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  • What does MySqlDataAdapter.Fill return when the results are empty?

    - by Brian
    I have a 'worker' function that will be processing any and all sql queries in my program. I will need to execute queries that return result sets and ones that just execute stored procedures without any results. Is this possible with MySqlDataAdapter.Fill or do I need to use the MySqlCommand.ExecuteNonQuery() method? Here is my 'worker' function for reference: private DataSet RunQuery(string SQL) { MySqlConnection connection; MySqlCommand command; MySqlDataAdapter adapter; DataSet dataset = new DataSet(); lock(locker) { connection = new MySqlConnection(MyConString); command = new MySqlCommand(); command = connection.CreateCommand(); command.CommandText = SQL; adapter = new MySqlDataAdapter(); adapter.Fill(dataset); } return dataset; }

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  • need help on my query.

    - by Dharmendra
    i have one table : nobel(yr, subject, winner) and i have this query : In which years was the Physics prize awarded but no Chemistry prize. this is what i tried : select distinct yr from nobel where subject='physics' and subject!='chemistry' but is not working where i am going wrong. see, i am not here to make my homework from someone. i am here to learn something. so, please give me suggetion.

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  • Keeping some choices in the Table for the Field of Type Dropdown

    - by Mercy
    Hi, i am having a Table named Attributes which has id form_id label size sequence_no Type 1 1 Name 200 1 Text 2 1 Age 150 2 Number 3 1 Address 300 3 Textarea 4 1 Gender 200 4 Dropdown I am having the doubt how can i keep the Choices of the Field of type "Dropdown" in the Table Eg. For Gender the choices will Male , Female.. Please give me the suggestions...

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  • getting sql records

    - by droidus
    when i run this code, it returns the topic fine... $query = mysql_query("SELECT topic FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; } but when I change it to this, it doesn't run at all. why is this happening? $query = mysql_query("SELECT topic, lock FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; $lockedThread = $row['lock']; echo "here: " . $lockedThread; }

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  • How to add condition on multiple-join table

    - by Jean-Philippe
    Hi, I have those two tables: client: id (int) #PK name (varchar) client_category: id (int) #PK client_id (int) category (int) Let's say I have those datas: client: {(1, "JP"), (2, "Simon")} client_category: {(1, 1, 1), (2, 1, 2), (3, 1, 3), (4,2,2)} tl;dr client #1 has category 1, 2, 3 and client #2 has only category 2 I am trying to build a query that would allow me to search multiple categories. For example, I would like to search every clients that has at least category 1 and 2 (would return client #1). How can I achieve that? Thanks!

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  • How to build a SQL statement when any combination of user input to the table is possible?

    - by Greg McNulty
    Example: the user fills in everything but the product name. I need to search on what is supplied, so in this case everything but productName= This example could be for any combination of input. Is there a way to do this? Thanks. $name = $_POST['n']; $cat = $_POST['c']; $price = $_POST['p']; if( !($name) ) { $name = some character to select all? } $sql = "SELECT * FROM products WHERE productCategory='$cat' and productName='$name' and productPrice='$price' "; EDIT Solution does not have to protect from attacks. Specifically looking at the dynamic part of it.

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  • PHP Multiple User Login Form - Navigation to Different Pages Based on Login Credentials

