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  • MySQL/PHP: How to insert logged in user id into another table that is gathering data from a form tha

    - by Lisa
    For the first time I am needing to join information from two tables and am quite nervous about doing it without any advice first. Basically, I am building a secure site that is accessed by authorised users. I have my login table with user_id, username, password Once the user is on the site, they have the option of inputting data into another table called input. At the moment this table only captures the information that is entered, not the user_id or username of the inputter. I would like the form to be able to input the user_id and/or username from the login table into the input table. Please could somebody talk me through this process? I am sure that once this is amended, I will then be able to use the table to only allow the logged in user to access the information that he or she have inputted, is that correct? Many thanks

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  • Moving from MySQL to MySQLi? I have the code here but I don't get it

    - by MuqMan
    I have posted the code there, please help me out as I am a newbie, I don't know much in terms of deprecation and stuff. <?php session_start(); include('settings.php'); $issub = $_POST['issub']; if($issub == "yes") { require('settings.php'); $dbcon = mysql_connect($dbhost, $dbuser, $dbpword); if(!dbcon) { die('Could not connect'.mysql_error()); } $selectdb = mysql_select_db($db, $dbcon); $formset = 'yes'; $val = 0; $user = trim($_POST['username'], ' '); $luser = mysql_real_escape_string($user); $password = $_POST['password']; $lpassword = mysql_real_escape_string($password); $selectdb; $userq = mysql_query("SELECT user FROM users WHERE user='".$luser."'"); $userresult = @mysql_result($userq, 0); //echo $userresult; if($userresult == $user) { $val = $val + 1; $usercorrect = 'yes'; } else { $usercorrect = 'no'; } $dbselect; $passwordq = mysql_query("SELECT password FROM users where user='".$luser."'"); $passresult = @mysql_result($passwordq, 0); if($passresult == sha1($password)) { $val = $val + 1; $passcorrect = 'yes'; } else { $passcorrect = 'no'; } if ($val == 2) { $_SESSION['loggedin'] = 'yes'; $_SESSION['uloggedin'] = $user; header('location: logged.php'); } }?> <?php ini_set('display_errors', 1); require('testinclude.php'); ?> <body> <div id="loginform"> <form action="/login.php" method="post" > <input type="hidden" name="issub" value="yes" /> <?php if($usercorrect == 'no') { echo '<span class="required"><i><small>The email address or password you entered is incorrect, please try again.</a></small></i></span>'; } ?> <br /> email: <?php if ($issub == 'yes') { if($user == null){ echo '<br /><span class="required"><i><small>Please enter your email address</a></small></i></span>'; } } ?> <br /><input type="text" name="username" id="usename" /> <br /> password: <br /><input type="password" name="password" id="password" /> <br /> <input type="submit" value="login" /> </form> <div> </body>

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  • Drupal 6: using too many Views module causing site to go down cos of too many mysql connection.

    - by artmania
    Hi friends, I have HostGator Baby Shared Plan . I develop Drupal site on. everything was fine at the beginning, then by the time i go further with development, site started ti work really slow. now it is not working at all. giving my sql errors like TOO many connections, etc... I created so many blocks, pages with View. so it makes my site to so much depend on database. should not I do that? can it be the reason of my site's no working now. appreciate helps!!!!

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  • php mysql search in 2 columns in 2 tables.

    - by andrew fishwick
    Hey, I have two tables in one DB, one called Cottages and one called Hotels. In both tables they have the same named fields. I basically have a search bar that i want it to search in both of the fields in both of the tables. (the two fields being called "Name" and "Location" SO far I have $sql = mysql_query("SELECT * FROM Cottages WHERE Name LIKE '%$term%' or Location LIKE '%$term%' LIMIT 0, 30"); But this only searches the Cottages table, how can I make it search both the cottages and hotel tables? Andy

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  • Are there drawbacks to altering MySQL table data types?

