PDO Exception 1049 when migrating PHP application from MySQL - php

I have been trying to migrate the following mysql connection function to PDO:
function validate_email_input($email)
{
$dbc = #mysql_connect('localhost', 'userName', '******');
mysql_select_db('users', $dbc);
$query = "SELECT email FROM user_list WHERE email='$email'";
if ($result = mysql_query($query, $dbc))
{
$row = mysql_fetch_array($result);
if ($row['email'] == $email)
{
return TRUE;
}
else
{
return FALSE;
}
}
else
{
echo '<p class="error">Could not retrieve he data because:<br />' . mysql_error($dbc) . '.</p>
<p>The query being run was: ' . $query . '</p>';
}
}
This works just fine. But I get an error when I try the following PDO function to achieve the same result:
function validate_email_input($email)
{
// Step 1: Establish a connection
$db = new PDO("mysql:host=localhost;dbname=users", "userName", "******");
// Step 2: Construct a query
$query = "SELECT * FROM user_list WHERE email = '$email'";
// Step 3: Send the query
$result = $db->query($query);
// Step 4: Iterate over the results
if ($result)
{
$row = $result->fetch(PDO::FETCH_ASSOC);
if ($row['email'] == $email)
{
return TRUE;
}
else
{
return FALSE;
}
}
// Step 5: Free used resources
$result->closeCursor();
$db = null;
}
And yes - I did try the above function with try-catch but that doesn't affect anything. I still get the error:
Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[HY000] [1049] Unknown database 'users''
I'm using the apache2 module included with Bitnami RubyStack (v. 1.9.3-25), which uses PHP version 5.4.3
Already checked php.ini file and all the PDO extensions are uncommented.
Please help! Thanks

Found the solution! Apparently it wanted me to specify the port number (localhost:3307) instead of just localhost when creating the PDO object. Anyone know how to change the default localhost port in Bitnami RubyStack for Windows?

Related

How do I start a PGSQL connection in PHP?

I am trying to start a pgsql connection in php but i get pg_last_error(): No PostgreSQL link opened yet Please forgive my amateur question as i am a bit new to php.
Here is my php code:
<?php
$connect = pg_connect("host=xxx.xx.xxx.21 dbname=d106 user=b16 password=bran") or die("Could not connect: " . pg_last_error());
$result = pg_query($connect,"SELECT distinct thestartgeom FROM bikes");
if (!$result)
{
echo "no results ";
}
while($row = pg_fetch_array($result))
{
$coor = $row['thestartgeom'];
echo $coor;
}
pg_close($connect);
?>
In your pgsql connection you have missed the port number.
Try this way.
<?php
$host = "host=xxx.xx.xxx.21";
$port = "port=5432";
$dbname = "dbname=d106";
$credentials = "user=b16 password=bran";
$connect= pg_connect( "$host $port $dbname $credentials" ) or die("Could not connect: " . pg_last_error());
$result = pg_query($connect,"SELECT distinct thestartgeom FROM bikes");
if (!$result)
{
echo "no results ";
}
while($row = pg_fetch_array($result))
{
$coor = $row['thestartgeom'];
echo $coor;
}
pg_close($connect);
?>
You can store PgSQL connection code in one PHP file to reuse
pgsql_db_connection.php file
<?php
$host = "host=xxx.xx.xxx.21";
$port = "port=5432";
$dbname = "dbname=d106";
$credentials = "user=b16 password=bran";
$connect= pg_connect( "$host $port $dbname $credentials" );
if(!$connect){
echo "Error : Unable to open database\n";
}
?>
Call pgsql_db_connection.php file in other php files to use your database connection.
<?php
require_once('pgsql_db_connection.php');
$result = pg_query($connect,"SELECT distinct thestartgeom FROM bikes");
if (!$result)
{
echo pg_last_error($connect);
exit;
}
while($row = pg_fetch_array($result))
{
$coor = $row[0];
echo $coor;
}
?>
When pg_connect fails, it returns FALSE and produces a PHP warning with the detailed information on why it couldn't initiate the connection. If you can see the other message:pg_last_error(): No PostgreSQL link opened yet that you're reporting, I'd expect you should be able to see the previous one too, which is the one normally telling the reason of the failure.
If the display_errors configuration setting is set to 0, the first message would not show up on the browser/screen but the second would not either.
Anyway, assuming you can't have access to pg_connect warnings for whatever reason such as custom error handler, what's wrong with your code is that pg_last_error() must have an already opened connection to work.
To access the detailed error message from a failed pg_connect, the built-in PHP function error_get_last() (returning an array) could be used.
<?
$connect= pg_connect("your-connect-string");
if (!$connect) {
print_r(error_get_last());
// for only the message:
// echo error_get_last()['message']
}
die("DB connection failed");
?>
See also how to catch pg_connect() function error? if you prefer exceptions.