    - by Zulu Irminger
    I am trying to create a login page that will send the user to a different index.php page based on their login credentials. For example, should a user with the "IT Technician" role log in, they will be sent to "index.php", and if a user with the "Student" role log in, they will be sent to the "student/index.php" page. I can't see what's wrong with my code, but it's not working... I'm getting the "wrong login credentials" message every time I press the login button. My code for the user login page is here: <?php session_start(); if (isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'IT Technician')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> <?php if (isset($_POST["username"]) && isset($_POST["password"]) && isset($_POST["role"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if (($existCount == 1) && ($role == 'Student')) { while ($row = mysql_fetch_array($sql)) { $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; $_SESSION["role"] = $role; header("location: http://www.zuluirminger.com/SchoolAdmin/student/index.php"); } else { echo 'Your login details were incorrect. Please try again <a href="http://www.zuluirminger.com/SchoolAdmin/index.php">here</a>'; exit(); } } ?> And the form that the data is pulled from is shown here: <form id="LoginForm" name="LoginForm" method="post" action="http://www.zuluirminger.com/SchoolAdmin/user_login.php"> User Name:<br /> <input type="text" name="username" id="username" size="50" /><br /> <br /> Password:<br /> <input type="password" name="password" id="password" size="50" /><br /> <br /> Log in as: <select name="role" id="role"> <option value="">...</option> <option value="Head">Head</option> <option value="Deputy Head">Deputy Head</option> <option value="IT Technician">IT Technician</option> <option value="Pastoral Care">Pastoral Care</option> <option value="Bursar">Bursar</option> <option value="Secretary">Secretary</option> <option value="Housemaster">Housemaster</option> <option value="Teacher">Teacher</option> <option value="Tutor">Tutor</option> <option value="Sanatorium Staff">Sanatorium Staff</option> <option value="Kitchen Staff">Kitchen Staff</option> <option value="Parent">Parent</option> <option value="Student">Student</option> </select><br /> <br /> <input type="submit" name = "button" id="button" value="Log In" onclick="javascript:return validateLoginForm();" /> </h3> </form> Once logged in (and should the correct page be loaded, the validation code I have at the top of the script looks like this: <?php session_start(); if (!isset($_SESSION["manager"])) { header("location: http://www.zuluirminger.com/SchoolAdmin/user_login.php"); exit(); } $managerID = preg_replace('#[^0-9]#i', '', $_SESSION["id"]); $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["manager"]); $password = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["password"]); $role = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["role"]); include "adminscripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM Users WHERE username='$manager' AND password='$password' AND role='$role' LIMIT 1"); $existCount = mysql_num_rows($sql); if ($existCount == 0) { header("location: http://www.zuluirminger.com/SchoolAdmin/index.php"); exit(); } ?> Just so you're aware, the database table has the following fields: id, username, password and role. Any help would be greatly appreciated! Many thanks, Zulu

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  • Enhancing an 'ORDER BY' clause to judge condition by more than 1 integer

    - by Yvonne
    Hi folks, I have some PHP code which allows me to sort a column into ascending and descending order (upon click of a table row title), which is good. It works perfectly for my D.O.B colum (with date/time field type), but not for a quantity column. For example, I have quantites of 10, 50, 100, 30 and another 100. The order seems to be only appreciating the 1st integer, so my sorting of the column ends up in this order: 10, 100, 100, 30, 50... and 50, 30, 100, 100, 10. This is obviously incorrect as 100 is bigger than 50, therefore both 100 values should appear at the end surely? It seems to me that 100 is only being taken into account as having the '1' value, then it appears before 10 because the system recognises it has another 0. Is this normal to happen? Is there any way I can solve this problem? Thanks for any help. P.S. I can show code if necessary, but would like to know if this is a common issue by default.

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  • Zend_Table_Db and Zend_Paginator num rows

    - by Uffo
    I have the following query: $this->select() ->where("`name` LIKE ?",'%'.mysql_escape_string($name).'%') Now I have the Zend_Paginator code: $paginator = new Zend_Paginator( // $d is an instance of Zend_Db_Select new Zend_Paginator_Adapter_DbSelect($d) ); $paginator->getAdapter()->setRowCount(200); $paginator->setItemCountPerPage(15) ->setPageRange(10) ->setCurrentPageNumber($pag); $this->view->data = $paginator; As you see I'm passing the data to the view using $this->view->data = $paginator Before I didn't had $paginator->getAdapter()->setRowCount(200);I could determinate If I have any data or not, what I mean with data, if the query has some results, so If the query has some results I show the to the user, if not, I need to show them a message(No results!) But in this moment I don't know how can I determinate this, since count($paginator) doesn't work anymore because of $paginator->getAdapter()->setRowCount(200);and I'm using this because it taks about 7 sec for Zend_Paginator to count the page numbers. So how can I find If my query has any results?

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  • Combine SQL statement

    - by ninumedia
    I have 3 tables (follows, postings, users) follows has 2 fields - profile_id , following_id postings has 3 fields - post_id, profile_id, content users has 3 fields - profile_id, first_name, last_name I have a follows.profile_id value of 1 that I want to match against. When I run the SQL statement below I get the 1st step in obtaining the correct data. However, I now want to match the postings.profile_id of this resulting set against the users table so each of the names (first and last name) are displayed as well for all the listed postings. Thank you for your help! :) Ex: SELECT * FROM follows JOIN postings ON follows.following_id = postings.profile_id WHERE follows.profile_id = 1

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  • Update multiple progress bar with gtk c++

