oci_parse error message - php

I get this error message, what does it mean and how to fix it?
Warning: oci_parse() expects parameter 1 to be resource, null given in /user_auth_fns.php on line 3
$conn = db_connect();
$result = oci_parse($conn, "select * from user where username='$username' and passwd = sha1('$password')");
if (!$result){
$err = oci_error();
echo "Could not log you in.";
exit;
}
$r = oci_execute($result);
if (!$r) {
$error = oci_error($conn);
echo "Could not log you in." . $error['message'];
exit;
}
function db_connect()
{
$db = "dbms";
if ($c=oci_connect("username", "password", $db)){
echo "Successfully connected to Oracle.\n";
OCILogoff($c);
} else {
$err = OCIError();
echo "Oracle Connect Error " . $err[text];
}
}
Edit 2 fixed the problem, another error message, what other compression function(other than SHA1) should use for oracle?
Warning: oci_execute() [function.oci-execute]: ORA-00904: "SHA1": invalid identifier in /user_auth_fns.php on line 10

I don't know what db_connect() is returning. Maybe it is just creating a connection by its own. Try this:
$conn = oci_connect("userName","password","hostName");
fill up the useName & password & hostName here. if you are having problem with hostName then try to put the whole connection string there. example:
$conn = oci_connect('userName', 'password', '(DESCRIPTION = (ADDRESS = (PROTOCOL = TCP)(HOST = )(PORT = )) (CONNECT_DATA = (SERVICE_NAME = ) (SID = )))');
then you can create a query like
$query="....";
then you can parse like this:
$result = oci_parse($conn, $query);
if you succeed in querying then $result holds Boolean value 'true'.

Your $conn variable is null. How do you instantiate that?
Edit
You instantiate $conn from db_connect(), which is not part of the standard PHP library so I can not tell you what's wrong with it other than it's returning null.
Edit 2
You're db_connect() doesn't return anything. Additionally, you close the connection immediately after opening it. Try this:
function db_connect()
{
$db = "dbms";
if ($c=oci_connect("username", "password", $db)){
echo "Successfully connected to Oracle.\n";
//OCILogoff($c); // You probably don't want to close the connection here
return $c;
} else {
$err = OCIError();
echo "Oracle Connect Error " . $err[text];
}
}

It means that $conn has no value, it is null. What did you want to have in $conn? Go back and check that.

Related

Returning id from access database

this is my code that is inserting data into an access database using php.
$conn = new COM ("ADODB.Connection") or die("Cannot start ADO");
$connStr = "PROVIDER=Microsoft.Ace.OLEDB.12.0;Data Source=" . realpath(‘my access path’) . ";";
// Open the connection to the database
$conn->open($connStr);
$query = “my insert query here which inserts into theaccess database fine”
$query2 = "select ##IDENTITY"
try{
$rs = $conn->execute($query);
$idReturned = $conn->lastInsertId();
echo json_encode($idReturned);
} catch(com_exception $e){
echo($e);
}
I’m trying to get the returned id but all I am getting is the below error :
exception 'com_exception' with message 'Source: ADODB.Connection
Description: Arguments are of the wrong type, are out of acceptable
range, or are in conflict with one another.' in
C:\inetpub\wwwroot\agency\createnewvaluation.php:132 Stack trace: #0
C:\inetpub\wwwroot\agency\createnewvaluation.php(132):
com->lastInsertId() #1 {main}
I went though the results manually and got the code myself
if($dbh->getAttribute(PDO::ATTR_DRIVER_NAME) == 'pgsql') {
} elseif($dbh->getAttribute(PDO::ATTR_DRIVER_NAME) == 'odbc') {
$sb = $dbh->prepare('SELECT ##IDENTITY AS lastID');
$sb->execute();
$row = $sb->fetch(PDO::FETCH_ASSOC);
$arr = array("ref" => $row["lastID"]);
echo json_encode($arr);
} else {
$arr = array("ref" => "error");
echo json_encode($arr);
}

