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  • How to make a table, which is wider than screen size, scrollable and keep header row fixed?

    - by understack
    I've a table with 2 columns and each column is 800px wide. I want to show this table in 800x50 window. So there should be horizontal and vertical scrollbar to view complete table. While I've found few related solutions (this and this) on SO, they only work if table width is smaller than screen size. In my case screen size is 1200px and total table width is 1600px. How could I do this? i want to achieve something like this. EDIT Oops, I forgot to add one more requirement. Sorry. I want the header of the table to remain fixed while user scrolls table. EDIT2 I've tried below mentioned solutions to wrap in a div, but in this case vertical scrollbar doesn't show up. Please see this table with wrapper div. It seems this problem only occurs if table width is bigger than screen size. I'm testing on FF3.6. EDIT3 current table code: <div style="overflow:scroll; width:800px;height:50px" > <table style="width:1600px" border="1"> <thead> <tr> <th style="width:800px">id_1</th> <th style="width:800px">id_1</th> </tr> </thead> <tbody style=""> <tr><td>1200</td><td>1200</td></tr> <tr><td>1200</td><td>1200</td></tr> <tr><td>1200</td><td>1200</td></tr> <tr><td>1200</td><td>1200</td></tr> <tr><td>1200</td><td>1200</td></tr> <tr><td>1200</td><td>1200</td></tr> </tbody> </table> </div>

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  • how to select just a row of table using jquery

    - by user1400
    hello i have created a table on my application same follwing code <table class="spreadsheet"> <tr> <td>1</td> <td>2</td> </tr> <tr> <td>3</td> <td>4</td> </tr> <tr> <td>5</td> <td>6</td> </tr> </table> i select a row by click with following jquery code $("tr").click(function(){ $(this).toggleClass("otherstyle"); } and my css is tr.otherstyle td { background-color: #d1db3e; color:#000; } i would like when i click on a row, other rows be unselected and just one row be selected how we could create this? thanks

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  • Table valued function only returns CLR error

    - by Anthony
    I have read-only access to a database that was set up for a third-party, closed-source app. Once group of (hopefully) useful table functions only returns the error: Failed to initialize the Common Language Runtime (CLR) v2.0.50727 with HRESULT 0x80131522. You need to restart SQL server to use CLR integration features. (severity 16) But in theory, the third-party app should be able to use the function (either directly or indirectly), so I'm convinced I'm not setting things up right. I'm very new to SQL Server, so I could be missing something obvious. Or I could be missing something really slight, I have no idea. Here is an example of a query that returns the above error: SELECT * FROM dbo.UncompressDataDateRange(4,'Apr 24 2010 12:00AM','Apr 30 2010 12:00AM') Where the function takes three parameters: The Data Set (int) -- basically the data has 6 classifications, and the giant table this should be pulling from has a column to indicate which is which. startDate (smalldatetime) endDate (smalldatetime) There are other, similar functions that expand on the same idea, all returning the same error. Quick Note: I'm not sure if this is relevant, but I was able to connect to the database via SQL Studio (but without the privs to script the functions as code), and a checked the dependency for the above sample function. It turns out that it is a dependent of a view that I have gotten to work, and that view is dependent of the larger, much-hairier data-table. This makes me think I should somehow be pointing the function at the results of the view, but I'm not seeing any documentation that shows how that is done.

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  • Cmd+Less Than (10.8.2) not working after Xcode (4.5.x) installed

    - by Felix Lieb
    I had to reinstall my MBP recently. I stress Cmd+Less Than a lot for switching between Xcode's main window and the Organizer for documentation. The standard OSX-shortcut for doing that is Cmd + Less Than. After installing Xcode it didn't work any longer. I saw, that Xcode uses Cmd+LT for "Edit Schemes", a rarely used option. Even after deleting the shortcut for "Edit Schemes" in Xcode, Cmd+LT didn't work. How can I get Cmd + Less Than to work again? Mac OS X Mount Lion 10.8.2 Xcode 4.5.2 I have less than 10 reputation on superuser (acutally first post here), so I can't post the answer to my question. Would yo be so kind and upvote this question, so I can officially answer the question? The question, as well as the answer is only correct, if you use German keyboard layout.