    - by Yadira Suazo
    I need to output the i progress bars and update them all. But only the last one updates i times. This is the code: static void calculaPi (GtkButton * boton, Datos * dDatos){ const char * threads; GtkWidget * barra, *bot2, *button, *progress, *vbox; threads = gtk_entry_get_text(GTK_ENTRY(dDatos->dthreads )); gint ithreads = 1; ithreads = atoi(threads); barra = gtk_window_new(GTK_WINDOW_TOPLEVEL); gtk_window_set_title((GtkWindow *) barra, "Loteria de Threads"); gtk_window_set_default_size(GTK_WINDOW(barra), 300, ithreads*30); gtk_window_set_position(GTK_WINDOW(barra), GTK_WIN_POS_CENTER); button = gtk_button_new_with_label ("Click me!"); vbox = gtk_vbox_new (FALSE, 5); gtk_box_pack_start (GTK_BOX (vbox), button, FALSE, FALSE, 5); gtk_container_add (GTK_CONTAINER (barra), vbox); for (gint i = 1 ; i <= ithreads; i++) { progress = gtk_progress_bar_new (); gtk_box_pack_start (GTK_BOX (vbox), progress, FALSE, FALSE, 5); g_object_set_data (G_OBJECT (barra), "pbar", (gpointer) progress); g_signal_connect (G_OBJECT (button), "clicked", G_CALLBACK (button_clicked), (gpointer) barra); } bot2 = gtk_button_new_with_label("Salir"); gtk_box_pack_start (GTK_BOX (vbox), bot2, FALSE, FALSE, 5); gtk_widget_set_size_request(bot2, 100, 35); g_signal_connect (G_OBJECT (bot2), "clicked", G_CALLBACK (destroy), G_OBJECT (barra)); gtk_widget_show_all(barra); gtk_main(); } static void button_clicked (GtkButton *button, GtkWidget *barra) { GtkProgressBar *progress; gdouble percent = 0.0; gtk_widget_set_sensitive (GTK_WIDGET (button), FALSE); progress = GTK_PROGRESS_BAR (g_object_get_data (G_OBJECT (barra), "pbar")); while (percent <= 100.0) { gchar *message = g_strdup_printf ("%.0f%% Complete", percent); gtk_progress_bar_set_fraction (progress, percent / 100.0); gtk_progress_bar_set_text (progress, message); while (gtk_events_pending ()) gtk_main_iteration (); g_usleep (500000); percent += 5.0; } }

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  • Resetting AUTO_INCREMENT on myISAM without rebuilding the table

    - by Artem
    Please help I am in major trouble with our production database. I had accidentally inserted a key with a very large value into an autoincrement column, and now I can't seem to change this value without a huge rebuild time. "ALTER TABLE tracks_copy AUTO_INCREMENT = 661482981" Is super-slow. How can I fix this in production? I can't get this to work either (has no effect): myisamchk tracks.MYI --set-auto-increment=661482982 Any ideas? Basically, no matter what I do I get an overflow: SHOW CREATE TABLE tracks CREATE TABLE tracks ( ... ) ENGINE=MYISAM AUTO_INCREMENT=2147483648 DEFAULT CHARSET=latin1

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  • PHP MySQLi isn't letting me alter a table (adding a new column)

    - by asdasd
    Well thats pretty much it. This is my query: $query = 'ALTER TABLE permissions ADD '.$name.' INT NOT NULL DEFAULT \'0\''; Where $name is already checked to exist with only lower case alpha letters, and not more than 20 length. Im just starting this out with very simple names. The next 4 lines of code after that one are: if($stmt = $db -> prepare($query)) { $success = $stmt -> execute(); $stmt -> close(); if(!$success) echo 'ERROR: Unsuccessful query: ',$db->error,PHP_EOL; } And I get back, every time ERROR: Unsuccessful query: And no error message. Is there a way to get more error messages so I can see what is failing? I can add new columns through phpmyadmin, but that really doesnt help me at all. The $db is fine, i do lots of stuff before and after this one section. It is only adding new column to the table that fails. side question: prepare() rejected my query every time when i tried to make those 2 variables, the $name and the 0 value as ? ? prepared statement values. Thats why they are in the real query and not bound later. If i could change that too I would like that.

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  • SQL - How to display the students with the same age?

    - by Cristian
    the code I wrote only tells me how many students have the same age. I want their names too... SELECT YEAR(CURRENT DATE-DATEOFBIRTH) AS AGE, COUNT(*) AS HOWMANY FROM STUDENTS GROUP BY YEAR(CURRENT DATE-DATEOFBIRTH); this returns something like this: AGE HOWMANY --- ------- 21 3 30 5 Thank you. TABLE STUDENTS COLUMNS: StudentID (primary key), Name(varchar), Firstname(varchar), Dateofbirth(varchar) I was thinking of maybe using the code above and somewhere add the function concat that will put the stundents' names on the same row as in

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  • Database Design for One to One relationships

    - by Greelmo
    I'm trying to finalize my design of the data model for my project, and am having difficulty figuring out which way to go with it. I have a table of users, and an undetermined number of attributes that apply to that user. The attributes are in almost every case optional, so null values are allowed. Each of these attributes are one to one for the user. Should I put them on the same table, and keep adding columns when attributes are added (making the user table quite wide), or should I put each attribute on a separate table with a foreign key to the user table. I have decided against using the EAV model. Thanks!

