|
PHP - Newb: Simple Mysql Query
The following query or while loop is only increasing the ArticleID variable every 3rd time the script is run, I've narrowed it down to the following code snippet. Can you spot a problem with this, I'm in my first week of PHP and MySQL and I can't see any problem with it.
Any help would be mighty appreciated by this idiot Code snippet: --- $result = mysql_query("SELECT ArticleID FROM test_top ORDER BY ArticleID ASC LIMIT 1") or die(mysql_error()); while($row = mysql_fetch_array($result)) { $ArticleID=$row['ArticleID']; } $ArticleID=intval($ArticleID); $ArticleID++; --- Similar TutorialsI have a database called "postvoting", It's basically to store when somebody votes on a particular posts. I store the post_id that the user votes on, the users id that voted on it, and the date. What I want to do is find the most popular posts in a given time. So if 4 people voted on the post with the id of 1, and 2 people voted on the post with the id of 2, I want to count the number of rows with post_id='1' A non working example what I want would look something like: $count_votes = mysql_query("SELECT * FROM postvoting WHILE post_id=post_id"); print mysql_num_rows($count_votes); result: 4 or $count_votes = mysql_query("SELECT * FROM postvoting GROUP BY post_id"); print mysql_num_rows($count_votes); result: 4 (counting the number of results in a group) Hope this isn't too confusing. (I've confused myself with this). Hello, I am trying to pick up php again and just exercising my skills. So I have it so that it fills my form with the values of what I want to edit, and when I click the edit button, it doesn't edit any of the information. When I echo out $result, I get a MYSQL query string that has the same values as the table, so its not getting the new values that are edited. <?php @mysql_connect('localhost', 'root', '') or die("Could not connect to Mysql Server. " . mysql_error()); @mysql_select_db('tutorials') or die("Could not connect to Database. " . mysql_error()); if(isset($_GET['edit'])) { $id = $_GET['edit']; $query = "SELECT `username`, `password` FROM `users` WHERE `id` = '$id'"; $result = mysql_query($query); $row = mysql_fetch_array($result); $name = $row['username']; $password = $row['password']; } if(isset($_POST['edit'])) { $id = $_GET['edit']; $query = "UPDATE `users` SET `username` = '$name', `password` = '$password' WHERE `id` = '$id'"; $result = mysql_query($query); echo $query; if(!$result) { echo mysql_error(); }else{ echo 'updated post'; } } ?> <form method="POST" action="" > <input type="text" name="name" value="<?php echo $name; ?>" /> First name <br /> <input type="text" name="password" value="<?php echo $password; ?>" /> Last name <br /> <input type="submit" name="edit" value="edit" /> </form> I believe it has something to do with the values of $name and $password in the form conflicting with the first if isset and the second if isset. Thanks for any help possible I'm attempting to implement a simple social networking system but at the moment am confused about how to create a multiple query which will display a certain user's friends list. The database contains four tables, the two tables that I'm using at the moment at 'usersTable' and 'friendshipsTable' are detailed below. usersTable | Table that stores all the user data UserID | Default primary key Forename | Surname | Username | Password | Email Address | friendshipTable | Table that stores information about friendships between users FriendshipID | Default primary key userID_1 | UserID userID_2 | UserID Status | Either Pending or Confirmed. The user's id is parsed into the url, and then saved into a variable. blah.com/userprofile.php?id=6 $id = $_GET['id']; I am familiar with creating simple queries, but can't quite work out how to set up multiple table queries. What the query needs to do is to check the userID that is parsed with the url, and then check the friendshipsTable by checking if either the userID_1 or userID_2 field matches the userID to grab the records from the table where there is a match. The next step is to check to see if the friendship is 'Confirmed' or 'Pending' and if it is 'Pending' to ignore it. Once the records have then been chosen I need the query to then check the value in either userID_1 or userID_2 that doesn't match userID and then pull the user's username