I am trying to display a record from my database, however the page appears blank and doesn't display the data I am expecting. The code follows below:
<?php
$mysqli = new mysqli(localhost, root, USERPASS, DBNAME);
$query = "SELECT * FROM usertable WHERE userID= '" . $_SESSION["sess_uid"] . "'";
$result = mysqli_query($mysqli, $query);
$row = mysqli_fetch_row($result);
echo $row['userQuestion'];
?>
Any help would be appreciated.
Thanks
<?php
// there need to be strings arguments here
$mysqli = new mysqli('localhost', 'root', USERPASS, DBNAME);
// sql injection friendly query
$query = "SELECT * FROM `usertable`
WHERE `userID`='{$_SESSION["sess_uid"]}' LIMIT 1;";
// do we have a result
if($result = mysqli_query($mysqli, $query)){
// fetch a single row
if($row = mysqli_fetch_row($result)){
// print the record
var_dump($row);
}
}
?>
You need to wrap 'localhost' and 'root' as strings.
mysqli_fetch_row returns a numerical array.
You can print the content of the record using var_dump or use mysqli_fetch_assoc instead.
I have a simple insert statement that should use $GLOBALS that has my connection string in. Problem is that it does not insert the data. Have I done this correctly?
Insert.php
<?php
require 'core/init.php';
$Name = $_REQUEST["Name"];
$Venue = $_REQUEST["Venue"];
$Category = $_REQUEST["Category"];
$query = "INSERT INTO bands (Name, Venue, Category)
VALUES ('$Name', '$Venue', '$Category')";
mysql_query ($query,$GLOBALS)
or die ("could not add to database");
echo("You have added: <br />");
$result = mysql_query("SELECT * FROM bands ORDER BY Band_id DESC LIMIT 1");
while($row = mysql_fetch_array($result))
{
echo $row['Name']. "" . $row['Venue']. "" . $row['Category'];
echo "<br />";
}
?>
You need to connect mysql with mysql_connect() function then select database with mysql_select_db() function then you can call mysql_query function
you just cant execute queries with query string parameter you need to execute that connection query first and then select your database
And i suggest you to use mysqli instead of mysql cuz mysql methods will be deprecated soon
You don't put whole $GLOBALS variable in your mysql_query() call. You put the specific variable with the connection string only:
mysql_query($query, $GLOBALS['connection']) // Assuming you called the connection var "connection"
or die ("could not add to database");
I use PHP for server side scripting and mysql server for database.
If I use mysql_insert_id() then it gives "0" and use of LAST_INSERT_ID() causes error "object returned empty description".This error I see when I debug on client-side in objective-C.
My table's id column is auto generated. I dont' pass id explicitly.
Below is the PHP code :
// Connect to our database
$db = Frapi_Database::getInstance();
$sql = "INSERT INTO userTrip
(userId, fromLat, fromLon, fromLoc, fromPOI,
toLat, toLon, toLoc, toPOI,
tripFinished, isMatched, departureTime, createdAt)
values
(".$userId.",".$fromLat.",".$fromLon.", GeomFromText('POINT($fromLat $fromLon)')".",'".$fromPOI."',".$toLat.","
.$toLon.", GeomFromText('POINT($toLat $toLon)')".",'".$toPOI."',0,0,'".
$departureTime."','".date('Y-m-d H:i:s')."')";
$stmt = $db->prepare($sql);
if (!$stmt->execute())
throw new Frapi_Error('ERROR_INSERTING_RECORD');
$lastId = LAST_INSERT_ID();
$this->data['tripId'] = $lastId;
$db = null;
Frapi Database extends from PDO, so you would use this:
$lastId = $db->lastInsertId();
See also: PDO::lastInsertId()
Try this (if you use mysqli):
$db->insert_id;
Or (if you use PDO):
$db->lastInsertId();
are you looking for this ?
to get the last inserted id
mysql_insert_id();
mysql_insert_id
Try with
$id = mysql_insert_id();
it will work for you,try this link mysql_insert_id
and this
If your table have AUTO INCREMENT column like UserID,Emp_ID,.. then you can use this query to get last inserted record
SELECT * FROM table_name where UserID=(select MAX(UserID)from table_name)
In PHP code:
$con = mysqli_connect('localhost', 'userid', 'password', 'database_name');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
$sql = "SELECT * FROM table_name where UserID=(select MAX(UserID)from table_name)";
$result = mysqli_query($con, $sql);
Then you can use fetched data as your requirement
I have been able to manually insert values in my table using phpmyadmin, and even if i end up using the same php code i get from php my admin to call the query it STILL won't add the value to the table. here is the code:
<?php
$link = mysql_connect('localhost', 'username', 'password');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
mysql_select_db('sc2broating1', $link);
$sql = "INSERT INTO `sc2broad_tesing1`.`Persons` (`re`) VALUES (\'hello11\')";
mysql_query($sql);
mysql_close($link);
?>
Don't escape value.