Changing Mysql functions to Mysqli breaks the script

I have this script that deletes a certain picture from the website. It's written with mysql functions so i wanted to update it to mysqli but doing so makes the script stop working. No die message from the script are shown no php errors and adding error_reporting(E_ALL); doesn't show any errors either.
Original script:
if(isset($_POST['F3Verwijderen']))
try
{
//delete the file
$sql = "SELECT PandFoto3 FROM tblpand WHERE `PK_Pand` = '".$pandid."'";
$con = mysql_connect('WEBSITE.mysql', 'WEBSITE', 'PASS');
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("WEBSITE");
$result = mysql_query($sql, $con);
while ($row = mysql_fetch_array($result)) {
if(file_exists($_SERVER['DOCUMENT_ROOT'].'/'.$row['PandFoto3'])) {
unlink($_SERVER['DOCUMENT_ROOT'].'/'.$row['PandFoto3']);
} else {
echo $row['PandFoto3'];
}
}
//delete the path url from the database field
mysql_query("UPDATE tblpand SET PandFoto3 = NULL WHERE `PK_Pand` = '".$pandid."'");
mysql_close($con);
header('Location: ../admin/pand-aanpassen.php?id='.$pandid);
}
Updated to mysqli:
try
{
//delete the file
$sql = "SELECT PandFoto3 FROM tblpand WHERE `PK_Pand` = '".$pandid."'";
$con = mysqli_connect('WEBSITE.mysql', 'WEBSITE', 'PASS');
if (!$con) {
die('Could not connect: ' . mysqli_error());
}
mysqli_select_db("WEBSITE");
$result = mysqli_query($sql, $con);
while ($row = mysqli_fetch_array($result)) {
if(file_exists($_SERVER['DOCUMENT_ROOT'].'/'.$row['PandFoto3'])) {
unlink($_SERVER['DOCUMENT_ROOT'].'/'.$row['PandFoto3']);
} else {
echo $row['PandFoto3'];
}
}
//delete the path url from the database field
mysqli_query("UPDATE tblpand SET PandFoto3 = NULL WHERE `PK_Pand` = '".$pandid."'");
mysqli_close($con);
header('Location: ../admin/pand-aanpassen.php?id='.$pandid);
}
Edit:
"no php errors and adding error_reporting(E_ALL); doesn't show any errors either."
That's because it isn't a PHP issue, it's a MySQL issue.
Those are two different animals altogether.
As I said in commments, you need to switch these variables ($sql, $con) around ($con, $sql).
Then this:
$con = mysqli_connect('WEBSITE.mysql', 'WEBSITE', 'PASS');
Just use the 4th parameter instead of mysqli_select_db("WEBSITE"); where you didn't pass the connection variable to.
$con = mysqli_connect('WEBSITE.mysql', 'WEBSITE', 'PASS', 'WEBSITE');
The syntax is:
host
username
password (if any)
database
You also could have done mysqli_select_db($con, "WEBSITE");
Sidenote: In mysql_ (see footnotes), the connection comes last, unlike in mysqli_ which comes first.
Do the same for your UPDATE and pass the connection parameter first.
mysqli_query($con, "UPDATE...
Sidenote: To verify that the update truly was successful, use affected_rows()
http://php.net/manual/en/mysqli.affected-rows.php.
Another thing, mysqli_error() requires a connection to it mysqli_error($con) and check for errors for your queries.
I.e.:
$result = mysqli_query($con, $sql) or die(mysqli_error($con));
References:
http://php.net/manual/en/mysqli.query.php
http://php.net/manual/en/mysqli.error.php
http://php.net/manual/en/mysqli.select-db.php
Sidenote:
You're using try() but no catch(). Either remove it, or consult the manual:
http://php.net/manual/en/language.exceptions.php
Example #4 pulled from the manual:
<?php
function inverse($x) {
if (!$x) {
throw new Exception('Division by zero.');
}
return 1/$x;
}
try {
echo inverse(5) . "\n";
} catch (Exception $e) {
echo 'Caught exception: ', $e->getMessage(), "\n";
} finally {
echo "First finally.\n";
}
try {
echo inverse(0) . "\n";
} catch (Exception $e) {
echo 'Caught exception: ', $e->getMessage(), "\n";
} finally {
echo "Second finally.\n";
}
// Continue execution
echo "Hello World\n";
?>
Final notes:
Your present code is open to SQL injection. Use prepared statements, or PDO with prepared statements, they're much safer.
Footnotes: (MySQL and MySQLi comparison)
In regards to mysql_query():
mixed mysql_query ( string $query [, resource $link_identifier = NULL ]
http://php.net/manual/en/function.mysql-query.php
For mysqli_query():
mixed mysqli_query ( mysqli $link , string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )
http://php.net/manual/en/mysqli.query.php

