysql_error());
mysql_select_db("Database_Name") or die(mysql_error());
while ( $mssqlRow = mssql_fetch_array($mssqlResult) ) {
$mssqlCode = $mssqlRow['code'];
$mssqlOnHand = $mssqlRow['onhand'];
mysql_query(
"UPDATE product_option_value SET quantity = $mssqlOnHand WHERE ob_sku = $mssqlCode"
// extra quotes may be required around $mssqlCode depending on the column type
);
}