$sql = "INSERT INTO `sc2broad_tesing1`.`Persons` (`re`) VALUES ('hello11')";
I would also consider using bound parameters, as seen in mysqli::prepare, if Mysqli is an option.
I have information spread out across a few databases and want to put all the information onto one webpage using PHP. I was wondering how I can connect to multiple databases on a single PHP webpage.
I know how to connect to a single database using:
$dbh = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
However, can I just use multiple "mysql_connect" commands to open the other databases, and how would PHP know what database I want the information pulled from if I do have multiple databases connected.
Warning : mysql_xx functions are deprecated since php 5.5 and removed since php 7.0 (see http://php.net/manual/intro.mysql.php), use mysqli_xx functions or see the answer below from #Troelskn
You can make multiple calls to mysql_connect(), but if the parameters are the same you need to pass true for the '$new_link' (fourth) parameter, otherwise the same connection is reused. For example:
$dbh1 = mysql_connect($hostname, $username, $password);
$dbh2 = mysql_connect($hostname, $username, $password, true);
mysql_select_db('database1', $dbh1);
mysql_select_db('database2', $dbh2);
Then to query database 1 pass the first link identifier:
mysql_query('select * from tablename', $dbh1);
and for database 2 pass the second:
mysql_query('select * from tablename', $dbh2);
If you do not pass a link identifier then the last connection created is used (in this case the one represented by $dbh2) e.g.:
mysql_query('select * from tablename');
Other options
If the MySQL user has access to both databases and they are on the same host (i.e. both DBs are accessible from the same connection) you could:
Keep one connection open and call mysql_select_db() to swap between as necessary. I am not sure this is a clean solution and you could end up querying the wrong database.
Specify the database name when you reference tables within your queries (e.g. SELECT * FROM database2.tablename). This is likely to be a pain to implement.
Also please read troelskn's answer because that is a better approach if you are able to use PDO rather than the older extensions.
If you use PHP5 (And you should, given that PHP4 has been deprecated), you should use PDO, since this is slowly becoming the new standard. One (very) important benefit of PDO, is that it supports bound parameters, which makes for much more secure code.
You would connect through PDO, like this:
try {
$db = new PDO('mysql:dbname=databasename;host=127.0.0.1', 'username', 'password');
} catch (PDOException $ex) {
echo 'Connection failed: ' . $ex->getMessage();
}
(Of course replace databasename, username and password above)
You can then query the database like this:
$result = $db->query("select * from tablename");
foreach ($result as $row) {
echo $row['foo'] . "\n";
}
Or, if you have variables:
$stmt = $db->prepare("select * from tablename where id = :id");
$stmt->execute(array(':id' => 42));
$row = $stmt->fetch();
If you need multiple connections open at once, you can simply create multiple instances of PDO:
try {
$db1 = new PDO('mysql:dbname=databas1;host=127.0.0.1', 'username', 'password');
$db2 = new PDO('mysql:dbname=databas2;host=127.0.0.1', 'username', 'password');
} catch (PDOException $ex) {
echo 'Connection failed: ' . $ex->getMessage();
}
I just made my life simple:
CREATE VIEW another_table AS SELECT * FROM another_database.another_table;
hope it is helpful... cheers...
Instead of mysql_connect use mysqli_connect.
mysqli is provide a functionality for connect multiple database at a time.