and name from the usersTable so it can be displayed on a webpage. I've no idea hoe much I may or may not be overcomplicating this, an example of the code that I've got so far for this query can be found below, but that's as far as I've got at the moment. $displayFriends = mysql_query("SELECT * FROM friendshipTable, usersTable WHERE friendshipTable.userID_1='$id' OR friendshipTable.userID_2='$id' "); Cheers for any help. I have an array I want to define in Php and then sort it. I am pretty new to Php and don't know the syntax or what funcs to use. Here's the array I want: user_email, user_name, hours_worked_per_week, total_earned There will be about 120 users, and I want to sort them ascending alphabetically on user_email I am new to php and starting to write my first programs. I have a laptop (my local pc) and I am wondering if there is some kind of php-interpreter available so I can test them out on my local machine without having to upload them to my website? Is there anyway I can avoid having to install a web-sever on my machine? Is there an all-in-one php interpreter? (someone suggested something called Xampp, - is this the best?) thanks,, Here is my code: // Start MySQL Query for Records $query = "SELECT codes_update_no_join_1b" . "SET orig_code_1 = new_code_1, orig_code_2 = new_code_2" . "WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)"; $results = mysql_query($query) or die(mysql_error()); // End MySQL Query for Records This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this??? Code: [Select] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1 If you also have any feedback on my code, please do tell me. I wish to improve my coding base. Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference. Code: Code: [Select] mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); Full code: Code: [Select] <?php include_once("includes/config.php"); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title><? $title; ?></title> <meta http-equiv="Content-Language" content="English" /> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <link rel="stylesheet" type="text/css" href="style.css" media="screen" /> </head> <body> <div id="wrap"> <div id="header"> <h1><? $title; ?></h1> <h2><? $description; ?></h2> </div> <? include_once("includes/navigation.php"); ?> <div id="content"> <div id="right"> <h2>Create</h2> <div id="artlicles"> <?php if(!$_SESSION['user']) { $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $name = mysql_real_escape_string($_POST['name']); $server_type = mysql_real_escape_string($_POST['type']); $description = mysql_real_escape_string($_POST['description']); if(!$username || !$password || !$server_type || !$description || !$name) { echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>"; echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'> <option value='Any'>Any</option> <option value='PvP'>PvP</option> <option value='Creative'>Creative</option> <option value='Survival'>Survival</option> <option value='Roleplay'>RolePlay</option> </select></td></tr> <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>"; echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>"; } elseif(strlen($password) < 8) { echo "Password needs to be higher than 8 characters!"; } elseif(strlen($username) > 13) { echo "Username can't be greater than 13 characters!"; } else { $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1"); if(mysql_num_rows($check1) < 0) { echo "Sorry, there is already an account with this username and/or server name!"; } else { $ip = $_SERVER['REMOTE_ADDR']; mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); echo "Server has been succesfully created!"; } } } else { echo "You are currently logged in!"; } ?> </div> </div> <div style="clear: both;"> </div> </div> <div id="footer"> <a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop </div> </div> </body> </html> I'm restarting this under a new subject b/c I learned some things after I initially posted and the subject heading is no longer accurate. What would cause this behavior - when I populate session vars from a MYSQL query, they stick, if I populate them from an MSSQL query, they drop. It doesn't matter if I get to the next page using a header redirect or a form submit. I have two session vars I'm loading from a MYSQL query and they remain, the two loaded from MSSQL disappear. I have confirmed that all four session vars are loading ok initially and I can echo them out to the page, but when the application moves to next page via redirect or form submit, the two vars loaded from MSSQL are empty. Any ideas? SELECT * FROM Sold WHERE substr(sold_date, 0, 4) = 2010 What I am trying to do is select all rows from the database from 2010, from the "sold_date" field. How is this accomplished with date values? (ex: "2010-08-11", "2009-01-15") i have a cell in my database like Code: [Select] Stats 100-10-3 and i want to update that with my code Code: [Select] $details = 100 . '-' . 