Issues with php connection to mySQL database - php

Hy everyone, I can't wrap my head around this. I'm trying to get some data from a table using PDO. this is my code:
//in db.php I have the connection:
$host = 'localhost';
$db = 'APL';
$dbuser = '';
$pass = ' ';
try{
$conn = new PDO("mysql:host=$host;dbname=$db", $dbuser, $pass);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch(PDOException $e)
{
echo "Connection failed: " . $e->getMessage();
}
//in my file I have this:
$id = $_GET['id'];
$sel_sql = "SELECT * FROM users WHERE id =:id";
$stmt = $conn ->prepare($sel_sql);
$stmt -> bindParam(':id', $id);
$stmt -> execute();
$result = $stmt -> fetchAll(PDO::FETCH_ASSOC);
The problem is that print_r($result) returns '1' (just the value 1, therefore I can't access any data stored in the table) as long as $_SESSION['user'] is set.
The whole data-retrieving worked just fine if the $_SESSION['user'] is not set.
Can someone please explain why this is happening? (I'm fairly new to all this and I'm really trying to understand why some issues occur).
Thank you!

The fetchAll function should be returning either an array, or a boolean FALSE.
You report that print_r($result) is displaying an integer value of 1.
I don't see how that's possible, unless you are assigning a different value to $result. Try relocating print_r($result) to immediately follow the assignment from fetchAll.
(My suspicion is that $result is being assigned a value of 1 elsewhere in your code, before you do the print_r. If there were "Issues with php connection to MySQL database", we'd be expecting to see a PDO error of some sort.)
NOTE: I don't think PDO::FETCH_ASSOC is a defined fetch style for the fetchAll function. (fetchAll has different fetch styles than fetch.)

Just in case someone else stumbles upon this, between the $result variable and the print_r($result) I had an include_once(); statement (which was wrongly put there in the first place).
Thank you everyone for your answers.

Related

MySQL not using PHP variables properly in Queries, replacing the variables with strings/integers works fine

