Add image column into an existing mysql table? - mysql

I want to add a new column called image in an existing mysql table.
The type is BLOB, the attributes are BINARY and NO NULL
What code should I use in PhpMyAdmin?

In PhpMyAdmin you can use the table editor to add a new column from clicking on the structure of a table.
The mysql command instead, would be:
ALTER TABLE table_name ADD image MEDIUMBLOB NOT NULL
MEDIUMBLOB has a maximum size of 16MB, use LONGBLOB (up to 4GB) for anything bigger
If you have problems uploading BLOBs, check the value of max_allowed_packet http://dev.mysql.com/doc/refman/5.5/en/packet-too-large.html

It is not recommended to use BLOBs for saving images.
Save the files somewhere on your webserver for example .../htdocs/images/picture.jpg (assuming you use xampp).
then create a simple Column in you Database for Strings of the image names. To show you why i will use PHP.
in your .../htdocs/index.php
you can then do something like the following:
<?php
//enter you PhpMyAdmin details here.
$conn = mysql_connect("localhost", "mysql_user", "mysql_password");
if (!$conn) {
echo "Unable to connect to DB: " . mysql_error();
exit;
}
if (!mysql_select_db("mydbname")) {
echo "Unable to select mydbname: " . mysql_error();
exit;
}
$sql = "SELECT imageName
FROM MyTable
;
$result = mysql_query($sql);
if (!$result) {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
// While a row of data exists, put that row in $row as an associative array
// Note: If you're expecting just one row, no need to use a loop
// Note: If you put extract($row); inside the following loop, you'll
// then create $userid, $fullname, and $userstatus
while ($row = mysql_fetch_assoc($result)) {
?>
<img src="images/<?=$row["userid"]?> <!-- this will get repeted for all the rows in you database -->
<?php
}
mysql_free_result($result);
?>
this is where i got the sample code from:
PHP Doc
i just modified it a bit .. but the php documentation is great btw.

Related

Create variable from MySQL datatable that can be used in HTML code

So I have a MySQL database and would like to create variables from that database that I can use in HTML code.
For example if I use <video blablabla some stuff src="phpVariable" /video>
I don't want to have a static URL in the HTML code. It should be assigned to a variable based on an SQL query. What is the best way to do this?
First, you need to retrieve the data from the database that you wish to use, using a database query. From there, you can loop through all the results and output them.
Replace 'tablename' with the name of the database table and 'columnName' with the column name.
$con should be used to connect to your database. See https://www.php.net/manual/en/pdo.connections.php
$stmt = $con->prepare("SELECT * FROM tablename");
$stmt->execute();
while($row = $stmt->fetch()) {
echo $row['columnName'];
}
If you only needed to retrieve one record, change the code to:
$stmt = $con->prepare("SELECT * FROM tablename");
$stmt->execute();
$row = $stmt->fetch();
echo $row['columnName'];
You can also store the retrieved data as a variable, like this:
$variableName = $row['columnName'];
// Then to use the variable in your code:
echo'<video src='.$variableName.'</video>';
This is very basic PHP though. Have you looked at some online guides?

How to save output table within the DB

I run a computation query and displayed an output table, instead of exporting it as csv file, i want is to save it or make it another table within my DB, is it possible?
Yes you can..
Create a table first to your database.
Then write a query to insert data to your created table.
$getTheDataFromYourPrevQuery = //set your data here
$sql = "INSERT INTO yourTable (name)"." VALUES('$getTheDataFromYourPrevQuery')";
try {
$db = $this->db;
$stmt = $db->prepare($sql);
$stmt->execute();
//$user->id = $db->lastInsertId();
$db = null;
echo json_encode($user);
} catch(PDOException $e) {
//error_log($e->getMessage(), 3, '/var/tmp/php.log');
echo '{"error":{"text":'. $e->getMessage() .'}}';
}

Admin panel - Creating an edit users button

I've created an admin panel on my website so when the admin logs in he can edit users. I'm trying to get it to create a table that displays a list of all the users on the database, however, when I run it I get the error:
No database selected
Here is the code in my editusers.php:
<?php
include 'adminpage.php';
include 'connection.php';
$sql = "SELECT * FROM Users";
$result = mysql_query($sql)or die(mysql_error());
echo "<table>";
echo "<tr><th>UserID</th><th>First Name</th><th>Last Name</th><th>Email</th><th>D-O-B</th></tr>Username</th><th>Password</th><th>";
while($row = mysql_fetch_array($result)){
$userid = $row['UserID'];
$firstname = $row['FirstName'];
$lastname = $row['LastName'];
$email = $row['Email'];
$dob = $row['DateofBirth'];
$username = $row['Username'];
$password = $row['Password'];
// Now for each looped row
echo "<tr><td style='width: 200px;'>".$userid."</td><td style='width: 200px;'>".$firstname."</td><td>".$scale."</td><td>".$lastname."</td><td>".$email."</td></tr>".$dob."</td></tr>".$username."</td></tr>".$password."</td></tr>";
} // End our while loop
echo "</table>"
?>
First of all it looks like you are using mysql which isn't a wise move. This is because Mysql is actually deprecated and was improved to mysqli. Your problem may be to do with your database connection. You also haven't set a database. Like I said you can set an active database in your connection script. It should or could look something like this.
<?php
$conn = mysqli_connect("localhost", "root", "password", "database");
// Evaluate the connection
if (mysqli_connect_errno()) {
echo mysqli_connect_error();
exit();
}
?>
After that, your sql query is correct by selecting all from you table 'users' but in order to proceed I recommend creating a query where you use mysqli_query an select the $sql and $conn as parameters. In all honesty it is much advised to stop and continue once you have adapted to mysqli. Alternatively you can use PDO which in some cases can be seen as better to use rather than mysqli but the choice is yours. I personally would get to grips with mysqli and then look at some answers on Stack Overflow to decide whether you should use PDO or not. Visit the PHP manual here. Enter all the mysql functions you know and it will show you how to use the new mysqli version of the functions. Don't think that it is as simple as just adding and 'i' to the end of a mysql function. That's what I initially thought but there is alot to do with extra parameters etc. Hope this helps :-)