10 . '-' . 3; $sql = "UPDATE usertable SET uSkillsMax=$details"; but my database updates instead of 100-10-3 it says 87? which is doing the math? any way to fix this? Code: [Select] $ids = implode (",", $ibforums->input['checkbox']); $time = time(); $ids2 = implode (",", $ibforums->input['pendingusers']); $DB->query("UPDATE friends_pending SET pending='0',date=$time WHERE id IN ($ids) AND toid IN ($ids2)"); weird thing is, it's not bring up any error's or nothing $ids2 spits out 2,30 and $ids spits out 9,7 for this particular project doesn't give me mysql error or nothing, script runs fine. I have mysql error enabling under $dbquery class so nn to worry, how to get this to work? Can i even use 2 IN's in 1 query or??? Hi -- This query seems to be problematic because the UPDATE is not being performed. Could you please take a gander and let me know what is the problem? BTW, the table "teamy" does contain 630 records. Thanks in advance! $sql = "SELECT COUNT( * ) AS records FROM teamy" ; $result = mysql_query( $sql ) ; if ( ! $result ) { die ( __line__ . "_teamy_" . mysql_error() ) ; } $a_count = mysql_fetch_assoc($result); if ( $a_count['records'] = 0 ) { echo "No records found in teamy." ; } else { /*** Update r_rost_rma ***/ $sql = "UPDATE r_rost_rma JOIN teamy ON r_rost_rma.student_id = teamy.student_id SET r_rost_rma.teachername = teamy.team WHERE RTRIM( UPPER( r_rost_rma.localcourse ) ) = 'HOMEROOM'" ; $result = mysql_query ( $sql ) ; if ( ! $result ) { die ( __line__ . "_update_r_rost_rma_" . mysql_error() ) ; } } I've got to be missing something pretty basic here.. considering the query is pretty basic. I'm trying to figure out how to pull a query as an array so I can compare it against another array (array_diff) I'm doing a mysql_fetch_array, and I'm getting an error ( mysql_fetch_array(): supplied argument is not a valid MySQL result resource): Quote $checker = "SELECT ID FROM edible_uses"; $result2 = mysql_fetch_array($checker) or die(mysql_error()); //echoing to see if I'm getting what I need. echo $row['ID']; I've done a mysql_query and I get results. The table name and all that is correct. I'm stumped. This seems like a pretty simple query? I tried mysql_fetch_assoc as well. Same result? I tried it with an extra set of parenthesis around it. nope. The following mysql query is not returning rows like I expect it to. '$update_field' is a variable, matching an actual field name in table 'users'. 'user_task[1]' is an integer value. What am I missing here? Code: [Select] $query_update_user = "UPDATE users SET ".$update_field." = 'Y' WHERE user_no = '".$user_task[1]."'"; This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=350117.0 OK, here it is. I have been trying to do this myself, but it has been driving me insane and I turn to professionals here for help.
I am a basic web developer for a company that I work for in other capacities. I have a reasonable understanding of HTML and that is about where my expertise ends. I am not typically a programmer, just a simple (extremely) part time designer that uses Muse and Dreamweaver when necessary.
However, recently my company has asked me to accomplish a task for their website. In plain English, they want a large database that exists currently as a CVS file made into a searchable web page. It is 21 columns by approximately 6,700 rows.
To explain what I need a little more technically, here are my ideas and where I have gotten to so far:
1. The company uses godaddy, into which I *believe* I have successfully imported the spreadsheet. I believe it is successful because through godaddy's SQL Control Panel (phpMyAdmin console), I can do the EXACT searches that the company needs, and it spits out the EXACT results that I need.