MySQL is not using the variables as it should. it is not taking any value from them it is incrementing the auto-increment numbers in the MYSQL table, however the row is not saved. I am not given any errors.
I have tried like this:
$sql = "INSERT INTO `tbl_bike` (`userID`, `ManuPartNo`, `BikeManufacturer`, `BikeModel`, `BikeType`, `BikeWheel`, `BikeColour`, `BikeSpeed`, `BrakeType`, `FrameGender`, `AgeGroup`, `DistFeatures`)
VALUES (“.$userID.”, “.$PartNo.”, “.$BikeManufacturer.”, “.$BikeModel.”, “.$BikeType.”, “.$BikeWheel.”, “.$BikeColour.”, “.$BikeSpeed.”, “.$BrakeType.”, “.$FrameGender.”, “.$AgeGroup.”, “.$DistFeatures.”)";
I have also tried replacing the " with ', Removing the . and even completely removing the ". Nothing has helped with this issue. When I use this query but remove the variables and instead put string, int etc in the correct places the query will function perfectly and put the results into the table. My variables are normally as follows:
$PartNo = $_POST['ManuPartNo’];
$BikeManufacturer = $_POST['BikeManufacturer’];
$BikeModel = $_POST['BikeModel’];
$BikeType = $_POST['BikeType’];
$BikeWheel = $_POST['BikeWheel’];
$BikeColour = $_POST['BikeColour’];
$BikeSpeed = $_POST['BikeSpeed’];
$BrakeType = $_POST['BrakeType’];
$FrameGender = $_POST['FrameGender’];
$AgeGroup = $_POST['AgeGroup’];
$DistFeatures = $_POST['DistFeatures’];
These variables normally take input from a separate PHP/HTML file with the '$_POST['DistFeatures’];'
I have tried removing the $_POST['DistFeatures’]; from the ends of each of them and just replacing the values with normal string or int values but still nothing helps. I am completely stuck and would appreciate any help with this.
This is all running on a plesk server.
Please stop using deprecated MySQL. I will suggest an answer using PDO. You can use this to frame your other queries using PDO.
// Establish a connection in db.php (or your connection file)
$dbname = "dbname"; // your database name
$username = "root"; // your database username
$password = ""; // your database password or leave blank if none
$dbhost = "localhost";
$dbport = "10832";
$dsn = "mysql:dbname=$dbname;host=$dbhost";
$pdo = new PDO($dsn, $username, $password);
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_WARNING);
// Include db.php on every page where queries are executed and perform queries the following way
// Take Inputs this way (your method is obsolete and will return "Undefined Index" error)
$userId = (!empty($_SESSION['sessionname']))?$_SESSION['sessionname']:null; // If session is empty it will be set to Null else the session value will be set
$PartNo = (!empty($_POST['ManuPartNo']))?$_POST['ManuPartNo']:null; // If post value is empty it will be set to Null else the posted value will be set
$BikeManufacturer = (!empty($_POST['BikeManufacturer']))?$_POST['BikeManufacturer']:null;
$BikeModel = (!empty($_POST['BikeModel']))?$_POST['BikeModel']:null;
$BikeType = (!empty($_POST['BikeType']))?$_POST['BikeType']:null;
$BikeWheel = (!empty($_POST['BikeWheel']))?$_POST['BikeWheel']:null;
// Query like this
$stmt = $pdo->prepare("INSERT INTO(`userID`, `ManuPartNo`, `BikeManufacturer`, `BikeModel`, `BikeType`)VALUES(:uid, :manuptno, :bkman, :bkmodel, :bktype)");
$stmt-> bindValue(':uid', $userId);
$stmt-> bindValue(':manuptno', $PartNo);
$stmt-> bindValue(':bkman', $BikeManufacturer);
$stmt-> bindValue(':bkmodel', $BikeModel);
$stmt-> bindValue(':bktype', $BikeType);
$stmt-> execute();
if($stmt){
echo "Row inserted";
}else{
echo "Error!";
}
See, it's that simple. Use PDO from now on. It's more secured. To try this, just copy the whole code in a blank PHP file and and run it. Your database will receive an entry. Make sure to change your database values here.
You should try this
$sql = "INSERT INTO tbl_bike (userID, ManuPartNo, BikeManufacturer, BikeModel, BikeType, BikeWheel, BikeColour, BikeSpeed, BrakeType, FrameGender, AgeGroup, DistFeatures) VALUES ('$userID', '$PartNo', '$BikeManufacturer', '$BikeModel', '$BikeType', '$BikeWheel', '$BikeColour', '$BikeSpeed', '$BrakeType', '$FrameGender', '$AgeGroup', '$DistFeatures')";
If this doesn't work, enable the null property in sql values. So you can find out where the error originated.

MySQL PHP Column Update Query

i would really appreciate if anyone can help me out with this mysql php problem of which i have no idea how to do it.
I Have a column named = 'x'
The text of that column 'x' is = "yz,zz,zy"
I want to edit the value of the column 'x' to = "yz,zyz,zy".
Now how do i add that 'y' in the middle term between yz and zy using CONCAT.
Regards.
You haven't provided any code or an attempt for us to go off of something so I'll give you a brief way of doing it. Look up PDO here This is a really easy to follow and secure way to manipulate data in your database using php. Again as you haven't given me much to go off i'm unsure if you want to just set something at specific count of characters along OR if you want to just update the entire thing, SO i'll help you with the latter as it will help give you some base understanding.
Please read into PDO further as this is just an example further down the line and will not run if you just blindly copy and paste it in.
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "test";
try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "UPDATE table SET x='yz,zyz,zy' WHERE id= 1"; // No clue if you've even given anything IDs
$stmt = $conn->prepare($sql);
$stmt->execute();
echo $stmt->rowCount() . " records UPDATED";
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage();
}
$conn = null;
?>
As you haven't really provided any Schema of the table you are updating this is the best I could provide based off of what you have given me.. try to include as much information as possible as I'm unsure if this is what you even want

How to fetch single row with PDO from href link, receives fatal error: Call to a member function prepare() on a non-object?

