I got this code to create a JSON file with the data I have in my database:
<?php
$connect = mysql_connect("localhost", "root", "123", "boxr") or die("Error connecting with the host.");
$sql = mysql_query("SELECT * FROM users") or die(mysql_error());
$response = array();
$users = array();
$result = mysql_query ($sql);
while ($row = mysql_fetch_array($result)) {
$gamertag = $row['gamertag'];
$banned = $row['banned'];
$users[] = array('gamertag' => $gamertag, 'banned' => $banned);
}
$response['users'] = $users;
$fp = fopen('users.json', 'w');
fwrite($fp, json_encode($response));
fclose($fp);
?>
The problem is when I run the script it shows the following:
No database selected
I checked my code twice but I can't figure out why its telling me that.
You have to select the database name with:
mysql_select_db ('your_db_name');
Well you can do it like that:
$connect = mysql_connect("localhost", "root", "123") or die("Error connecting with the host.");
mysql_select_db("boxr", $connect);
And later, correct it:
$sql = mysql_query("SELECT * FROM users") or die(mysql_error());
while ($row = mysql_fetch_array($result))...
By:
$sql = mysql_query("SELECT * FROM users") or die(mysql_error());
while ($row = mysql_fetch_array($sql))...
If you are calling the function with the name $sql, you can't later put the name of $result.
mysql_* functions are deprecated but you could change your query
$sql = mysql_query("SELECT * FROM dbname.users") or die(mysql_error());
Use :
$db_selected = mysql_select_db('db_name', $connect);
if (!$db_selected) {
die ('Can\'t use db_name : ' . mysql_error());
}
Related
I got this error while adding this code. Would appreciate some help. It's for a CS jackpot site.
$sitename = "website.com"; // YOUR DOMAIN
$link = mysql_connect("localhost", "db_user", "db_pass"); // MYSQL , LOCALHOOST , USERNAME , PASSWORD
$db_selected = mysql_select_db('db_name', $link); // MYSQL DATABASE
mysql_query("SET NAMES utf8");
function fetchinfo($rowname,$tablename,$finder,$findervalue) {
if($finder == "1") $result = mysql_query("SELECT $rowname FROM $tablename");
else $result = mysql_query("SELECT $rowname FROM $tablename WHERE `$finder`='$findervalue'");
while($row = mysql_fetch_assoc($query))
return $row[$rowname];
}
Some tips:
Use mysqli better than mysql
Split the vars in the query, like "SELECT ".$rowname." FROM ".$tablename;
Hope this help...
<?php
//Open a new connection to the MySQL server
$mysqli = new mysqli('host','username','password','database_name');
if ($mysqli->connect_error) {
die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);
}
//MySqli Select Query
$result = $mysqli->query("SELECT id, product_name FROM products");
while($row = $results->fetch_assoc()) {
echo $row["id"].' - '.$row["product_name"].'<br>';
}
$results->free();
$mysqli->close();
<?php
$username = "root";
$password = "password";
$database = "xxxxxx";
$link = mysql_connect("localhost", $username, $password);
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
mysql_select_db('xxxxxx', $link);
$result = mysql_query($query) or die(mysql_error($link));
$num = mysql_num_rows($result);
mysql_close();
$rows = array();
$result = mysql_query($query) or die(mysql_error());
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
print_r($rows);
?>
This is my code i want to display the roll number of the currently logged in user and
when i run this code i get no database selected.
First of all, you should code properly, means database connection and database selection should be on top:
<?php
$username = "root";
$password = "password";
$database = "xxxx";
$link = mysql_connect("localhost", $username, $password);
mysql_select_db('xxxx', $link);
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
$result = mysql_query($query) or die(mysql_error($link));
$num = mysql_num_rows($result);
$rows = array();
$result = mysql_query($query) or die(mysql_error());
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
print_r($rows);
mysql_close();
?>
Also moved mysql_close(); on last
One other main point was, now mysql_ is deprecated, please use mysqli_
Remove this statement from line number 10
mysql_close();
just remove these two lines before while that will solve ur problem
$result = mysql_query($query) or die(mysql_error());
$rows = array();
Use this code and use mysql_close function in the last.
