PHP - Phpmyadmin Picking Up Strange Value From Somewhere

i have been working with a website but only locally. i am now trying to put it online which is ok. when i was working with it locally i have been able to create multiple users which have different privileges to each page. for example i said one of my users can only read and the other can read write.
the problem is when i have went to create these users online i can't seem to find the create user option in phpmyadmin.

i am missing the obvious or will i have to code them in and if so can you direct me to a tutorial which shows me how.

thanks in advance     

I have a form where it ask the user to select the student ID, course ID, enter the grades and comments.
The Student ID and CourseID is selected from a drop down menu, but when the data is sent to PHPMYADMIN it enters a 0 into the SID and CID

How to I get it save the numbers which has been selected

Hi guys

I have two questions:

Question 1: phpmyadmin
If my host does not provide phpmyadmin and I want to install it myself, can I just download phpmyadmin and copy it to my public_html folder or is there another way to install it on my site??

Question 2: Email Script
I have a script that sends an email to a system that sends out multiple sms's. The system works like this, I send an email to an address with an attachment called cellnumbers.txt with a list of all the cellnumbers. I must add a specific subject. When I send the email to my own email address, it comes correctly through. But when it is sent to the address of the server, no sms are received. I attached my script. Please take a look. Your help will be greatly appretiated.

Thank you

Each time i try to open WAMP's phpmyadmin so i can create a database it has this ERROR


#1045 - Access denied for user 'root'@'localhost' (using password: NO)

how do i fix it

I have a db.php and inside it I filled out all the info needed:

<?php
define('DB_HOST', 'example.com');
define('DB_NAME', 'database_name');
define('DB_USERNAME', 'user_name');
define('DB_PASSWORD', '*******');

$odb = new PDO('mysql:host=' . DB_HOST . ';dbname=' . DB_NAME, DB_USERNAME, DB_PASSWORD);
?>

My problem is with the DB_HOST, is that my direct URL to the site? for example, google.com or is it like an IP?  

Hi,

I am trying to create a drop down list in php and I want the data to come from a table that I have created in phpmyadmin. The code that I have created allows me to select values from the drop down list and insert the rest of the data. However when I check the the table the SID and Cid are set to 0 and the grade field is empty and the comments field contains the grade. The SID and Cid are both composite keys.


<?php
        $sql = "SELECT Cid FROM course";
        $db1 = new DBStudent_Course();
        $db1->openDB();
        $result = $db1->getResult($sql);

        echo"<select name = Cid>";
        while ($row = mysql_fetch_object($result)) {
            echo "<option value = '" . $row->Cid . "'>$row->Cid</option>";
        }

        echo"</select>";
        echo "</p>";
        ?>

        <?php
        $sql = "SELECT SID FROM student";
        $db1 = new DBStudent_Course();
        $db1->openDB();
        $result = $db1->getResult($sql);

        echo"<select name = SID>";
        while ($row = mysql_fetch_object($result)) {
            echo "<option value = '" . $row->SID . "'>$row->SID</option>";
        }

        echo"</select>";
        echo "</p>";
        ?>
        <?php
        if (!$_POST) { //page loads for the first time
        ?>
            <form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post">
                grade:<input type="text" name="grade"/><br/>
                comments:<input type="text" name="comments" /><br />
                <input type="submit" value="Save" />
            </form>
<?php
        } else {
            $Cid = $_POST["Cid"];
            $SID = $_POST["SID"];
            $grade = $_POST["grade"];
            $comments = $_POST["comments"];
            $db1 = new DBStudent_Course();
            $db1->openDB();
            $numofrows = $db1->insert_student_course("", $SID, $Cid, $grade, $comments);
            echo "Success. Number of rows affected:
<strong>{$numofrows}<strong>";
            $db1->closeDB();
        }
?>


 

Hi

How can I make the data in phpmyadmin accept Arabic Language. Now Arabic language shows as question markss ?????????????.

Is there anyway to do this via php?

Hi. I having trouble with counting rows in phpmyadmin.

It works fine this way:
Code: [Select]
$result_rows = mysql_query("SELECT * FROM events");

But when i modify the code to this it doesnt work at all.
Code: [Select]
$result_rows = mysql_query("SELECT * FROM events WHERE category = 'adults' ORDER BY 'date'");

Any idea what is wrong?