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  • Get table row based on radio button using prototype/javascript

    - by David Buckley
    I have an html table that has a name and a radio button like so: <table id="cars"> <thead> <tr> <th>Car Name</th> <th></th> </tr> </thead> <tbody> <tr> <td class="car">Ford Focus</td> <td><input type="radio" id="selectedCar" name="selectedCar" value="8398"></td> </tr> <tr> <td class="car">Lincoln Navigator</td> <td><input type="radio" id="selectedCar" name="selectedCar" value="2994"></td> </tr> </tbody> </table> <input type="button" value="Select Car" onclick="selectCar()"></input> I want to be able to select a radio button, then click another button and get the value of the radio button (which is a unique ID) as well as the car name text (like Ford Focus). How should I code the selectCar method? I've tried a few things like: val1 = $('tr input[name=selectedCar]:checked').parent().find('#cars').html(); val1 = $("td input[name='selectedCar']:checked").parents().find('.cars').html(); val1 = $('selectedCar').checked; but I can't get the proper values. I'm using prototype, but the solution can be plain javascript as well.

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  • DAO design pattern and using it across multiple tables

    - by Casey
    I'm looking for feedback on the Data Access Object design pattern and using it when you have to access data across multiple tables. It seems like that pattern, which has a DAO for each table along with a Data Transfer Object (DTO) that represents a single row, isn't too useful for when dealing with data from multiple tables. I was thinking about creating a composite DAO and corresponding DTO that would return the result of, let's say performing a join on two tables. This way I can use SQL to grab all the data instead of first grabbing data from one using one DAO and than the second table using the second DAO, and than composing them together in Java. Is there a better solution? And no, I'm not able to move to Hibernate or another ORM tool at the moment. Just straight JDBC for this project.

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  • Rows added to table are not showing up

    - by Lars
    This has been asked before, but I have found no solution that worked for me yet: When adding rows to a table from code, the rows are not showing up in the application. There is one row I specified in XML, that one is showing up, but nothing below it. This is the code: public void addRow(LocationMessage locationMsg){ View messageView = theInflater.inflate(R.layout.homepage, null); TableLayout table = (TableLayout)messageView.findViewById(R.id.distanceTable); TextView senderNameTextView = new TextView(thisContext); senderNameTextView.setText(locationMsg.getSenderName()); TableRow tr = new TableRow(thisContext); tr.addView(distanceTextView); table.addView(tr); rows.addFirst(messageView); } homepage.xml contains this, I removed some elements and parameters: <?xml version="1.0" encoding="utf-8"?> <LinearLayout> <TabHost> <TabWidget /> <FrameLayout> [..] <LinearLayout> [..] <TableLayout android:id="@+id/distanceTable" android:layout_height="wrap_content" android:layout_width="wrap_content" android:layout_gravity="center" android:background="#DDDDDD" android:stretchColumns="1" > <TableRow> <TextView android:textColor="#000000" android:text="@string/label_device" android:layout_gravity="center" android:padding="3dip" android:textSize="18sp" /> <TextView android:textColor="#000000" android:text="@string/label_distance" android:layout_gravity="center" android:padding="3dip" android:textSize="18sp" /> <TextView android:textColor="#000000" android:text="@string/label_time" android:layout_gravity="center" android:padding="3dip" android:textSize="18sp" /> </TableRow> </TableLayout> </LinearLayout> </FrameLayout> </TabHost> </LinearLayout> Unfortunately hierarchyviewer.bat doesn't work for me in order to check if the rows are there but just not visible. In the debugger it looks fine to me.