Converting from mysql to mysqli and getting Fatal error: Cannot use object of type mysqli as array

I am working on converting some PHP code from mysql to mysqli. I have created an error and am unable to understand how to fix it. Any suggestions would be greatly appreciated.
The code looks like this:
<?php
include ("admin/includes/connect.php");
$query = "select * from posts order by 1 DESC LIMIT 0,5";
$run = mysqli_query($conn["___mysqli_ston"], $query);
while ($row=mysqli_fetch_array($run)){
$post_id = $row['post_id'];
$title = $row['post_title'];
$image = $row['post_image'];
?>
The error produced is: Fatal error: Cannot use object of type mysqli as array
The error is being called out on this line:
$run = mysqli_query($conn["___mysqli_ston"], $query);
In the line above $conn is a variable from the database connect file which has this code:
<?php
// Stored the db login credentials in separate file.
require("db_info.php");
// Supressing automated warnings which could give out clues to database user name, etc.
mysqli_report(MYSQLI_REPORT_STRICT);
// Try to open a connection to a MySQL server and catch any failure with a controlled error message.
try {
$conn=mysqli_connect ('localhost', $username, $password) or die ("$dberror1");
} catch (Exception $e ) {
echo "$dberror1";
//echo "message: " . $e->message; // Not used for live production site.
exit;
}
// Try to Set the active MySQL databaseand catch any failure with a controlled error message.
try {
$db_selected = mysqli_select_db($conn, $database) or die ("$dberror2");
} catch (Exception $e ) {
echo "$dberror2";
//echo "message: " . $e->message; // Not used for live production site.
exit;
// We want to stop supressing automated warnings after the database connection is completed.
mysqli_report(MYSQLI_REPORT_OFF);
}
?>
This line
$run = mysqli_query($conn["___mysqli_ston"], $query);
should be
$run = mysqli_query($conn, $query);
If you're migrating to mysqli, you should really read these docs at least.
The proper way to use a mysqli connection:
<?php
$mysqli = new mysqli("example.com", "user", "password", "database");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
$res = $mysqli->query("SELECT id FROM test ORDER BY id ASC");
while ($row = $res->fetch_assoc()) {
echo " id = " . $row['id'] . "\n";
}
?>
you should also consider utilizing mysqli's prepared statements

Convert mysql_* to PDO

I have made a program using PHP and trying to store data into Local Server Xampp, but whenever i run my php script using this url:
http://127.0.0.1/test.php
Getting error message: {"StatusID":"0","Error":"Cannot save data!"}
Please someone help me in this how can i make it useful for me, please check below PHP Script:
<?php
$objConnect = mysql_connect("localhost","root","");
mysql_error($ObjConnect);
$objDB = mysql_select_db("registration_login");
mysql_error($ObjDB);
$strUsername = $_POST["sUsername"];
$strPassword = $_POST["sPassword"];
$strName = $_POST["sName"];
$strEmail = $_POST["sEmail"];
$strTel = $_POST["sTel"];
/*** Insert ***/
$strSQL = "INSERT INTO member (Username,Password,Name,Email,Tel)
VALUES (
'".$strUsername."',
'".$strPassword."',
'".$strName."',
'".$strEmail."',
'".$strTel."'
)
";
$objQuery = mysql_query($strSQL);
mysql_error($ObjQuery);
if(!$objQuery)
{
$arr["Status"] = "0";
$arr["Message"] = "Cannot Save Data!";
echo json_encode($arr);
exit();
}
else
{
$arr["Status"] = "1";
$arr["Message"] = "Register Successfully!";
echo json_encode($arr);
exit();
}
mysql_close($objConnect);
?>
Note: I have created registration_login database and member table under this DB..
Why don't you return the error reported by mysql or log it somewhere?
$objConnect = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
You forgot to check the return the return value to see if this was successful - if it failed, the reason is in mysql_error()
$objDB = mysql_select_db(DB_DATABASE);
You forgot to check the return the return value to see if this was successful - if it failed, the reason is in mysql_error()
$objQuery = mysql_query($strSQL);
At least this time you check the return value - but you don't check what the error was.
BTW your script is wide open to SQL injection.
Convert mysql_* to PDO
What has that got to do with your post?