I am starting to learn php PDO because I've read that it is more efficient and secure.
I could do the following with simple mysqli but am having trouble making it work with PDO.
PID stands for an id number.
fname stands for: first name.
lname stands for: last name.
age stands for ... age.
Basically I have an index.php that contains links from a test table called "persons" inside of the database drinks. When I click on the link which shows the fname of every row, it goes to insertcarbonated.php which is then supposed to $_GET['fname']; of the link and search up that specific row. However, my code in insertcarbonated.php is not working and I am not familiar enough with PDO to know exactly why, I would like some enlightenment on this because I literally begun learning PDO yesterday. :(
Here is my insertcarbonated.php:
<html>
<?php
/*** mysql hostname ***/
$hostname = 'localhost';
/*** mysql username ***/
$username = 'theusername';
/*** mysql ***/
$password = 'thepass';
try {
$dbh = new PDO("mysql:host=$hostname;dbname=drinks", $username, $password);
/*** echo a message saying we have connected ***/
echo 'Connected to database';
/*** The SQL SELECT statement ***/
$fname = $_GET['fname'];
//is _GET even working with PDO?
$STH = $dbh-> prepare( "SELECT * FROM persons WHERE fname LIKE '$fname'" );
/***as Joachim suggested, I had actually two different variables here, however, it
did not solve the issue **EDITED** from ($DBH to $dbh)****/
$STH -> execute();
$result = $STH -> fetch(0);
//$result should print out the first column correct? which is the person's ID.
}
catch(PDOException $e)
{
echo $e->getMessage();
}
?>
<head>
</head>
<body>
<p><?php print $result; ?></p>
//me trying to print out person's ID number here.
</body>
</html>
As previously mentioned, I'm not sure where my error is, I get fatal error:
Call to a member function prepare() on a non-object?
and If I try to not use that function, my page is simply blank and nothing prints out.
Basically I would just like to print out different bits of information from that row (that is from it's relevant link in index.php). I would like to know how to solve this using PDO.
Here is the previous question I asked, and it was solved but not with PDO.
Previous question
You could do something like this...
try {
$dbh = new PDO("mysql:host=$hostname;dbname=drinks", $username, $password);
$fname = $_GET['fname'];
$sth = $dbh->prepare("SELECT * FROM persons WHERE fname LIKE ?");
$sth->execute( array($fname) );
$result = $sth->fetch(PDO::FETCH_OBJ); // or try PDO::FETCH_ASSOC for an associative array
}
catch(PDOException $e)
{
die( $e->getMessage() );
}
In the HTML part you can do print_r($result) and you will see the exact structure of your results.
Comments: one of the best reasons to use PDO is the automatic escaping of the dynamic user inputs, like $fname here, so you should use it. Also, with $sth->fetch($param) the $param is not the column number but the type of the fetch method PDO will use (see PHP manual). Depending the method, you can get the PID of the result by $result->PID in case of PDO::FETCH_OBJ or by $result['PID'] when using PDO::FETCH_ASSOC. I hope this helps.

Update datebase by php

I want to update MySQL database by below code, but it doesn't work why?
<?php
mysql_connect("localhost","root","");
mysql_select_db("timer");
$update=$_COOKIE['name'];
mysql_query("UPDATE user SET password='2' WHERE username=$update");
?>
String values should be quoted: MySQL String Literals.
mysql_query("UPDATE user SET password='2' WHERE username='$update'");
$dsn = 'mysql:dbname=timer;host=127.0.0.1';
$user = 'root';
$password = '';
$update=$_COOKIE['name'];
$query = 'UPDATE user SET password='2' WHERE username=(?)';
try {
$dbh = new PDO($dsn, $user, $password);
$dbh->prepare( $query )->execute( array($query) );
} catch (PDOException $e) {
echo 'Connection failed ' . $e->getMessage();
}
Can't be sure if it works because I don't know what you have in your $_COOKIE array.
But the code should look like this.
For security, you can add some text transformation to the variable $update like addslashes or other ones.
Use prepared queries, the way you do it isn't a good practice. Check also if $update isn't empty, otherwise the query will not work, Finally, use mysqli_* functions instead of mysql_* :)
<?php
mysql_connect("localhost","root","");
mysql_select_db("timer");
$update=$_COOKIE['name'];
mysql_query("UPDATE ´user´ SET password='2' WHERE username='$update'");
?>
Try to mention the name of your table in this query like this: ´user´
Let me recommend you that if your query doesn't work then you could put the whole query in an echo (or print) to see whats wrong.
For example:
echo "mysql_query(\"UPDATE user SET password='2' WHERE username=$update\";