$Db1 = new mysqli($hostname,$username,$password,$db_name1);
// this is connection 1 for DB 1
$Db2 = new mysqli($hostname,$username,$password,$db_name2);
// this is connection 2 for DB 2
Try below code:
$conn = mysql_connect("hostname","username","password");
mysql_select_db("db1",$conn);
mysql_select_db("db2",$conn);
$query1 = "SELECT * FROM db1.table";
$query2 = "SELECT * FROM db2.table";
You can fetch data of above query from both database as below
$rs = mysql_query($query1);
while($row = mysql_fetch_assoc($rs)) {
$data1[] = $row;
}
$rs = mysql_query($query2);
while($row = mysql_fetch_assoc($rs)) {
$data2[] = $row;
}
print_r($data1);
print_r($data2);
$dbh1 = mysql_connect($hostname, $username, $password);
$dbh2 = mysql_connect($hostname, $username, $password, true);
mysql_select_db('database1', $dbh1);
mysql_select_db('database2',$dbh2);
mysql_query('select * from tablename', $dbh1);
mysql_query('select * from tablename', $dbh2);
This is the most obvious solution that I use but just remember, if the username / password for both the database is exactly same in the same host, this solution will always be using the first connection. So don't be confused that this is not working in such case. What you need to do is, create 2 different users for the 2 databases and it will work.
Unless you really need to have more than one instance of a PDO object in play, consider the following:
$con = new PDO('mysql:host=localhost', $username, $password,
array(PDO::ATTR_PERSISTENT => true));
Notice the absence of dbname= in the construction arguments.
When you connect to MySQL via a terminal or other tool, the database name is not needed off the bat. You can switch between databases by using the USE dbname statement via the PDO::exec() method.
$con->exec("USE someDatabase");
$con->exec("USE anotherDatabase");
Of course you may want to wrap this in a catch try statement.
You might be able to use MySQLi syntax, which would allow you to handle it better.
Define the database connections, then whenever you want to query one of the database, specify the right connection.
E.g.:
$Db1 = new mysqli('$DB_HOST','USERNAME','PASSWORD'); // 1st database connection
$Db2 = new mysqli('$DB_HOST','USERNAME','PASSWORD'); // 2nd database connection
Then to query them on the same page, use something like:
$query = $Db1->query("select * from tablename")
$query2 = $Db2->query("select * from tablename")
die("$Db1->error");
Changing to MySQLi in this way will help you.
You don't actually need select_db. You can send a query to two databases at the same time. First, give a grant to DB1 to select from DB2 by GRANT select ON DB2.* TO DB1#localhost;. Then, FLUSH PRIVILEGES;. Finally, you are able to do 'multiple-database query' like SELECT DB1.TABLE1.id, DB2.TABLE1.username FROM DB1,DB2 etc. (Don't forget that you need 'root' access to use grant command)
if you are using mysqli and have two db_connection file. like
first one is
define('HOST','localhost');
define('USER','user');
define('PASS','passs');
define('**DB1**','database_name1');
$connMitra = new mysqli(HOST, USER, PASS, **DB1**);
second one is
define('HOST','localhost');
define('USER','user');
define('PASS','passs');
define(**'DB2**','database_name1');
$connMitra = new mysqli(HOST, USER, PASS, **DB2**);
SO just change the name of parameter pass in mysqli like DB1 and DB2.
if you pass same parameter in mysqli suppose DB1 in both file then second database will no connect any more. So remember when you use two or more connection pass different parameter name in mysqli function
<?php
// Sapan Mohanty
// Skype:sapan.mohannty
//***********************************
$oldData = mysql_connect('localhost', 'DBUSER', 'DBPASS');
echo mysql_error();
$NewData = mysql_connect('localhost', 'DBUSER', 'DBPASS');
echo mysql_error();
mysql_select_db('OLDDBNAME', $oldData );
mysql_select_db('NEWDBNAME', $NewData );
$getAllTablesName = "SELECT table_name FROM information_schema.tables WHERE table_type = 'base table'";
$getAllTablesNameExe = mysql_query($getAllTablesName);
//echo mysql_error();
while ($dataTableName = mysql_fetch_object($getAllTablesNameExe)) {
$oldDataCount = mysql_query('select count(*) as noOfRecord from ' . $dataTableName->table_name, $oldData);
$oldDataCountResult = mysql_fetch_object($oldDataCount);
$newDataCount = mysql_query('select count(*) as noOfRecord from ' . $dataTableName->table_name, $NewData);
$newDataCountResult = mysql_fetch_object($newDataCount);
if ( $oldDataCountResult->noOfRecord != $newDataCountResult->noOfRecord ) {
echo "<br/><b>" . $dataTableName->table_name . "</b>";
echo " | Old: " . $oldDataCountResult->noOfRecord;
echo " | New: " . $newDataCountResult->noOfRecord;
if ($oldDataCountResult->noOfRecord < $newDataCountResult->noOfRecord) {
echo " | <font color='green'>*</font>";
} else {
echo " | <font color='red'>*</font>";
}
echo "<br/>----------------------------------------";
}
}
?>