mysqli connection and query

I am new to mysqli and was going through a tutorial from: http://www.binpress.com/tutorial/using-php-with-mysql-the-right-way/17#comment1
I was able to connect to my database using this:
$config = parse_ini_file('../config.ini');
$connection = mysqli_connect('localhost',$config['username'],$config['password'],$config['dbname']);
if($connection === false) {
die('Connection failed [' . $db->connect_error . ']');
}
echo("hello"); //this worked!
But then I tried wrapping it in a function (as discussed in the tutorial)... I saw that you call the connection function from another function... in the tutorial each function keeps getting called from another and another... and I never quite found where the initial call started from to get the domino effect of functions calling eachother.. so anyway, I tried to stop it at two just to test and teach myself.. but it's not working and I don't know why:
function db_connect() {
static $connection;
if(!isset($connection)) {
$config = parse_ini_file('../config.ini');
$connection = mysqli_connect('localhost',$config['username'],$config['password'],$config['dbname']);
}
if($connection === false) {
return mysqli_connect_error();
}
return $connection;
echo("hello2");
}
function db_query($query) {
$connection = db_connect();
$result = mysqli_query($connection,$query);
return $result;
echo("hello1");
}
db_query("SELECT `Q1_Q`,`Q1_AnsA` FROM `Game1_RollarCoaster`"); //this didn't work :(
Well I ended up taking it out of the functions and made the code super simple (sticking with procedural instead of OOP even though a lot of tutorials use OOP - thought it was better to start this way):
<?php
$config = parse_ini_file('../config.ini');
$link = mysqli_connect('localhost',$config['username'],$config['password'],$config['dbname']);
if(mysqli_connect_errno()){
echo mysqli_connect_error();
}
$query = "SELECT * FROM Game1_RollarCoaster";
$result = mysqli_query($link, $query);
while ($row = mysqli_fetch_array($result)) {
echo $row[Q1_Q] . '<-- Here is your question! ' . $row[Q1_AnsA] . '<-- Here is your answer! ';
echo '<br />';
}
mysqli_free_result($result);
mysqli_close($link);
?>
Here's a simple mysqli solution for you:
$db = new mysqli('localhost','user','password','database');
$resource = $db->query('SELECT field FROM table WHERE 1');
$row = $resource->fetch_assoc();
echo "{$row['field']}";
$resource->free();
$db->close();
If you're grabbing more than one row, I do it like this:
$db = new mysqli('localhost','user','password','database');
$resource = $db->query('SELECT field FROM table WHERE 1');
while ( $row = $resource->fetch_assoc() ) {
echo "{$row['field']}";
}
$resource->free();
$db->close();
With Error Handling: If there is a fatal error the script will terminate with an error message.
// ini_set('display_errors',1); // Uncomment to show errors to the end user.
if ( $db->connect_errno ) die("Database Connection Failed: ".$db->connect_error);
$db = new mysqli('localhost','user','password','database');
$resource = $db->query('SELECT field FROM table WHERE 1');
if ( !$resource ) die('Database Error: '.$db->error);
while ( $row = $resource->fetch_assoc() ) {
echo "{$row['field']}";
}
$resource->free();
$db->close();
With try/catch exception handling: This lets you deal with any errors all in one place and possibly continue execution when something fails, if that's desired.
try {
if ( $db->connect_errno ) throw new Exception("Connection Failed: ".$db->connect_error);
$db = new mysqli('localhost','user','password','database');
$resource = $db->query('SELECT field FROM table WHERE 1');
if ( !$resource ) throw new Exception($db->error);
while ( $row = $resource->fetch_assoc() ) {
echo "{$row['field']}";
}
$resource->free();
$db->close();
} catch (Exception $e) {
echo "DB Exception: ",$e->getMessage(),"\n";
}

sometimes my sql connection keep saying "PHP Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result"