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  • Passing dynamic string in hyperlink as parameter in jsp

    - by user3660263
    I am trying to pass a dynamic string builder variable in jsp I am generating a string through code. String Builder variable has some value but i am not able to pass it in at run time.It doesn't get the value. CODE FOR VARIABLE <% StringBuilder sb=new StringBuilder(""); if(request.getAttribute("Brand")!=null) { String Brand[]=(String[])request.getAttribute("Brand"); for(String brand:Brand) { sb.append("Brand="); sb.append(brand); sb.append("&&"); } } if(request.getAttribute("Flavour")!=null) { String Flavour[]=(String[])request.getAttribute("Flavour"); for(String flavour:Flavour) { sb.append(flavour); sb.append("&&"); } sb.trimToSize(); pageContext.setAttribute("sb", sb); } out.print("this is string"+sb); %> CODE FOR HYPERLINK <a href="Filter_Products?${sb}page=${currentPage + 1}" style="color: white;text-decoration: none;">Next</a></td>

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  • database connection OK,result not appear

    - by klox
    hi..all.for now i'm already connected to database but the result not appear at Tuner range is"+res+"this is my code: var str=data[0]; var matches=str.match(/[EE|EJU].*D/i); $.ajax({ type:"post", url:"process1.php", data:"tversion="+matches+"&action=tunermatches", cache:false, async:false, success: function(res){ $('#value').replacewith("<div id='value'><h6>Tuner range is"+res+".</h6></div>"); } }); }); and this is my process file: //connect to database $dbc=mysql_connect(_SRV,_ACCID,_PWD) or die(_ERROR15.": ".mysql_error()); $db=mysql_select_db("qdbase",$dbc) or die(_ERROR17.": ".mysql_error()); switch(postVar('action')) { case 'tunermatches' : tunermatches(postVar('tversion')); break; function tunermatches($tversion)){ $Tuner=mysql_real_escape_string($tversion); $sql= "SELECT remark FROM settingdata WHERE itemname='Tuner_range' AND itemdata='".$Tunermatches."'"; $res=mysql_query($sql) or die (_ERROR26.":".mysql_error()); $dat=mysql_fetch_array($res,MYSQL_NUM); if($dat[0]>0) { echo $dat[0]; } mysql_close($dbc); }

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  • How to structure this query...?

    - by SpikETidE
    Hi Everyone... Consider the following table.... hotel facilities 1 internet 1 swimming pool 1 wi-fi 1 parking 2 swimming pool 2 sauna 2 parking 3 toilets 3 bungee-jumping 3 internet 4 parking 4 swimming pool I need to select only the hotels that have parking, swimming pool and internet....? I worked out the following.... SELECT hotel FROM table WHERE facilties IN(internet, swimming pool, parking) This query selects the hotels that has atleast one among the choices. But what i need is a query that selects the hotels that has ALL of the selected facilities... Thanks for your suggestions....

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  • sql statement question. Need to query 3 tables in one go!

    - by Stefan
    Hey there, I have an sql database. In this database is 3 tables I need to query. The first table has all the item info called item and the other two tables has data for votes and comments called userComment and the third for votes called userItem I currently have a function which uses this sql query to get the latest more popular (in terms of both votes and comments): $sql = "SELECT itemID, COUNT(*) AS cnt FROM ( SELECT `itemID` FROM `userItem` WHERE FROM_UNIXTIME( `time` ) >= NOW() - INTERVAL 1 DAY UNION ALL SELECT `itemID` FROM `userComment` WHERE FROM_UNIXTIME( `time` ) >= NOW() - INTERVAL 1 DAY AND `itemID` > 0 ) q GROUP BY `itemID` ORDER BY cnt DESC"; I know how to change this for either by votes alone or comments.... HOWEVER - I need to query the database to only return the itemID's of the ones which have specific conditions in only the item table these are WHERE categoryID = 'xx' AND typeID = 'xx' If the sql ninja could please help me on this one? Do I have to first return the results from the above query and the for each in the array fetched then check each against the item table and see if it fits the conditions to build a new array - or is that overkill? Thanks, Stefan

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  • I can't insert data into my database

    - by Ken
    I don't know why, but my data doesn't go into my database 'users' with the table 'data'. <html> <body> <?php date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "g00dfor@boy"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db(`users`, $con) or die(mysql_error()); mysql_query(INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password')) or die(mysql_error()); mysql_close($con) echo("Thank you for registering!"); ?> </body> </html> All i get is a blank page.

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