I am a complete database newbie. So far, I know that I can connect to MySQL using PHP's mysql_connect() command, but apart from that, I really don't see how to take that data and put it onto a web page.
a) are there ways other than mysql_connect()
b) lets say I had a table of data in mysql and all I wanted was for that table (for example: list of names and telephone numbers) to now appear on my web page. I can't for the life of me find a tutorial for this.
<?
$database_name = "dbname";
$mysql_host = "localhost"; //almost always 'localhost'
$database_user = "dbuser";
$database_pwd = "dbpass";
$dbc = mysql_connect($mysql_host, $database_user, $database_pwd);
if(!$dbc)
{
die("We are currently experiencing very heavy traffic to our site, please be patient and try again shortly.");
}
$db = mysql_select_db($database_name);
if(!$db)
{
die("Failed to connect to database - check your database name.");
}
$sql = "SELECT * FROM `table` WHERE `field`= 'value'";
$res = mysql_query($sql);
while($row = mysql_fetch_assoc($res)) {
// code here
// access variables like the following:
echo $row['field'].'<br />';
echo $row['field2'];
}
?>
Check out mysql_fetch_assoc mysql_fetch_array and mysql_fetch_object
This is the very basics, you will want to search for tutorials. There are many about.
Related
I'm relatively new to any type of programming or coding. I'm not quite understanding why no matter what adjustments I make to my php file I can't seem to pull any data from a table.
Here is a link to the table: https://i.gyazo.com/4ad5e860895014c49dbe0539c38cdec2.png
Above is the test table I have been trying to use. From what I can understand I'm connecting to the database okay, but all of my problems come after the connection. Also, I'm using php 7.0 so a lot of the information I'm finding online has not been helpful.
If there is something glaringly wrong with my table or in my code, please let me know.
Here is my code:
'''
//Set Variables
$serverName = "localhost";
$userName = "root";
$password = "";
$databaseName = "test";
//Create Connection
$connection = mysqli_connect($serverName,$userName,$password,$databaseName);
//Check Connection
if(!$connection){
die("Connection failed: ".mysqli_connect_error());
}
echo "Connected successfully <br>";
//Fetch Data
$query = "SELECT * from table1";
$result = mysqli_query($connection, $query);
$row = mysqli_fetch_array($result, MYSQLI_NUM);
printf ($row[1], $row[2]);
mysqli_free_result($result);
mysqli_close($connection);
I figured out the issue a few hours ago. The code I had posted would have worked perfectly fine if I was connecting to the port for MySQL rather than MariaDB. Didn't realize that the port that MariaDB was connected to was the default.
MariaDB by default was port 3306, but MySQL was 3308. After specifying 'localhost:3308' I was able to start properly pulling rows from my tables.
I'm trying to figure out how can I put two differents MYSQL database connections name on the same page, one is for local PC and one is for hosting server.
Does is possible to have two different databases servers name on the same page so that way not to change connection database name in a script before upload.
I have code like this for local PC
<?php
$username = "your_name";
$password = "your_password";
$hostname = "localhost";
//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");
echo "Connected to MySQL<br>";
?>
Can can i add another connection name in the server in same page without changing the connection database!
alidad
Yes, a single page may query as many different servers as needed:
<?php
$server1 = new mysqli("server1.example.com", "user1", "password1", "database1");
$server2 = new mysqli("server2.example.com", "user2", "password2", "database2");
$result = $server1->query("SELECT 'Hello user' AS _message FROM DUAL");
$row = $result->fetch_assoc();
echo htmlentities($row['_message']);
$result = $server2->query("SELECT 'Hello user' AS _message FROM DUAL");
$row = $result->fetch_assoc();
echo htmlentities($row['_message']);
The way is to have a config.ini file in your localhost, and a different one in your production environment. This way you will parse this file and get your credentials.
A better way to do that is using this library: https://github.com/vlucas/phpdotenv
It works basically on the same way, but is easy to maintain.
Can anyone see what the problem with my code is / where im going wrong?
I know i have the correct host,database,user and password.
This is the code in the php file, it should get all the details available on the players from my sql database, however if i go on the page it just gives me a white page. Im using go daddy as a host and my database is also on there.
