mikosiko
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Everything posted by mikosiko
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ok... one more try... do this modifications in your code (commented with ADD THIS LINE ... 2 lines) <?php $list = "SELECT * FROM section_main"; $result = mysql_query($list) or die ("Query failed"); $numofrows = mysql_num_rows($result); echo "Select 1 Rows :" . $numofrows; // ADD THIS LINE echo "<table border='1' id='section_list'>"; echo "<tr><th>section_id</th><th>section_title</th></tr>"; for($j = 0; $j < $numofrows; $j++) { echo '<tr>'; $row = mysql_fetch_array($result); echo "<tr><td>". $row['section_id'] . "</td><td>". $row['section_title'] . "</td></tr>"; $answer = $row['section_title']; $query2 = "SELECT * FROM section_sub WHERE section_title = ".$answer." ORDER BY section_sub_title"; $result2 = mysql_query($query2) or die("Select Error :" . mysql_error()); $numofrows2 = mysql_num_rows($result2); echo "Select 2 Rows :" . $numofrows2; // ADD THIS LINE for($i = 0; $i<$numofrows2; $i++){ $row2 = mysql_fetch_array($result2); echo '<tr><td>'.$row2['section_sub_title'].'</td></tr>'; } } echo '</table>'; ?> and analyze what numbers those echoes gave to you... and after that look your real data for both tables and try to understand what is happening.
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no even close to the error that your code is giving to you... nor similar... but well... change this line... in your code $query2 = "SELECT * FROM section_sub WHERE section_title = ".$answer." ORDER BY section_sub_title"; to this $query2 = "SELECT * FROM section_sub WHERE section_title = '".$answer."' ORDER BY section_sub_title"; notice that I just added a single ' before and after the " surrounding $answer
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well.. look again... first your tables and then your select... and then the error " Select Error: Unknown column 's.section_id' in 'order clause' " do you see any section_id field in your section_sub table ?
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post the structure of your tables (both)... the error is telling you exactly what is happening.. just check your tables and look if the select is consistent with their definition.
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$result = mysql_query($query) or die("Select Error: " . mysql_error()); and check the error.
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for your future reference (if you don't already have it) http://dev.mysql.com/doc/refman/5.1/en/reserved-words.html
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same answer than in the other forum - In your $query definition... what is the value of $newdbname ? do you have that variable defined in your dbinfo.inc.php file? - Suppress the "@" before your @mysql_select_db if there are an error you are suppressing the display of it. - write the line mysql_query($query) at least in this way mysql_query($query) or die("Query Error: " . mysql_error()); that will show that you have an error in your UPDATE sentence $query="UPDATE * FROM $newdbname SET username ='$ud_username', password= PASSWORD('$ud_password') WHERE id='$ud_id'" is not a valid syntax for the UPDATE clause http://dev.mysql.com/doc/refman/5.0/en/update.html
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article_number is ambiguous... you have that column in both tables... you must specify which one you want to use in your group by
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you have an incorrect ")" after the first item in your VALUES ($item => title),... etc..etc
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fortnox is incorrect based on the format of the sentence. this sentence is incomplete $sql =("SELECT `customer_id`, `nic`, `full_name`, `name_with_initials`, `address`, `contact_number`, `gender` FROM `customer` WHERE `nic`='%s'", mysql_real_escape_string($_POST['nic']) ) ; should be write this way $sql = sprintf("SELECT `customer_id`, `nic`, `full_name`, `name_with_initials`, `address`, `contact_number`, `gender` FROM `customer` WHERE `nic`='%s'", mysql_real_escape_string($_POST['nic']) ); there '%s' is formating mysql_real_escape_string($_POST['nic']) as a string. http://www.php.net/manual/en/function.sprintf.php
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check the values for your system variables auto_increment_increment and auto_increment_offset maybe is something there http://dev.mysql.com/doc/refman/5.0/en/server-system-variables.html
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good... looks this lines: //Function to sanitize values received from the form. Prevents SQL injection function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } //Sanitize the POST values $login = clean($_POST['login']); $password = clean($_POST['password']); Do you see the reference to a "form" and then the usage of $_POST['login'] and $_POST['password'] ?? if you don't have the form (another php/html file) which is supposed to show you a screen to allow you enter your login and password and then call the script that you posted, then I'm afraid that you need to read/study a lot more before make it works. and forget about the "mysql.sql" document that you have.. the purpose of that was only create the table and import a row of data on it ... which according to your post you already did. just read/try some more examples and if you have doubts post again
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well.. try to isolate the problem first... test again that MYSQL is working try this replacing the "your_etc.etc" with the username and password for your database and see if that works. <?php $link = mysql_connect('localhost', 'your_username', 'your_username_password'); if (!$link) { die('Could not connect: ' . mysql_error()); } echo 'Connected successfully'; mysql_close($link); ?>
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Sorry if I'm insulting your intelligence here (no my intention).. but.. after read again your post here: I'm wondering if you really have MYSQL installed, up an running.. could you please clarify it you have it and how did you tested that it is in working conditions?
