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PHP - Phpmyadmin Won't Update
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 Similar Tutorialsi 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 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 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 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. 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. 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? 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? Using phpMyAdmin I loaded 6 test records with the id set to auto_increment and it loaded all the data correctly with id # 1-6. Then from somewhere it got the number 333353 and auto_increments it as the value for the id. So now I have id's 1-6 and 333353, 333354, ect. For every record I add it increments it. I deleted all but records 1-6 and tried again but it has the last value of 3333xx stored somewhere and increments it. Deleted them again, closed the program, came back and it still does it. Hi, I cannot find where to add foreign key constraints in phpmyadmin! Any help is appreciated! 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(); } ?> 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, Any other way to import the native xlsx excel file into PHPmyadmin. I know this function was supported in the older versions prior to phpMyAdmin 3.4.5. I don't want to convert the excel file to csv format. Thanks Hi I am creating an e commerce website and I have been using a php dummies book to help me as I am a beginner. I am creating an e commerce website and I have created a page which consists of radio buttons where a user will be able to choose what type of shoe they would like to view. Once they have chosen the radio button and then clicked onto the show me button the type of shoe shown should be the same as the type they have chosen but no shoes are showing on the webpage. I am unsure if i have not entered the format correct in the image section of the database. Here is the code which i have created so far: /* Select shoes the type specified*/ $query = "SELECT * FROM products WHERE prod_type=\"{$_POST['interest']}\""; $result = mysqli_query($cxn,$query) or die ("Couldn't execute query."); /* Display results in a table */ echo "<table cellspacing='10' border='1' cellpadding='87' width='100%'>"; echo "<tr><td colspan='5' style='text-align: right'> Click on any picture to see a larger version. <hr /></td></tr>\n"; while($row = mysqli_fetch_assoc($result)) { $f_price = number_format($row['product_price'],2); /* display row for each shoe */ echo "<tr>\n"; echo " <td>{$row['product_id']}</td>\n"; echo " <td style='font-weight: bold; font-size: 1.1em'>{$row['product_name']}</td>\n"; echo " <td>{$row['product_description']}</td>\n"; echo "<tr><td colspan='5'><hr /></td></tr>\n"; } echo "</table>\n"; echo "<div style='text-align: center'> <a href='shoes_list.php'> <h3>See more shoes</h3></a></div>"; ?> in my database the product_image column the field just contains the file name e.g: heels.jpg. please can anyone help me to get the images displayed.. thank you This topic has been moved to PHP Applications. http://www.phpfreaks.com/forums/index.php?topic=352409.0 I have pretty much the same problem.
I am just starting without any previous experience with xampp, php or MySQL, I have been coding in C for a few years and have used Java before.
Just wanted to play with a local website design using a Joomla template as my method for learning php etc.
I installed XAMPP 1.8.3 and then a Joomla template into htdocs.
Xampp runs fine and I have localhost and other pages without issue.
localhost/myphpadmin is blank.
I have not yet created a DB as I cannot access myphpadmin and have no code yet either.
The error reporting is set as you suggest but no errors are reported.
For error_reporting I used a - 1, phpinfo shows it as 22527 while display_error is On
The source for the phpmyadmin is blank but using FF dev tools the error console shows:
Error: The character encoding of the HTML document was not declared. The document will render with garbled text in some browser configurations if the document contains characters from outside the US-ASCII range. The character encoding of the page must be declared in the document or in the transfer protocol. Source File: http://localhost/phpmyadmin/ Also tried with Opera and IE I have installed Xampp on a USB drive labelled as V: as my SS C: drive has limited space. I installed the Java SDK as well (Version 8 update 20 build 1.8.0_20-b26 php is reported as 5.5.15 in xampp. This all happened straight from install, so no editing of files by me. All I have edited since is the error entries in php.ini I have 5 tables that I changed on my development server (laptop) that I'd like to upload to my test server. I can export just those tables from PhpMyAdmin but on import it drops all of the other tables first. Short of editing the SQL before I upload it (finding the drop and deleting it) is there an easy way to uncheck a box and change that behavior that I am just not seeing?
This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=355485.0 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 PHP Applications. http://www.phpfreaks.com/forums/index.php?topic=359282.0 |
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