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  • Database Table design for Weekdays and Values

    - by Ved
    Guys, I would love to here your ideas on this. I am creating a table which will store weekly shift hours for employees. for example: Jon works 9:00am tp 9:00pm Monday , 10:00 AM to 5:00PM on Tuseday etc.... How should i go on designing this table. I thought of 2 options. (1) For every record I can have two lines for AM and PM and have columns for Monday to Sunday ID | ParentID | Type | Monday | Tuseday ......Sunday 1 | 1 | AM | 9:00 | 9:00 12:00 2 | 1 | PM | 9:00 | 5:00 02:00 3 | 2 | AM | 10:00 | 4 | 2 | PM | 10:00 | (2) I can save Preference in XML format in one column ID | Info 1 | |Hours|Monday|9:00 AM - 9:00PM|Monday|Tuseday........|Hours Any better ideas ? Thanks

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  • jQuery adding a row to a table with click() handler for tr

    - by Dave
    I'm having an issue with a control I'm building that contains a table where the body is scrollable. The rows have a click() function handler established like: /** * This function is called when the user clicks the mouse on a row in * our scrolling table. */ $('.innerTable tr').click (function (e) { // // Only react to the click if the mouse was clicked in the DIV or // the TD. // if (event.target.nodeName == 'DIV' || event.target.nodeName == 'TD' ) { // // If the user wasn't holding down the control key, then deselect // any previously selected columns. // if (e.ctrlKey == false) { $('.innerTable tr').removeClass ('selected'); } // // Toggle the selected state of the row that was clicked. // $(this).toggleClass ('selected'); } }); There is a button that adds rows to the table like: $('#innerTable > tbody:last').append('<tr>...some information...</tr>'); While the rows ARE added successfully, for some reason, the static rows work with the click handler, but the newly added rows do not. Is there something I'm missing? Thanks in advance!

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  • Need to find number of new unique ID numbers in a MySQL table

    - by Nicholas
    I have an iPhone app out there that "calls home" to my server every time a user uses it. On my server, I create a row in a MySQL table each time with the unique ID (similar to a serial number) aka UDID for the device, IP address, and other data. Table ClientLog columns: Time, UDID, etc, etc. What I'd like to know is the number of new devices (new unique UDIDs) on a given date. I.e. how many UDIDs were added to the table on a given date that don't appear before that date? Put plainly, this is the number of new users I gained that day. This is close, I think, but I'm not 100% there and not sure it's what I want... SELECT distinct UDID FROM ClientLog a WHERE NOT EXISTS ( SELECT * FROM ClientLog b WHERE a.UDID = b.UDID AND b.Time <= '2010-04-05 00:00:00' ) I think the number of rows returned is the new unique users after the given date, but I'm not sure. And I want to add to the statement to limit it to a date range (specify an upper bound as well).

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  • Stretching a Table's Width to fit web browser

    - by Techn0guy
    Okay so I have sowered the internet for a long time trying to solve my problem. I have a table with my whole site in it and would like for it to stretch to fit the user's web browser size. Here is all the site's code atm <body> <table width="100%" border="0"> <tr> <td width="525" rowspan="2"><img src="My Images/Speaker.png" width="525" height="772" ondragstart="return false" /> </td> <td width="697" height="16"> *Flash Video Here* </td> </tr> <tr> <td><img src="My Images/Vymil.jpg" width="822" height="597" ondragstart="return false"/></td> </tr> </table> </body> </html> Any helpful solutions would be greatly appreciated

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  • Design question - loading info from DB

    - by eriks
    I need to build a class that will represent a row in some table in DB (lets say the table is 'Subscriber' and so is the class). I can have the class Subscriber which constructor receives the Objectkey of subscriber, retrieves info from DB and initializes its members. I add another class - SubscriberLoader which have a static method 'LoadSubscriber'. This method will receive the subscriber objectkey, retrieve info from DB, crate a Subscriber object and initialize its members. Subscriber constructor will be private and SubscirberLoader will be friend class of Subscriber - this way, client could build a Subscriber only using the loader. which of the two in better? any other suggestions?