2. The end result needs to be a .php that I can upload to the website's root folder that can be then inserted into premade pages using:
<iframe src="SMQ.php" scrolling="yes" width="950" height="800"></iframe>3. On the .php page, I need to have a way to log in to the SQL server and a simple search box built in that will allow the user to input a very simple search string consisting of no more than 4 numbers and 3 letters at a time. No buttons, no check boxes, just a search box. 3. This query then needs to be output as a nice data table, similar to this: This in fact is a screenshot of a search I performed out of my SQL database, in phpMyAdmin using the column "Scott" for the search, and the number 226 as the search term. All column names are visible with the exception of the first column, entitled LINEID, made to be the key, and the output should not include the key but have everything else as above. 4. I can see what the simple line of php is that performed this task: SELECT * FROM `SMQSQL` WHERE `Scott` = '226' ORDER BY `LINEID` ASCbut I can't figure out how the hell to get this incorporated to a .php search. To sum it up, I need a .php page written that can connect to a SQL database, perform a data based search, and spit out a clean table when it is done. I had accomplished this in the past using an import into google docs and using it to perform a search and result display via the following code built into a php called SMQ.php:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Example of Google Spreadsheet Data Visualisation</title>
</head>
<body>
<form id="form1" method="post" action ="<?php echo $_SERVER['PHP_SELF']; ?>"> <label>
<input id="search" name="search" type="text" />
</label>
<label>
<input id="Scott #" name="Scott #" type="submit" value="Scott #" />
</label>
<img src="loading.gif" width="16" height="11" />
</form>
<p>
<?php
$search= $_REQUEST['search'];
if ($search > ''){ $search = $search;} else { $search = '';}
?>
<script type="text/javascript" src="http://www.google.com/jsapi"></script>
<script type="text/javascript">
google.load('visualization', '1', {packages: ['table']});
</script>
<script type="text/javascript">
var visualization;
function drawVisualization() {
var query = new google.visualization.Query(
'https://docs.google.com/spreadsheet/ccc?key=0AronCwm9QPefdGpIUllscGgtLUJod2pOazc0bjU0cUE&usp=sharing');
query.setQuery('SELECT A, B, C, D, E, F, G, H, I, J, K, L, M, N, O ,P ,Q ,R ,S ,T WHERE (A) LIKE ("<?php echo $search; ?>") order by A asc label A "Scott #", B "Den", C "Color", D "Cond", E "40", F "60", G "70", H "70J", I "75", J "75J", K "80", L "80J", M "85", N "85J", O "90", P "90J", Q "95", R "95J", S "98", T "98J"');
query.send(handleQueryResponse);
}
function handleQueryResponse(response) {
if (response.isError()) {
alert('Error in query: ' + response.getMessage() + '' + response.getDetailedMessage());
return;
}
var data = response.getDataTable();
visualization = new google.visualization.Table(document.getElementById('table'));
visualization.draw(data, { page: 'enable', page: 16, pageSize: 16, legend: 'bottom'});
}
google.setOnLoadCallback(drawVisualization);
</script>
<div id="table"></div>
</div>
</body>
</html>
But as you can see, this may not be the most secure thing in the world, plus we want to be able to expand it in the future and not be so simplistic, hence the need to switch to SQL.
Please let me know right away by contacting me at disead@gmail.com if this is something YOU might be able to help with. I'm sure for an experienced programmer, once you have the details from me that you need, it would take maybe 10 minutes to write. I don't have much, but I can pay a little bit for this one time job. If it ends up working out, I may be able to pay more down the line for more advanced options such as being able to do drop-down searches based on the column titled "ISSUE", as well as more things down the line as it grows.
Thank you so much, I hope to hear from someone soon!!!
Im sure this is simple, but I cannot see what my problem is! I am hitting an error on my insert query Code: [Select] Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1 I know the output of $user_id is 1, so my error is on $mysavepath Code: [Select] $mysavepath = $folder.'/'.$worldname.'_'.date("dMjY"); echo $mysavepath; $savepath = mysql_query("INSERT INTO saves (user_id,savepath) VALUES ('$user_id','$mysavepath')"); echo '<br>'.$savepath; if(!mysql_query($savepath)) { die('<br>Error: ' . mysql_error()); } however it all echos out ok? Code: [Select] 188ea678f0dcdc8252aeb15e3c910408/world_15Jan152012 1 Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1 Can anyone see the problem? Cheers Dave This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=314391.0 This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=329660.0 |
|||||||