Trying to get MySQL to connect via php into JSON, but returning errors connecting to MySQL from a php script

Essentially, I am trying to print out information in JSON so that I can communicate with my app, but I cannot connect to the MySQL database from a php script for some odd reason. What could it be that causes the error:
Warning: mysql_connect() [function.mysql-connect]: Lost connection to MySQL server during query in /srv/disk11/1158855/www/(myphpwebsite)/lib.php on line 13
Could not connect: Lost connection to MySQL server during query.
Also, line 13 is indicating the line in lib.php:
mysql_connect ( $dbhost, $dbuser, $dbpass) or die("Could not connect: ".mysql_error());
It should also be noted that this is a followup to a previous question in case anyone wanted to track down the source: MySQL issue connecting to site with php.
Lastly, I get the same error from both a localhost and a remote server using mysql
lib.php
<?
//Database Information
$dbhost = "31.170.160.76";
$dbname = "testdatabase";
$dbuser = "(personalinformation)";
$dbpass = "tested123";
//Connect to database
mysql_connect ( $dbhost, $dbuser, $dbpass) or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());
//executes a given sql query with the params and returns an array as result
function query() {
global $link;
$debug = false;
//get the sql query
$args = func_get_args();
$sql = array_shift($args);
//secure the input
for ($i=0;$i<count($args);$i++) {
$args[$i] = urldecode($args[$i]);
$args[$i] = mysqli_real_escape_string($link, $args[$i]);
}
//build the final query
$sql = vsprintf($sql, $args);
if ($debug) print $sql;
//execute and fetch the results
$result = mysqli_query($link, $sql);
if (mysqli_errno($link)==0 && $result) {
$rows = array();
if ($result!==true)
while ($d = mysqli_fetch_assoc($result)) {
array_push($rows,$d);
}
//return json
return array('result'=>$rows);
} else {
//error
return array('error'=>'Database error');
}
}
//loads up the source image, resizes it and saves with -thumb in the file name
function thumb($srcFile, $sideInPx) {
$image = imagecreatefromjpeg($srcFile);
$width = imagesx($image);
$height = imagesy($image);
$thumb = imagecreatetruecolor($sideInPx, $sideInPx);
imagecopyresized($thumb,$image,0,0,0,0,$sideInPx,$sideInPx,$width,$height);
imagejpeg($thumb, str_replace(".jpg","-thumb.jpg",$srcFile), 85);
imagedestroy($thumb);
imagedestroy($image);
}
?>
Index.php
<?
session_start();
require("lib.php");
require("api.php");
header("Content-Type: application/json");
switch ($_POST['command']) {
case "login":
login($_POST['username'], $_POST['password']); break;
case "register":
register($_POST['username'], $_POST['password']); break;
}
exit();
?>
api.php
<?php
function errorJson($msg){
print json_encode(array('error'=>$msg));
exit();
}
function register($user, $pass) {
//check if username exists
$login = query("SELECT username FROM login WHERE username='%s' limit 1", $user);
if (count($login['result'])>0) {
errorJson('Username already exists');
//try to register the user
$result = query("INSERT INTO login(username, pass) VALUES('%s','%s')", $user, $pass);
if (!$result['error']) {
//success
login($user, $pass);
} else {
//error
errorJson('Registration failed');
}
}
}
function login($user, $pass) {
$result = query("SELECT IdUser, username FROM login WHERE username='%s' AND pass='%s' limit 1", $user, $pass);
if (count($result['result'])>0) {
//authorized
$_SESSION['IdUser'] = $result['result'][0]['IdUser'];
print json_encode($result);
} else {
//not authorized
errorJson('Authorization failed');
}
}
?>
As this is on the "connect" line, the server has been found (otherwise you get a different message) but you've not negotiated your log in.
Straight from the manual:
More rarely, it can happen when the client is attempting the initial connection to the server. In this case, if your connect_timeout value is set to only a few seconds, you may be able to resolve the problem by increasing it to ten seconds, perhaps more if you have a very long distance or slow connection.
If that isn't it, it's either a network problem or your connection has been terminated mid-authentication. Check that your mysql host doesn't have some weird validation that you're coming from a particualr IP (I say weird, as there are more standard ways of managing it than killing the authentication mid-flow), or try your PHP script from a a server that is closer to the MySQL server (closer in terms of network speed).
I figured out what was the matter. It turns out that my php code further down was conflicting with the login. That's why it wouldn't authenticate on multiple remote MySQL's and my own Localhost