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  • Struts2 Hibernate Login with User table and group table

    - by J2ME NewBiew
    My problem is, i have a table User and Table Group (this table use to authorization for user - it mean when user belong to a group like admin, they can login into admincp and other user belong to group member, they just only read and write and can not login into admincp) each user maybe belong to many groups and each group has been contain many users and they have relationship are many to many I use hibernate for persistence storage. and struts 2 to handle business logic. When i want to implement login action from Struts2 how can i get value of group member belong to ? to compare with value i want to know? Example I get user from username and password then get group from user class but i dont know how to get value of group user belong to it mean if user belong to Groupid is 1 and in group table , at column adminpermission is 1, that user can login into admincp, otherwise he can't my code: User.java /* * To change this template, choose Tools | Templates * and open the template in the editor. */ package org.dejavu.software.model; import java.io.Serializable; import java.util.Date; import java.util.HashSet; import java.util.Set; import javax.persistence.CascadeType; import javax.persistence.Column; import javax.persistence.Entity; import javax.persistence.FetchType; import javax.persistence.GeneratedValue; import javax.persistence.Id; import javax.persistence.JoinColumn; import javax.persistence.JoinTable; import javax.persistence.ManyToMany; import javax.persistence.Table; import javax.persistence.Temporal; /** * * @author Administrator */ @Entity @Table(name="User") public class User implements Serializable{ private static final long serialVersionUID = 2575677114183358003L; private Long userId; private String username; private String password; private String email; private Date DOB; private String address; private String city; private String country; private String avatar; private Set<Group> groups = new HashSet<Group>(0); @Column(name="dob") @Temporal(javax.persistence.TemporalType.DATE) public Date getDOB() { return DOB; } public void setDOB(Date DOB) { this.DOB = DOB; } @Column(name="address") public String getAddress() { return address; } public void setAddress(String address) { this.address = address; } @Column(name="city") public String getCity() { return city; } public void setCity(String city) { this.city = city; } @Column(name="country") public String getCountry() { return country; } public void setCountry(String country) { this.country = country; } @Column(name="email") public String getEmail() { return email; } public void setEmail(String email) { this.email = email; } @ManyToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL) @JoinTable(name="usergroup",joinColumns={@JoinColumn(name="userid")},inverseJoinColumns={@JoinColumn( name="groupid")}) public Set<Group> getGroups() { return groups; } public void setGroups(Set<Group> groups) { this.groups = groups; } @Column(name="password") public String getPassword() { return password; } public void setPassword(String password) { this.password = password; } @Id @GeneratedValue @Column(name="iduser") public Long getUserId() { return userId; } public void setUserId(Long userId) { this.userId = userId; } @Column(name="username") public String getUsername() { return username; } public void setUsername(String username) { this.username = username; } @Column(name="avatar") public String getAvatar() { return avatar; } public void setAvatar(String avatar) { this.avatar = avatar; } } Group.java /* * To change this template, choose Tools | Templates * and open the template in the editor. */ package org.dejavu.software.model; import java.io.Serializable; import javax.persistence.Column; import javax.persistence.Entity; import javax.persistence.GeneratedValue; import javax.persistence.Id; import javax.persistence.Table; /** * * @author Administrator */ @Entity @Table(name="Group") public class Group implements Serializable{ private static final long serialVersionUID = -2722005617166945195L; private Long idgroup; private String groupname; private String adminpermission; private String editpermission; private String modpermission; @Column(name="adminpermission") public String getAdminpermission() { return adminpermission; } public void setAdminpermission(String adminpermission) { this.adminpermission = adminpermission; } @Column(name="editpermission") public String getEditpermission() { return editpermission; } public void setEditpermission(String editpermission) { this.editpermission = editpermission; } @Column(name="groupname") public String getGroupname() { return groupname; } public void setGroupname(String groupname) { this.groupname = groupname; } @Id @GeneratedValue @Column (name="idgroup") public Long getIdgroup() { return idgroup; } public void setIdgroup(Long idgroup) { this.idgroup = idgroup; } @Column(name="modpermission") public String getModpermission() { return modpermission; } public void setModpermission(String modpermission) { this.modpermission = modpermission; } } UserDAO /* * To change this template, choose Tools | Templates * and open the template in the editor. */ package org.dejavu.software.dao; import java.util.List; import org.dejavu.software.model.User; import org.dejavu.software.util.HibernateUtil; import org.hibernate.Query; import org.hibernate.Session; /** * * @author Administrator */ public class UserDAO extends HibernateUtil{ public User addUser(User user){ Session session = HibernateUtil.getSessionFactory().getCurrentSession(); session.beginTransaction(); session.save(user); session.getTransaction().commit(); return user; } public List<User> getAllUser(){ Session session = HibernateUtil.getSessionFactory().getCurrentSession(); session.beginTransaction(); List<User> user = null; try { user = session.createQuery("from User").list(); } catch (Exception e) { e.printStackTrace(); session.getTransaction().rollback(); } session.getTransaction().commit(); return user; } public User checkUsernamePassword(String username, String password){ Session session = HibernateUtil.getSessionFactory().getCurrentSession(); session.beginTransaction(); User user = null; try { Query query = session.createQuery("from User where username = :name and password = :password"); query.setString("username", username); query.setString("password", password); user = (User) query.uniqueResult(); } catch (Exception e) { e.printStackTrace(); session.getTransaction().rollback(); } session.getTransaction().commit(); return user; } } AdminLoginAction /* * To change this template, choose Tools | Templates * and open the template in the editor. */ package org.dejavu.software.view; import com.opensymphony.xwork2.ActionSupport; import org.dejavu.software.dao.UserDAO; import org.dejavu.software.model.User; /** * * @author Administrator */ public class AdminLoginAction extends ActionSupport{ private User user; private String username,password; private String role; private UserDAO userDAO; public AdminLoginAction(){ userDAO = new UserDAO(); } @Override public String execute(){ return SUCCESS; } @Override public void validate(){ if(getUsername().length() == 0){ addFieldError("username", "Username is required"); }if(getPassword().length()==0){ addFieldError("password", getText("Password is required")); } } public String getPassword() { return password; } public void setPassword(String password) { this.password = password; } public String getRole() { return role; } public void setRole(String role) { this.role = role; } public User getUser() { return user; } public void setUser(User user) { this.user = user; } public String getUsername() { return username; } public void setUsername(String username) { this.username = username; } } other question. i saw some example about Login, i saw some developers use interceptor, im cant understand why they use it, and what benefit "Interceptor" will be taken for us? Thank You Very Much!