Any ideas? thanks
<?php
$host = "abc12345"; //Your database host server
$db = "abc12345"; //Your database name
$user = "abc12345"; //Your database user
$pass = "abc12345"; //Your password
$connection = mysql_connect($host, $user, $pass);
//Check to see if we can connect to the server
if (!$connection) {
die("Database server connection failed.");
} else {
//Attempt to select the database
$dbconnect = mysql_select_db($db, $connection);
//Check to see if we could select the database
if (!$dbconnect) {
die("Unable to connect to the specified database!");
} else {
$query = "SELECT * FROM Player";
$resultset = mysql_query($query);
$records = array();
//Loop through all our records and add them to our array
while ($r = mysql_fetch_assoc($resultset)) {
$records[] = $r;
}
//Output the data as JSON
echo json_encode($records);
}
}
?>
The script is all right, I checked it with a different query.
Assuming that the table Player is not empty, the error can be either with your raw sql query (as pointed by #Sharikov in comments), otherwise the error is with the way you have configured your godaddy server.
For debugging that, I would suggest inserting dummy print_r or echo statements before you execute the query, and going through your apache logs at /var/log/apache2/access.log.
Also make sure that you don't have any core php package missing on your server (like php5-mysql if you use mysql).
I am trying to make a blog site.For this purpose I need to use a specific data from a specific field from my database table.To do that I wrote these code.
<?php
$host = "localhost";
$user = "root";
$pass = "12345";
$db = "bnsb";
$conn = mysql_connect($host, $user, $pass) or die("Connection Failed!");
mysql_select_db($db, $conn) or die("Database couldn't select!");
$img = "select image from news where uid=1";
echo $img;
?>
My database connection is OK.It should print like this user_img1.jpg. But it prints the whole sql query like select image from news where uid=1. I run this code on phpmyadmin. It works! But it does not work in my php script.How can I do now?
You can not give the query as it is and expect result like in phpadmin.
For this first of all you have to connect to your DB like this
$con = mysqli_connect("localhost","my_user","my_password","my_db");
execute required query like this
$query22 = "select image from news where uid = 1";
$result22 = mysqli_query($con, $query22) or die (mysqli_error());
Get the result and display like this
while($rows = mysqli_fetch_array($result22, MYSQLI_BOTH))
{
echo "<br>Values in db: " . $rows['columnname'];
}
Also i advice you to take a look at these tutorials
http://codular.com/php-mysqli
http://www.dreamincode.net/forums/topic/54239-introduction-to-mysqli-and-prepared-statements/
Please read some PHP 101 kind of tutorials on how to use PHP.
To get data from DB (in almost any language)
You need to connect to a DB. The connection gets you some sort of resource
You formulate your query (which you seem to have done)
You execute the query against the DB that you connected to (step #1)
You get a result (set)
You iterate over the result set to get the individual result(s); in your case the result set would be just one result (or row).
The examples to do this in PHP are very basic; please do your own lookup on net. This one seems good enough to get you started - http://www.w3schools.com/php/php_mysql_intro.asp
Try this,
<?php
$host = "localhost";
$user = "root";
$pass = "12345";
$db = "bnsb";
$conn = mysql_connect($host, $user, $pass) or die("Connection Failed!");
mysql_select_db($db, $conn) or die("Database couldn't select!");
$img = "select image from news where uid=1";
$result=mysql_query($img);
while($row=mysql_fetch_array($result)){
echo '<img src="your_path_to_image/'.$row['image'].'" /> - '.$row['image'];
}
?>
I have done a fair bit of research into what i want to do, although i haven't found anything. I am not too sure if i am looking for the right thing :( I am also a little bit new to PHP and MySQL syntax, so please be kind.
I wish to perform the following in this order:
Connect to a database (DONE)
Query for a specific string (I think im done)
From here is gets a bit fuzzy :(
If a match is found for the variable, copy the whole row (I need other variables).
Assign the values from the SQL query to a PHP variables.
From there i will be right to carry on.
I have established the connection to the database with the following:
function connect() {
$dbname = 'database';
$dbuser = 'username';
$dbpass = 'password';
$dbhost = 'localhost';
mysql_connect($dbhost, $dbuser, $dbpass) or die("Unable to connect to database");
}
And then calling the function connect();
I then wish to query the database for a particular value, for the sake of this argument i will use a static value. This is what i have:
mysql_select_db(DATABASENAME) or die( "Unable to select database");
$query = "SELECT * FROM `TABLE` WHERE `COLUMN` LIKE 'VAULE'";
$result=mysql_query($query);
From here i am not too sure how to compare the query result to see if it is a match (something along the lines of mysql rows?).