    - by Tower
    Hi, I'm wondering that how much worried I should be about data types. I can easily jump from TINYINT to SMALLINT and from SMALLINT to INT, but are there any drawbacks to this? Obviously situations like from text to int will have consequences, but I'm talking about situations like INT-BIGINT, TINYTEXT-TEXT, etc.

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  • mysql : possible to put IF statment in LEFT JOIN ?

    - by Haroldo
    Ok so i want to get an artists info from the db, but i want to know if they have any forthcoming events. To do this i need to traverse 2 tables, where events_artists is a 2 col link table... (this doesnt work but is what id like to do) SELECT art.*, events.event_id FROM art LEFT JOIN events_artists ON art.art_id = events_artists.art_id LEFT JOIN events ON events_artists.event_id = events.event_id IF ( {criteria} ) what should i be doing here to get this to work?!!

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  • PHP mysql - ...AND column='anything'... ?

    - by Nike
    Is there any way to check if a column is "anything"? The reason is that i have a searchfunction that get's an ID from the URL, and then it passes it through the sql algorithm and shows the result. But if that URL "function" (?) isn't filled in, it just searches for: ...AND column=''... and that doesn't return any results at all. I've tried using a "%", but that doesn't do anything. Any ideas?

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  • Mysql Query is not working in edited jTable code, why?

    - by Furkan Kadioglu
    I'm using this example: www.jtable.org I've downloaded the jTable PHP version. I then edited the script. The jTable simple version is working, but my edited version isn't. I can create a list, but I can't add a row; this code is causing problems. However, PHP doesn't display any error messages. else if($_GET["action"] == "create") { //Insert record into database $result = mysql_query("INSERT INTO veriler(bolge, sehir, firma, adres, tel, web) VALUES('" . $_POST["bolge"] . "', '" . $_POST["sehir"] . "', '" . $_POST["firma"] . "', '" . $_POST["adres"] . "', '" . $_POST["tel"] . "', '" . $_POST["web"] . "'"); //Get last inserted record (to return to jTable) $result = mysql_query("SELECT * FROM veriler WHERE id = LAST_INSERT_ID();"); $row = mysql_fetch_array($result); //Return result to jTable $jTableResult = array(); $jTableResult['Result'] = "OK"; $jTableResult['Record'] = $row; print json_encode($jTableResult); } What is the problem?

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  • Google calendar query returns at most 25 entries

    - by Dean Hill
    I'm trying to delete all calendar entries from today forward. I run a query then call getEntries() on the query result. getEntries() always returns 25 entries (or less if there are fewer than 25 entries on the calendar). Why aren't all the entries returned? I'm expecting about 80 entries. As a test, I tried running the query, deleting the 25 entries returned, running the query again, deleting again, etc. This works, but there must be a better way. Below is the Java code that only runs the query once. CalendarQuery myQuery = new CalendarQuery(feedUrl); DateFormat dfGoogle = new SimpleDateFormat("yyyy-MM-dd'T00:00:00'"); Date dt = Calendar.getInstance().getTime(); myQuery.setMinimumStartTime(DateTime.parseDateTime(dfGoogle.format(dt))); // Make the end time far into the future so we delete everything myQuery.setMaximumStartTime(DateTime.parseDateTime("2099-12-31T23:59:59")); // Execute the query and get the response CalendarEventFeed resultFeed = service.query(myQuery, CalendarEventFeed.class); // !!! This returns 25 (or less if there are fewer than 25 entries on the calendar) !!! int test = resultFeed.getEntries().size(); // Delete all the entries returned by the query for (int j = 0; j < resultFeed.getEntries().size(); j++) { CalendarEventEntry entry = resultFeed.getEntries().get(j); entry.delete(); } PS: I've looked at the Data API Developer's Guide and the Google Data API Javadoc. These sites are okay, but not great. Does anyone know of additional Google API documentation?

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  • Can I lock tables in an IF statement in MySQL?

