MySQL and PHP gives me null var - php

I make a random quote app, in which by pressing a button, i can load one random phrase. I created for this a MySQL database, and two php code.
I upload my two code in a web hosting, and the app is running!
But sometime it gives me "null" instead of the phrase. I don't know why.
I'm not very good with this.
What is probably the problem?
This is my index.php
require_once 'db.php';
$query = "SELECT * FROM quotes ORDER BY rand() LIMIT 1";
$result = mysqli_query($con, $query);
while($row = mysqli_fetch_array($result)) {
print(json_encode($row['quote']));
}
And this is my db.php
// Create connection
$con=mysqli_connect("host","user","pass","a6361246_phrases");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

You can use the first three lines in your db.php file like this...
<?php
$host = 'localhost'; $db = 'database-name'; $user = 'database-user'; $pw = 'database-password';
$conn = new PDO('mysql:host='.$host.';dbname='.$db.';charset=utf8', $user, $pw);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
?>
Don't forget to change database-name, database-user & database-password to your specific credentials.
Then using PDO get your phrase like this...
<?php
require_once 'db.php';
try {
$sql = "SELECT * FROM Quotes ORDER BY rand() LIMIT 1";
$query = $conn->prepare($sql);
$query->execute();
$row = $query->fetch(PDO::FETCH_ASSOC);
} catch (PDOException $e) {
die("Could not get the data: " . $e->getMessage());
}
?>

Related

Updating data in DataBase with PHP

I have a simple script that should Update variable in a column where user login equals some login.
<?PHP
$login = $_POST['login'];
$column= $_POST['column'];
$number = $_POST['number'];
$link = mysqli_connect("localhost", "id3008526_root", "12345", "id3008526_test");
$ins = mysqli_query($link, "UPDATE test_table SET '$column' = '$number' WHERE log = '$login'");
if ($ins)
die ("TRUE");
else
die ("FALSE");
?>
but it doesn't work. It gives me - FALSE. One of my columns name is w1 and if I replace '$column' in the code with w1 it works fine. Any suggestions?
Simply remove quotes: '$column' = should be $column =
Your code is open for SQL Injection, use prepared statements.
change this "UPDATE test_table SET '$column' = '$number' WHERE log = '$login'"
to this "UPDATE test_table SET '".$column."' = ".$number." WHERE log = '".$login."'"
It's possible that your error is to do with the $column being set as a string with single quotation marks? Because it returns false, it suggests that you have a MySQL error of some sort.
To find out what the error message is, on your else block, rather than dying with a "FALSE" message, try use mysqli_error($link) - this should give you your error message
If removing the quotes surrounding the $column doesn't work, you could try the PDO method. Here's the snippet:
function insertUser($column, $number, $login) {
try
{
$connect = getConnection(); //db connection
$sql = "UPDATE test_table SET $column = '$number' WHERE log = '$login'";
$connect->exec($sql);
$connect = null;
} catch (Exception $ex) {
echo "EXCEPTION : Insert failed : " . $ex->getMessage();
}
}
function getConnection() {
$servername = "localhost";
$dbname = "my_db";
$username = "root";
$password = "12345";
try {
$connection = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
$connection->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (Exception $ex) {
echo "EXCEPTION: Connection failed : " . $ex->getMessage();
}
return $connection;
}
Regarding $number, I'm not sure the datatype for the $number whether quotes or not is needed so experiment with or without quotes to see which one works.
And the getConnection() function is in separate PHP file where it will be included in any PHP files that calls for database connection.

How do I reconnect my web pages on my website after updating to PHP 7 with a MySQL database 5.0.0?<?

I added the i updates to communicate with the database & now the page links don't work.
<?php
// Connect to database
$link=mysqli_connect('localhost', 'xxxxx', 'xxxxx');
mysqli_select_db($link, 'waddellc_PHRDB');
$sql = "SELECT * FROM quotes ORDER BY id";
$result = mysqli_query($link, $sql) or die(mysql_error());
$tenant_quotes = array();
$owner_quotes = array();
while($row = mysqli_fetch_array($result)) {
This should do the work, using PDO :
$servername = "localhost";
$username = "username";
$password = "password123";
$conn = null;
try {
$conn = new PDO("mysql:host=$servername;dbname=databaseName", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
echo "Connected successfully";
}
catch(PDOException $e)
{
echo "Connection failed: " . $e->getMessage();
}
if(!is_null($conn)){
$stmt = $conn->prepare("SELECT * FROM quotes ORDER BY id");
if ($stmt->execute()) {
while ($row = $stmt->fetch()) {
print_r($row);
}
}
}
I also think you need to update your database, it's quite old now.