If there is a match, then i would like to obtain the entire row, and assign each value to a php variable.
I am not asking for you to do it for me, simply i kick in the right direction should be fine!
Hope it explains it enough :)
Thanks for your kind guidance
Ok. You will want to keep the connection to the mysql database somewhere. A common use is $conn.
So you would have
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die("Unable to connect to database");
Then, either from the URL or Post, or just some variables you have sitting in your php file, you can query the database by putting the variables in the query itself. Also, here you can use $conn so that you have one place to connect to the database, in an include for example, and you won't have to make all of the connection string in each place you need to connect to the DB.
$query = "SELECT * FROM `TABLE` WHERE `COLUMN` LIKE '%" . $varToCompare . "%'";
$result=mysql_query($query,$conn);
Above you are using a like. You may want to just look at doing .. Where column=$var.
Then you can use php to spin through the results into an array (for queries where would get multiple rows).
Where the hell you learned how to use MySQL in PHP ? The mysql_* functions are more then 10 years old and not maintained anymore. Community has already begun to work on deprecating them.
You should be using PDO or MySQLi for that.
// connection to database
$db = new PDO('mysql:host=localhost;dbname=datadump_pwmgr;charset=UTF-8',
'datadump_pwmgr',
'kzddim05xrgl');
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
// setting up prepared statement for the query
$statement = $db->prepare('SELECT * FROM table WHERE column LIKE :value');
$statement->bindParam(':value', $some_variable, PDO::PARAM_STR, 127);
// executing query and fetching first result
if ( $statement->execute())
{
$data = $statement->fetch(PDO::FETCH_OBJ);
var_dump( $data );
}
This should give you something like what you needed. Though, I would recommend to try this tutorial. And learning more about prepared statements could be useful too.
Also , if you are working with objects, then it is possible to create a single DB connection object , and pass it to multiple other classes to use it:
$pdo = new PDO('sqlite::memory:');
$a = new Foo( $pdo );
$b = new Bar( $pdo, 'something');
This way you pass both objects the same database connection, and you do not need to reinitialize it.
I think you're looking for something like this:
$count = mysql_num_rows($result);
//if there is more then 1 record retrieved from the database
if($count > 0)
{
//Do what ever you want to do here, which I think you want to be
while ($row = mysql_fetch_assoc($result))
{
echo $row["Columnname1"];
echo $row["Columnname2"];
echo $row["Columnname3"];
}
}
else
{
echo "There are no matches for this specific value";
}
You can get the queried data by rows as an associated array using mysql_fetch_array():
$row = 0;
$data = mysql_query("SELECT name1,name2 FROM ....");
while(($result = mysql_fetch_array($data)) !== false)
{
echo "row = $row, name1 = " . $result["name1"] . ", name2 = " . $result["name2"];
$row ++;
}
... or as an objects using mysql_fetch_object():
$row = 0;
$data = mysql_query("SELECT name1,name2 FROM ....");
while(($result = mysql_fetch_object($data)) !== false)
{
echo "row = $row, name1 = $result->name1, name2 = $result->name2";
$row ++;
}
I'm not too sure of what you want, but I can see one probable bug here: you're using LIKE in a way which means =: in order to have LIKE to behave like a like, you need some joker chars :
"SELECT * FROM `TABLE` WHERE `COLUMN` LIKE 'VAULE'" // This will return all rows where column='VAUL'
"SELECT * FROM `TABLE` WHERE `COLUMN` LIKE '%VAULE%'" // This will return all rows where column='%VAUL%' // This will return any row containing 'VAUL' in column
"SELECT * FROM `TABLE` WHERE `COLUMN` LIKE '%VAULE'" // This will return all rows where column='%VAUL' // this will return all rows ending by VAUL. I guess you get it now :)
An to retrieve the actual results:
$query = "SELECT * FROM `TABLE` WHERE `COLUMN` LIKE '%VAULE%'";
$result=mysql_query($query);
while (false !== ($row = mysql_fetch_assoc($result))) {
//here $row is an array containing all the data from your mysql row
}
Try to write the database connection in another page no need to use function and include that page in where ever you need.
ex: require_once 'dbConnect.php';
dbConnect.php consists:
<?php
$dbname = 'datadump_pwmgr';
$dbuser = 'datadump_pwmgr';
$dbpass = 'kzddim05xrgl';
$dbhost = 'localhost';
mysql_connect($dbhost, $dbuser, $dbpass) or die("Unable to connect to database");
?>