    - by MalcomTucker
    This is throwing a syntax error - --from body of a stored proc IF (name = in_name) SET out_id = temp; ELSE LOCK TABLE People WRITE; INSERT INTO People (Name) VALUES (in_name); UNLOCK TABLE; SELECT LAST_INSERT_ID() INTO out_id END IF do I have to lock any tables I need at the start of the SP?

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  • MySQL stored procedure to INSERT DELAYED but CREATE TABLE first if needed?

    - by dkamins
    I'm planning on doing a lot of INSERT DELAYED into MyISAM tables. But the tables might not exist yet. Let's say e.g. a new table will exist for each day. So instead of detecting absence of table in my client and creating then retrying the insert, it seems like a good case for a stored procedure ("function"). Is this possible, and what would it look like? Are there any downsides to this approach?

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  • Which of the two ways should I use to insert tags into mysql?

    - by ggfan
    For each ad, I allow users to choose up to 5 tags. Right now, in my database, I have it like... Posting_id TagID 5 1 5 2 5 3 6 5 6 1 But i was thinking if I should make it like... Posting_id TagID 5 1 2 3 6 5 1 Then first option is much easier to insert and retrieve data. But if I have 100 posts with 3 tags each, that's 300 rows...so ALOT more rows The second option requires using explode() impode(), etc but it is much cleaner. Which option should I do and why? thanks!

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  • Changing the current count of an Auto Increment value in MySQL?

    - by RD
    Currently every time I add an entry to my database, the auto increment value increments by 1, as it should. However, it is only at a count of 47. So, if I add a new entry, it will be 48, and then another it will be 49 etc. I want to change what the current Auto Increment counter is at. I.e. I want to change it from 47 to say, 10000, so that the next value entered, will be 10001. How do I do that?

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  • Why is this loop over mysql resultset slow? (1.4ms per cycle)

    - by pawpro
    The $res contains around 488k rows the whole loop takes 61s! that's over 1.25ms per cycle! What is taking all that time? while($row = $res->fetch_assoc()) { $clist[$row['upload_id']][$row['dialcode_id']][$row['carrier_id']]['std'] = $row['cost_std']; $clist[$row['upload_id']][$row['dialcode_id']][$row['carrier_id']]['ecn'] = $row['cost_ecn']; $clist[$row['upload_id']][$row['dialcode_id']][$row['carrier_id']]['wnd'] = $row['cost_wnd']; $dialcode_destination[$row['upload_id']][$row['carrier_id']][$row['dialcode_id']]['other_destination'] = $row['destination_id']; $dialcode_destination[$row['upload_id']][$row['carrier_id']][$row['dialcode_id']]['carrier_destination'] = $row['carrier_destination_id']; } Now resultset of 10 rows, smaller arrays and performance 30 times higher (0.041ms) not the fastest still but better. while($row = $res->fetch_assoc()) { $customer[$row['id']]['name'] = $row['name']; $customer[$row['id']]['code'] = $row['customer']; }

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  • How to search for a specific word in a row in MYSQL?

    - by user220755
    I have a row that has keywords in this way (keyword1, keyword2, keyword3) separated by commas as shown. When a user signs in, you know that he wants information about (keyword1, keyword3). Now, I have another table that has bunch of information related to different keywords, this table has a row called (keywords) which indicates if this information is suitable for which keyword. How do I render for the user the information he needs depending on the keywords. In other words, if the user wants information about (keyword3, keyword1) how do I go to the (information) table and find all the information that has the word (keyword1) or the word (keyword3) in the row (keyword)? Sorry if this is complicated (my explanation) but I tried my best to explain it. Thank you in advance :)

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  • Making a function for selecting from MySQL, how is mine?

    - by Doug
    This is my first time. I will appreciate any thoughts, tips, and what not. How can I improve this? Ultimately, I don't want so many selects in my script. function mysqlSelectCodes($table, $where, $order, $limit) { $sql = "SELECT * FROM $table WHERE $where ORDER BY $order LIMIT $limit" or die(mysql_error()); }

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