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at first glance (I didn't read all the code) this line: //Connect to mysql server $link = mysql_connect(host, user, pass); doesn't look good at all... that should be something like: //Connect to mysql server $link = mysql_connect($host, $user, $pass); and obviously you have to define the values for each variable in some place
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and ?... it is working or not?.... if not: - Check that your first line in the code that you posted is not a blank line (nothing before your first "<?php" - Check the same for the file that you are including (dbconnect.php). hope it helps
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Store form data into mysql database, then email same form data
mikosiko replied to lightningrod's topic in PHP Coding Help
a few basic questions.... which platform is your server?... *NIX (any unix flavor) or Windows? ... - if *NIX... did you test that sendmail is installed/configured and working? - if Windows... did you configure your php.ini with the proper SMTP server? -
what part is not working?.... works for me.... what datatype is your column 'date' is a DATE or DATETIME?
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this is the reason why it works and your previous intent don't.... read the allowed "types" below CAST in the function CONVERT http://dev.mysql.com/doc/refman/5.0/en/cast-functions.html#function_cast
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what Zanus said... but the main reason for your error is that you didn't define the alias (b) for your table products as you did for the table basket (a). +1 in the way that Zanus wrote the sentence
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Using PHP (or anything) to format SQL results on a webpage
mikosiko replied to kaisaj's topic in PHP Coding Help
could you post your table simple_search description?.... or tell us what datatype/length the column sdescription has... the example that you posted is exactly 254 characters long.... could be your sdescription field a VARCHAR(255) ? -
yes, you can ... the only thing that you need to do is add a field that represent the difference... easy and more simple.... but.. If you want to do in your way and pay the price is ok. you can solve your search creating a VIEW with the UNION of all the tables (horrible... but works)... just google for examples
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do you want to solve it just with a query?... I don't think it is possible. I can think in 2 approachs: - Use a combination of PHP and mysql PREPARED INSERT. or if you are more comfortable with MYSQL - a MySQL Stored procedure that include the PREPARED INSERT below is a few lines that could help you as an example (incomplete example of course) /* Prepare Insert Statement */ $stmt = $mysqli->prepare("INSERT INTO test1 (ip,posdate,obs) VALUES (?, ?, ?)") or die("Error " . $mysqli->error); $stmt->bind_param('sss', $val1, $val2, $val3) or die($mysqli->error); /* execute prepared statement */ $stmt->execute() or die($mysqli->error); printf("%d Row inserted.\n", $stmt->affected_rows); /* close statement and connection */ $stmt->close();
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Reconstruct ER diagram of MySQL Database from the sql code
mikosiko replied to binhngoc17's topic in MySQL Help
MySql Workwench should solve your issue. -
In case that you didn't notice... the name of the function that was provided to you is a link to the manual ... here is the whole link http://www.php.net/mysql_num_rows