Trying to execute a SELECT statement in MYSQL but it is not working

I believe I have the syntax correct, at least according to my textbook. This is just a piece of the file as the other info is irrelevant to my problem. The table name is user, as well as the column name is user. I don't believe this to be the problem, as other sql statements work. Though it isn't the smartest thing to do I know :) Anyone see an error?
try {
$db=new PDO("mysql:host=$db_host;dbname=$db_name",
$db_user,$db_pass);
} catch (PDOException $e) {
exit("Error connecting to database: " . $e->getMessage());
}
$user=$_SESSION["user"];
$pickselect = "SELECT game1 FROM user WHERE user='$user' ";
$pickedyet = $db->prepare($pickselect);
$pickedyet->execute();
echo $pickselect;
if ($pickedyet == "0")
{
echo '<form method="post" action="makepicks.php">
<h2>Game 1</h2>......'
Since you're seemingly using prepared statements, I'd recommend using them to their fullest extent so that you can avoid traditional problems like SQL injection (this is when someone passes malicious SQL code to your application, it's partially avoided by cleansing user inputs and/or using bound prepared statements).
Beyond that, you've got to actually fetch the results of your query in order to display them (assuming that's your goal). PHP has very strong documentation with good examples. Here are some links: fetchAll; prepare; bindParam.
Here is an example:
try
{
$db = new PDO("mysql:host=$db_host;dbname=$db_name",
$db_user, $db_pass);
}
catch (PDOException $e)
{
exit('Error connecting to database: ' . $e->getMessage());
}
$user = $_SESSION['user'];
$pickedyet = $db->prepare('SELECT game1 FROM user WHERE user = :user');
/* Bind the parameter :user using bindParam - no need for quotes */
$pickedyet->bindParam(':user', $user);
$pickedyet->execute();
/* fetchAll used for example, you may want to just fetch one row (see fetch) */
$results = $pickedyet->fetchAll(PDO::FETCH_ASSOC);
/* Dump the $results variable, which should be a multi-dimensional array */
var_dump($results);
EDIT - I'm also assuming that there is a table called 'user' with a column called 'user' and another column called 'game1' (i.e. that your SQL statement is correct aside from the usage of bound parameters).
<?php
session_start();
$db_user = 'example';
$db_pass = 'xxxxx';
try
{
// nothing was wrong here - using braces is better since it remove any confusion as to what the variable name is
$db=new PDO( "mysql:host={$db_host}dbname={$db_name}", $db_user, $db_pass);
}
catch ( Exception $e ) // catch all exceptions here just in case
{
exit( "Error connecting to database: " . $e->getMessage() );
}
// this line is unecessary unless you're using it later.
//$user = $_SESSION["user"];
// no need for a new variable here, just send it directly to the prepare method
// $pickselect = '...';
// also, I changed it to a * to get the entire record.
$statement = $db->prepare( "SELECT * FROM user WHERE user=:user" );
// http://www.php.net/manual/en/pdostatement.bindvalue.php
$statement->bindValue( ':user', $_SESSION['user'], PDO::PARAM_STR );
$statement->execute();
// http://www.php.net/manual/en/pdostatement.fetch.php
// fetches an object representing the db row.
// PDO::FETCH_ASSOC is another possibility
$userRow = $statement->fetch( PDO::FETCH_OBJ );
var_dump( $userRow );
echo $userRow->game1;
Change this user=$user with this user='$user'. Please, note the single quotes.
Moreover, you are executing the query $pickedyet->execute(); but then you do echo $pickselect; which is nothing different from the string that contains the query.
Little hints:
You've to retrieve the result of the query execution.
You're using prepared statement which are very good but you're not really using they because you're not doing any binding.

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