Echo a mysql column by php as an array [duplicate] - php

This question already has answers here:
Can I mix MySQL APIs in PHP?
(4 answers)
Closed 6 years ago.
i want to fetch a mysql table named "my_table" column named "Email" contents as an array by php , so this is my code :
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "my_table";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$result = mysql_query("SELECT * FROM my_table");
if ($result === FALSE) {
die(mysql_error()); // TODO: better error handling
}
$data = array();
while ($row = mysql_fetch_array($result)) {
$data[] = $row['Email'];
}
echo join($data, ',');
?>
but this code returns me this error :
No database selected
but i've selected my table and database ...
and i know this code have some problems as mixing mysql and mysqli content but i dont know how to fix it i just want that array echo , if this code need to be fixed just guid me ,
how to solve this problem ? thanks in advance
Thanks to #Martin
my problem has solved i just changed the code by this way :
<?php
$servername = "localhost"; $username = "root"; $password = ""; $dbname = "my_db";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname); // Check connection
if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error);}
$result = mysqli_query($conn, "SELECT * FROM my_table");
if($result === FALSE) {
die(mysql_error()); // TODO: better error handling
}
$data = array();
while ($row = mysqli_fetch_array($result))
{
$data[] = $row['Email'];
}
echo join($data, ',')
?>

You are conencting to a database here called "my_table":
$dbname = "my_table";
And then, in your SQL statement, you try connecting to a table called the same:
$result = mysql_query("SELECT * FROM my_table");
Are you sure this is the correct name for your database?
On PHPMyAdmin you can click "Databases" to view the Database names and then, when clicking on the db, it will give you a list of tables:
Image file of getting database views from tables in PHPMyAdmin

Related

How to use PHP SQL with where Clause? [duplicate]

This question already has answers here:
How to include a PHP variable inside a MySQL statement
(5 answers)
Closed 2 years ago.
I want to select an email from the DB which needs to return the related Account_ID
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "connected accounts details";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT acc_id FROM users
WHERE
email = $user_email";
$result = $conn->query($sql);
echo "$result";
$conn->close();
This code returns nothing, just blank.
While using the same connection/db data is being inserted in db successfully.
Try this code
// Php Sql Select From DB....
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "connected accounts details";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT `acc_id` FROM `users` WHERE `email` = ?";
$result->$conn->prepare($sql);
$result->bind_param('s',$user_email);
$result->execute();
$data = $result->get_result();
// Fetch all
$data->fetch_all(MYSQLI_ASSOC);
print_r($data);
$mysqli -> close();
// #link http://php.net/manual/en/mysqli-result.fetch-all.php
$sql = "SELECT acc_id FROM users
WHERE
email LIKE '%$user_email%' ";
$result = mysqli_query($conn, $sql);
$row = $result->fetch_assoc();
$id = $row['acc_id'];
echo "$id";
$conn->close();
Got the wanted Record, trying this way.
Thanks everyone answering on this post.

MySQL table value be equal with php var (then display it)

I have the following MySQL table and I would like to do that RVR11 (500) be equal with a php var $RVR11
I have this code but it doesnt work.
$servername = "localhost";
$username = "root";
$password = "";
$conn = new mysqli($servername, $username, $password);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
$price = mysql_query("SELECT RVR11 FROM tablename");
$result = mysql_fetch_array($RVR11);
echo $result['RVR11'];
Sorry about this beginner question.
You should choose a database before any queries, you can pass the fourth arguments to new mysqli constructor.

When I add a column, I can't retrieve it from mysql

After I add a column to my database, I want to retrieve it but not expected.
In PHP, I try reopening apache and mysql still not work.
Does anyone know how to resolve it? Thanks!
your question is not fully explanatory but with what I could try to understand you want to retrieve data or records from your database
you could try the code below and tweak it to work your way
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM databaseName";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data
while($row = $result->fetch_assoc()) {
print $row"<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>

How To Get a Single Number From a Database in PHP [duplicate]

This question already has answers here:
Can I mix MySQL APIs in PHP?
(4 answers)
Closed 6 years ago.
I need to echo the number 1 or 0 whatever the database has in that cell. I cant seem to get it to work. What am I doing wrong people of the internet?
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "testfp";
$conn = new mysqli($servername, $username, $password, $dbname);
$sql = "SELECT hookup FROM testfp WHERE name = 100100" ;
$bob = mysql_query($sql);
echo $bob;
?>
You can not use mysql and mysqli together. So you can try the following code (using mysqli only, since mysql is deprecated) :
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "testfp";
$conn = new mysqli($servername, $username, $password, $dbname);
//Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT hookup FROM testfp WHERE name = 100100" ;
$result = $conn->query($sql);
if($result->num_rows>0)
{
while($row = $result->fetch_assoc()) {
echo $row["hookup"];
}
}
else
{
echo "No results found!";
}
?>

phpMyAdmin: one table in database work, another doesn't

I've created an app were you can register as a user. You can sign up and then you're in the database "myAppDataBase" in "firsttable". A second table contains a list of lets say other important users that I manually created in the PHPmyAdmin-Website/"App". This table is called "secondtable".
My code to get the data is as follows:
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "mydatabas";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}else{
//Print ("successfully connected");
}
$query = "SELECT * FROM firsttable";
$result = mysqli_query($conn, $query) or die("Error: " . mysqli_error($query));
$num = mysqli_num_rows($result);
$rows = array();
while ($r = mysqli_fetch_assoc($result))
{
$rows[] = $r;
Print ("sf");
}
Print json_encode($rows);
mysqli_close($conn);
?>
The only thing i changed was this line: THIS WORKS
$query = "SELECT * FROM firsttable";
But when I change it to this it won't work anymore.
$query = "SELECT * FROM secondtable";
Any help?
Change this:
mysqli_error($query)
With this:
mysqli_error($conn) // with your connection
Explanation:
mysqli_error() function needs connection link identifier not your query as param.
Mysqli_error PHP Manual
I SOLVED IT! Somehow, my second wasn't encoded the right way. I simply added this coder and it worked:
mysqli_set_charset($conn, 'utf8mb4');
Thanks for all you help though. ;)

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