Insert data into MySql table strange exception

I am using this code to insert some values in MySql table:
<?php
mysql_connect("localhost","root","root");
mysql_select_db("bib");
$id = "12";
$titlu = "Joe";
$query = "INSERT INTO carte SET id='$id', titlu='$titlu'";
$result = mysql_query($query);
// Display an appropriate message
if ($result)
echo "<p>Product successfully inserted!</p>";
else
echo "<p>There was a problem inserting the Book!</p>";
mysql_close();
?>
After running it into browser, the following error occurs:
"Apache HTTP Server has encountered a problem and needs to close. We are sorry for the inconvenience."
It seems that mysql_select_db("bib") statement causes it. Database is create , also table...
I am running php 5.3 and mysql 5.1 on windows xp sp 2.
Please any ideas are welcomed...
Thanks...
Any of the mysql_* functions can fail for various reasons. You have to check the return values and if a function indicates an error (usually by returning FALSE) your script has to react appropriately.
mysql_error($link) and mysql_errno($link) can give you more detailed information about the cause. But you don't want to show all the details to just any arbitrary user, see CWE-209: Information Exposure Through an Error Message.
If you don't pass the connection resource returned by mysql_connect() to subsequent mysql_* functions calls, php assumes the last successfully established connection. You shouldn't rely on that; better pass the link resource to the functions. a) If you ever have more than one connection per page you must pass it anyway. b) If there is no valid db connection the php-mysql modules tries to establish the default connection which is usually not what you want; it only takes up more time to fail ..again.
<?php
define('DEBUGOUTPUT', 1);
$mysql = mysql_connect("localhost","root","root");
if ( !$mysql ) {
foo('query failed', mysql_error());
}
$rc = mysql_select_db("bib", $mysql);
if ( !$rc) {
foo('select db', mysql_error($mysql));
}
$id = "12";
$titlu = "Joe";
$query = "INSERT INTO carte SET id='$id', titlu='$titlu'";
$result = mysql_query($query, $mysql);
// Display an appropriate message
if ($result) {
echo "<p>Product successfully inserted!</p>";
}
else {
foo("There was a problem inserting the Book!", mysql_error($mysql), false);
}
mysql_close($mysql);
function foo($description, $detail, $die=false) {
echo '<pre>', htmlspecialchars($description), "</pre>\n";
if ( defined('DEBUGOUTPUT') && DEBUGOUTPUT ) {
echo '<pre>', htmlspecialchars($detail), "</pre>\n";
}
if ( $die ) {
die;
}
}
try this to connect to database:
$mysqlID = mysql_connect(DB_HOST, DB_USERNAME, DB_PASSWORD) or die("Unable to connect to database");
mysql_select_db(DB_DATABASE) or die("Unable to select database ".DB_DATABASE);
also, try this as your insert query:
$query = "INSERT INTO carte (id, title) values ('".$id."', '".addslashes($titlu)."')
$result = mysql_query($query) or die(mysql_error());
By using die(), it will tell you where it has failed and why

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