I have used the XAMPP installer to install php and MySQL locall on my computer. I also succeeded in setting the security for XAMPP pages, the MySQL admin user root and phpMyAdmin login. When I enter phpMyAdmin via the link in the XAMPP initial page I do however receive a red notification:

phpMyAdmin configuration storage is not fully configured; some extensions are not activated. To find out click here.   I have attached a screenshot showing three items which are not OK, shown in red. I looked up in the documentation, but could not find out. I hope someone can help. I don't even know if it is important to fix this problem.   Regards,   Erik           Attached Files  XAMPP2.jpg   41.08KB   0 downloads

Hi guys, I would like to seek help on inserting data whenever the switch is on or off to my sensor mySQL database in phpMyAdmin from my control.php. I'm using Raspberry PI as my hardware and follow a few tutorials to create my own Web Control Interface, it works perfectly without insert method. After I implemented insert method to my control.php and execute it, it cannot works and cannot store.

Hi, I cannot find where to add foreign key constraints in phpmyadmin!

Any help is appreciated!

Sorry for the caps, but this is relatively time sensitive.  I am trying to make a register form, but when I click the submit button, nothing happens.  It doesn't add to the table, it doesn't bring me home, doesn't even display the errors if the PWD's don't match or the fields are blank.  Here's my code, thanks guys !  PS:  The DB name is phptest, and the table is called users. 
Code: [Select]
<?php
error_reporting(0);

require_once('connector.php');

$errors = array();

if ($_POST["submit"]) {

if (empty($_POST['username'])) { array_push($errors, 'You did not submit a username.');}
if (empty($_POST['email'])) { array_push($errors, 'You did not submit a email.');}
if (empty($_POST['password1'])) { array_push($errors, 'You did not submit a password.');}

$old_usn = mysql_query("SELECT id FROM users WHERE name = '".$_POST['username']."' LIMIT 1") or die (mysql_error());
if (mysql_num_rows($old_usn) > 0) { array_push($errors, 'This username is already registered.');}

$old_email = mysql_query("SELECT id FROM users WHERE email = '".$_POST['email']."' LIMIT 1") or die (mysql_error());
if (mysql_num_rows($old_email) > 0) { array_push($errors, 'This email is already registered.');}

if ($_POST['password1'] != $_POST['password2']) { array_push($errors,'You entered two different passwords');}

if(sizeof($errors) == 0) { 


$username = $_POST['username'];
$email = $_POST['email'];
$password = sha1 ($_POST['password1']);


mysql_query("INSERT INTO users (name, hashed_psw, email, joined)
VALUES ('{$username}', '{$password1}', '{$email}', NOW());") or die (mysql_error());

 

header ('Location: index.php?msg=1');

}
}
?>
<html>
<head>
<title>register</title>
</head>

<body>
<?php
foreach($errors as $e) {
echo $e;
echo "<br/>\n";
}


?>

<form action="register.php" method="post">
<h4>
Username:
<br />
<input name="username" type="text" value="" size="10" maxlength="16" />
<br />
<br />
Email:
<br />
<input name="email" type="text" value="" size="10" maxlength="100" />
<br />
<br />
Password:
<br />
<input name="password1" type="password" value="" size="10" maxlength="16" />
<br />
<br />
Confirm Password:
<br />
<input name="password2" type="password" value="" size="10" maxlength="16" />
<br />
<br />
<input name="submit" type="button" value="Register" />
</h4>
</form>

</body>
</html>

And heres the connector.php script:

Code: [Select]
<?php

mysql_connect("localhost", "***", "***") or die (mysql_error());
mysql_select_db("phptest") or die (mysql_error());

?>(yes, the asterisks have the name and pw, just put them just in caseys!

Hi all!

I created a simple form using html. I created a database, a table and some records in the table with the code in localhost/phpmyadmin.

I wanted to know how you I can link the form to my database. Do I need a processing php code to link to the database?

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=355485.0

This topic has been moved to PHP Applications.

http://www.phpfreaks.com/forums/index.php?topic=352409.0