<?php
$username = "root";
$password = "password";
$database = "xxxxxx";
$link = mysql_connect("localhost", $username, $password);
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
mysql_select_db($database, $link);
$result = mysql_query($query) or die(mysql_error($link));
$num = mysql_num_rows($result);
$rows = array();
$result = mysql_query($query) or die(mysql_error());
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
print_r($rows);
mysql_close();
?>
You have closed the connection from MySQL before the mysql_query mysql_close();
try this
<?php
$username = "root";
$password = "password";
$database = "dfsdftwsdgdfgdfsgsdf";
$link = mysql_connect("localhost", $username, $password);
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
mysql_select_db('xxxxxxx', $link);
$result = mysql_query($query) or die(mysql_error($link));
$num = mysql_num_rows($result);
$rows = array();
$result = mysql_query($query) or die(mysql_error());
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
mysql_close();
print_r($rows);
?>
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO, or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.
The order should be like this
mysql_select_db('meipolytechnic', $link);
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";
mysql_select_db('meipolytechnic', $link);
I want to get data from a table in my MySQL database.
$linkID = mysql_connect($host, $user, $pass) or die("Could not connect to host.");
mysql_select_db($database, $linkID) or die("Could not find database.");
$search='flu';
$query = mysql_query("SELECT * From ProviderDxCptCodes WHERE CodeType='CPT'");
$rows = array();
while($row = mysql_fetch_assoc($query)) {
$rows[] = $row;
}
echo json_encode($rows);
I want this query to show me all CPT codes matching the search string flu. How can I do that?
Do you mean something like that?
SELECT * From ProviderDxCptCodes WHERE CodeType='CPT' AND name LIKE '%flu%';
$linkID = mysql_connect($host, $user, $pass) or die("Could not connect to host.");
mysql_select_db($database, $linkID) or die("Could not find database.");
$flu = "CPT";
$query = mysql_query("SELECT * From ProviderDxCptCodes WHERE CodeType='%".mysql_real_escape_string($flu)."%'");
$rows = array();
while($row = mysql_fetch_assoc($query)) {
$rows[] = $row;
}
echo json_encode($rows);
I must be doing something wrong
I have a very simple script and a very simple database
No idea why it's not working..
Please help
<?php
error_reporting(E_ALL);
$link = mysql_connect('localhost', 'root', 'password');
if(!$link)
{
die('Could not connect: ' . mysql_error());
}
$database = mysql_select_db('test_db', $link);
if(!$database)
{
die('Could not connect to database: ' . mysql_error());
}
$result = mysqli_query($link, "SELECT forename FROM users WHERE id='1'");
if(!$result)
echo 'PROBLEM';
$row = mysqli_fetch_array($result);
echo $row[0];
?>
It's not even giving any errors, just echoing 'Problem'...
The database connects fine, and there is 1 user in the database with an ID of 1 and forename is Cristian.
In $result = mysqli_query try just useing $result = mysql_query same in $row = mysqli_fetch_array should help
Try replaceing you code from $result onward with this
$result = mysql_query("SELECT forename FROM users WHERE id='1'");
if(!$result)
echo 'PROBLEM';
$row = mysql_fetch_array($result);
echo $row[0];
?>
I am sure this is a stupid question but I can't figure out how to do it. I need to retrieve a large block of text that is stored in a database. I know how to connect to the database and submit the sql query. What I can't figure out how to do is have it store that text in a string so I can have it echo later on the page.
What I have so far
$db_name = "j_db";
$table_name = "contacts";
$connection = #mysql_connect("localhost", "*****", "****") or die(mysql_error());
$db = #mysql_select_db($db_name, $connection) or die(mysql_error());
$sql = "SELECT meaning
FROM $table_name
WHERE card = (php variable here)";
$result = #mysql_query($sql, $connection) or die(mysql_error());
Using mysql_fetch_array()
while( $row = mysql_fetch_array($result) ){
$meaning = $row["meaning"];
echo $meaning;
}
Look at the mysql_fetch_* functions. mysql_fetch_assoc is a popular choice.
<?php
$url= '' ;
$username = '' ;
$password = '' ;
$dbname = '' ;
$connection = mysql_connect('$url' , '$username', '$password');
mysql_select_db('$dbname' , $connection);
$result = mysql_query("SELECT textColumnName FROM tableName", $connection);
while($row = mysql_fetch_array($result)){
echo $row['textColumnName'];
}
?>