List all the tables in your MySQL database using PHP

I have a hard time trying to list all the tables in my database.
I tried
<?php
//Configuration
$dbname = 'local';
$user = 'root';
$host = '127.0.0.1';
$pass = '';
$date = date('Y-m-d');
$export_type = 'mysql'; // option : mysql | psql
$file_name = $date.'-portal';
$file_path = $file_name;
// Create connection
$conn = mysqli_connect($host, $user, $pass);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$sql = "SHOW TABLES FROM $dbname";
$res = mysqli_query($conn, $sql);
if($res != false){
echo "Connected successfully";
$FILE = fopen("output.csv", "w");
$table = array();
while($row = mysql_fetch_array($res)){
$table[] = $row['0'];
}
foreach($tables as $table) {
$columns = array();
$res = mysqli_query($conn, "SHOW COLUMNS FROM $table");
while($row = mysql_fetch_array($res, MYSQL_NUM)) {
$columns[] = "$row[0]";
}
fwrite($FILE, implode(",", $columns)); fwrite("\n");
$resTable = mysqli_query($conn, "SELECT * FROM $table");
while($row = mysql_fetch_array($resTable, MYSQL_NUM)) {
fwrite($FILE, implode(",", $row)); fwrite("\n");
}
}
}else{
die(mysql_error());
}
?>
Result
if($res != false){
//.. everything in here never get executed
}
`$res` kept returning `false`.
What did I do wrong that could have lead to this ?
You can always execute the query:
Show tables;
After you selected database.
By the way you also can execute:
Show databases;
To list all of the databases your current user has permission to view.
Use db in your connection
mysqli_connect($host, $user, $pass, $dbname);
And use query like this
$sql = "SHOW TABLES";
You need to pass through your database in the connection script.
Like so:
$conn = mysqli_connect($host, $username, $pass, $dbname);
Then, when you want to pull rows from a table, you do it like this:
mysqli_query($conn, "SELECT rows FROM table");
One of the reasons this wasn't working for you was because you weren't passing through your database name through the connection. Also, rather than doing the above query, you selected a table from a database; rather than a row from a table.
Also, I noticed that you're using the mysql_* error output on the last line.
Here is a working version
Changes:
Added $dbname to mysqli_connect function
Added the backtick ` char between the table names, to avoid errors with reserved keyword from MySQL
Changed mysql_ functions to mysqli_
Close the file
Close the connection
Here is the code
NOTE: sorry I don't why, but when I pasted the code in the answer, all de code identation was messed up, even trying to indent it properly, I wasted like 10 minutes :(

IS NOT NULL giving null values

I have the following PHP code using PDO. I want rows with empty values to not appear in the results. How do I achieve this, and what am I doing wrong below?
<?php
require_once 'dbconfig.php';
try {
$conn = new PDO("mysql:host=$host;dbname=$dbname", $username, $password);
echo "Connected to $dbname at $host successfully.";
$sql = 'SELECT *
FROM as_questions
WHERE Answer IS NOT NULL';
$q = $conn->query($sql);
$q->setFetchMode(PDO::FETCH_ASSOC);
} catch (PDOException $pe) {
die("Could not connect to the database $dbname :" . $pe->getMessage());
}
As above suggested by two persons (Fred -ii and Alex Anderi) please change your query like below:-
$sql = "SELECT * FROM as_questions WHERE Answer !=''";

Query MySQL with PHP

I am trying to query a MySQL database with PHP and return the results as JSON. I'm new to PHP and web development so I'm not sure what I'm doing wrong. I've set up the database using MAMP. My parameters are being printed but I'm not getting the JSON. I've gotten this far with the help of a tutorial.
EDIT: I just went into phpMyAdmin to make sure it was working and when I click on Server:localhost:8889, a window pops up that says Error in processing request. Error code 404.
I'm thinking this is the problem, I'm just not sure why it isn't working. I may reinstall MAMP.
<?php
$user = 'root';
$password = 'root';
$db = 'TestDB';
$host = '127.0.0.1';
$port = '8889';
$first_name = filter_input(INPUT_GET, 'first_name');
$last_name = filter_input(INPUT_GET, 'last_name');
$membership_number = filter_input(INPUT_GET, 'membership_number');
echo $first_name;
echo $last_name;
echo $membership_number;
// Create connection
// $con = mysqli_connect("localhost", "root", "root", "TestDB");
// $con = mysqli_connect("localhost", "root", "root", "TestDB", "8889", $socket);
$link = mysqli_init();
$con = mysqli_real_connect($link, $host, $user, $password, $db, $port);
// Check connection
if(mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "SELECT * FROM NAME WHERE FIRST_NAME = \'$first_name\' and LAST_NAME = \'$last_name\' and MEMBERSHIP_NUMBER = \'$membership_number\'";
$result = mysqli_query($con, $sql);
if(!$result) {
die('Query failed: ' . mysqli_error());
}
// Check for results
// if ($result = mysqli_query($con, $sql)) {
if($result) {
// If there are results, create results array and a temporary one to hold the data
$resultArray = array();
$tempArray = array();
// Loop through each row in the result set
// while($row = $result->fetch_object()) {
while($row = mysqli_fetch_object($result)) {
// Add each row to the results array
$tempArray = $row;
array_push($resultArray, $tempArray);
}
echo $tempArray;
echo $resultArray;
echo $result;
echo json_encode($resultArray);
}
// Close connections
mysqli_close($con);
?>
You need to change you $sql variable to remove the escapes on the single quotes. They register as part of the string because you are using double-quotes to wrap it. Basically, you're telling the database to run the query "SELECT * FROM NAME WHERE FIRST_NAME = \'John\' and LAST_NAME = \'Smith\' and MEMBERSHIP_NUMBER = \'VRX78435\'". This will error if you run it directly because the escape characters are not escaping.
$sql = "SELECT * FROM NAME WHERE FIRST_NAME = '$first_name' and LAST_NAME = '$last_name' and MEMBERSHIP_NUMBER = '$membership_number'";
That should fix it for you.
There may also be an issue with your connection to the server. mysqli_query() uses the results of mysqli_connect() to run the query. mysqli_real_connect() only returns a boolean value, so it is invalid for this particular use (at least it failed to work on my server).
This would be a simple matter of replacing the $con and then you can drop the $link variable.
$con = mysqli_connect($host, $user, $password, $db, $port);
These changes, and assuming the $first_name, $last_name, and $membership_number are all valid, allowed your script to run for me, so I hope this helps.
Seems you are using procedural style coding
Instead of
while($row = $result->fetch_object()) {
You need mysqli_fetch_object in procedural style
while($row = mysqli_fetch_object($result)) {

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