Actually, I connected to 3 databases in different server, but I got problem in mysqli_query.
sometimes it worked fluently without any error, but sometimes it showed "PHP Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result"..
I don't know what happen, it ocurred sometimes.
here is my connection :
connection.php
<?php
$DatabaseName1 = "db_name1";
$DbHostName1 = "host1";
$DbUserName1 = "username";
$DbPassWord1 = "password";
$DatabaseName2q = "db_name2";
$DbHostName2 = "host2";
$DbUserName2 = "username2";
$DbPassWord2 = "password2";
$DatabaseName3q = "db_name3";
$DbHostName3 = "host3";
$DbUserName3 = "username3";
$DbPassWord3 = "password3";
$mysqli1 = mysqli_connect("$DbHostName1", "$DbUserName1", "$DbPassWord1", "$DatabaseName1");
$mysqli2 = mysqli_connect("$DbHostName2", "$DbUserName2", "$DbPassWord2", "$DatabaseName2q");
$mysqli3 = mysqli_connect("$DbHostName3", "$DbUserName3", "$DbPassWord3", "$DatabaseName3q");
$server[1] = $mysqli1;
$server[2] = $mysqli2;
$server[3] = $mysqli3;
$count_db = 3;
if(!$mysqli1){
echo "error to connect server 1st";
die();
}
if(!$mysqli2){
echo "error to connect server 2nd";
die();
}
if(!$mysqli3){
echo "error to connect server 3rd";
die();
}
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
?>
task1.php
<?php
include("connection.php")
for($i=1;$i<=$count_db;$i++){
$conn = $server[$i];
$array = mysqli_fetch_array(mysqli_query($conn,"select * from `table`"));
$task = $array["field"];
}
?>
anyone guys can help?
Do it like this
for($i=1;$i<=$count_db;$i++){
$conn = $server[$i];
$resultset = mysqli_query($conn,"select * from `table`")
if(!$resultset){
//do something
continue; //if you don't want to break it
}
$array = mysqli_fetch_array($resultset);
$task = $array["field"];
}

PHP: connected to database, but mysql_query fails

I have very strange problem with PHP which I am starting to learn .. I have created tables in MySQL database with some data, and now I want to show them in webpage.
This is my source where I have this problem:
<?php
// Here I open connection
$con = mysql_connect("localhost","root","aaaaaa");
// set the mysql database
$db = mysql_select_db("infs", $con);
// I check the connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
else {
// It always goes here
echo "Connected to database!";
}
// I am testing very simple SQL query.. there should be no problem
$result = mysql_query("SELECT * FROM cathegories", $con, $db);
if (!$result) {
// but it always dies
$message = 'Invalid query: ' . mysql_error() . "\n";
$message .= 'Whole query: ' . $query;
die($message);
}
mysql_close($con);
?>
What is wrong?
Thanks in advance!
You are mixing mysql and mysqli.
Try something like:
<?php
$con= new mysqli("localhost","user","passwd","database");
if ($con->connect_errno){
echo "could not connect";
}
$select = "SELECT * FROM tablename";
if($result = $con->query($select)){
while($row = $result->fetch_object()){
echo $row->rowname."<br>";
}
}
else { echo 'no result'; }
$con->close();
?>
// Here I open connection
$con = mysql_connect("localhost","root","aaaaaa");
// set the mysql database
$db = mysql_select_db("infs", $connection);
change to
// Here I open connection
$con = mysql_connect("localhost","root","aaaaaa");
// set the mysql database
$db = mysql_select_db("infs", $con);
mysql_query only takes two parameters - the actual SQL and then the link identifier (I assume in your case that's stored in $con; therefore remove $db from the third parameter).
You don't even need the second $con parameter really.
Where's the actual logic to connect to the database initially? Just because mysqli_connect_errno() doesn't return an error it doesn't mean the connection actually exists and that $con is available in the current scope.
I'd var_dump($con) before the mysql query to make sure it's a valid connection.

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