Parse json object from table via php connected to phpmyadmin

I'm using a json-parser in Xcode to fetch a table from phpmyadmin. The parser gets (or should get) the json-formated document via a php-file uploaded on my ftp-server. The file is successfully parsed but it doesn't recognize any objects. I think this is because there are multiple arrays in the json-document.
When there's only one entry the document looks like this:
[{"id":"1","Name":"Eric","Message":"first from web"}]
but when i add an entry it looks like this:
[{"id":"1","Name":"Eric","Message":"first from web"}]
[{"id":"1","Name":"Eric","Message":"first from web"},{"id":"6","Name":"Claes","Message":"Hurrburr"}]
As you can see the old array (containing only the single entry) is still there in the second array with both entries.
I suspect the problem is that the old arrays are still there when i update the database because when i tried parsing the json document with only one entry (only one array) it worked.
So my question is first if thereĀ“s something i missed in my code, or if you know why the old arrays are still there when I update the database or how to remove all previous arrays when the document is updating.
Here is my .php-file:
<?php
$username = "perhaps not sharing this information";
$password = "or this";
$database = "nah";
mysql_connect("the server url",$username, $password);
#mysql_select_db($database) or die("Error here");
$query = "SELECT * FROM debug_db";
$result = mysql_query($query) or die(mysql_error());
$num = mysql_numrows($result);
mysql_close();
$rows = array();
while($r = mysql_fetch_assoc($result))
{
$rows[] = $r;
echo json_encode($rows);
}
?>
And check out the json-document at: http://app.levinnovation.se/getjson.php
Thank you!

Need help connecting to my Mysql database on local server

I'm quite new to databases and have no idea where I have gone wrong. Please help me find out why I can't connect to my database.
I'm getting an error with Dreamweaver - Dynamicaly-related files cannot be discovered because there is no site definition for this document
My Site root is located in htdocs.
the main file I'm trying to run is "dataquery.php" - htdocs/LetsPlays/dataquery.php
<?php
include 'includes/databaseform.php';
$query = "SELECT * FROM userchanel";
$result = mysql_query($query);
while($person = mysql_fetch_array($result));
{
echo $person['chanelurl'];
}
?>
Dataquery is connected to "databaseform.php" - htdocs/LetsPLays/includes/databaseform.php
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass ='';
$db = 'mysql_tut';
$conn = mysql_connect($dbhost,$dbuser,$dbpas);
mysql_select_db($userchanel);
?>
So I'm trying to connect to userchanel table through user tbl
screenshot: http://imageshack.us/f/23/usertbl.png/
Files run through wordpress are set up to run on 127.0.0.1
Please help me.
Sorry for the noobishness! I have no idea what I'm missing!
When I run the html file all I get is a blank page!
Replace the line
$conn = mysql_connect($dbhost,$dbuser,$dbpas);
with
$conn = mysql_connect($dbhost,$dbuser,$dbpass);
notice the double 's' on $dbpass, also if I may you should not rely on deprecated features when writing new code, use PDO instead, also don't login to db as root, create a user, don't be lazy, with that your database connection code should like:
<?php
$db = new PDO('mysql:dbname=databasename', 'username', 'password',
array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8"));
the init command is not actually necessary I just included it cause I always use it, while the querying could be done as
<?php
include 'includes/databaseform.php';
$query = $db->prepare("SELECT * FROM userchanel");
$query->execute();
while(($person = $query->fetch(PDO::FETCH_ASSOC)) !== false);
{
echo $person['chanelurl'];
}
If you're only using one column you should fetch only that as below:
<?php
include 'includes/databaseform.php';
$query = $db->prepare("SELECT chanelurl FROM userchanel");
$query->execute();
while(($channelurl = $query->fetch(PDO::COLUMN)) !== false);
{
echo $channelurl;
}
I didn't include closing braces for php code as they are not necessary also do some error checking var_dump($db->errorInfo()); and var_dump($query->errorInfo()); - didn't include this in the code as I only use them in checking any issues with my code, good luck!
instead of htdocs/LetsPlays/dataquery.php
try localhost/dataquery.php
I recommend you to make and additional file for database connection and include it anywhere you need. Because when you need to push project live you need to change in every file incase file is includeed you need to change just at one place and it effects every where connection for localhost is
<?php
// Replace the variable values below
// with your specific database information.
$host = "localhost";
$user = "root";
$pass = "";
$db = "yourdatabase";
// This part sets up the connection to the
// database (so you don't need to reopen the connection
// again on the same page).
$ms = mysql_pconnect($host, $user, $pass);
if ( !$ms )
{
echo "Error connecting to database.\n";
}
// Then you need to make sure the database you want
// is selected.
mysql_select_db($db);
?>
Save all the above code in one file save it as dbConfig.php and include it any where
like
include ("dbConfig.php");
Now in that file you are connected to db you can interact with database.