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  • ssh Password-less login to multiple machines when you already have one

    - by tandu
    I'm a little bit confused about setting up a password-less login for multiple machines to begin with, but I think I could do it from scratch. The problem is I already have it set up for one machine and I don't want that to be blown away when I try to set it up for the other machine. Let's clarify: Machine A: the machine I'm connecting from Machine B: the machine I'm connecting to. Password required Machine C: the machine I'm connecting to. Password-less ssh I have read some tutorials on setting up password-less ssh to a certain site, but they usually start with "move id_rsa out of the way so it doesn't get blown away," but then at the end of the tutorial it's not moved back. If I had no help at all, here is what I would do: Log into B ssh-keygen -t rsa -f ~/id_rsa.other scp id_rsa.other.pub A:~/.ssh echo "Host A \n Identity File ~/.ssh/id_rsa.other" > ~/.ssh/config (Note that I realize these commands may not be exactly correct, but this is just the idea). What I'm not quite clear on is if I need to update the config for A, B, or both. I'm fairly certain to do a password-less login from A to B, it is A that needs the public key .. but I also suppose I need B to use the correct id_rsa file for that public key. Finally, I don't want the password-less login for C to be affected at all .. it's using id_rsa. Am I going wrong anywhere?

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  • Working with PHP and MySQL - need a good and secure design with OO design

    - by Andrew
    I am new to PHP- first time developer. I am working on my web application and it is nearly done; nevertheless, most of my sql was done directly via code using direct mysql requests. This is the way I approached it: In classes_db.php I declared the db settings and created methods that I use to open and close DB connections. I declare those objects on my regular pages: class classes_db { public $dbserver = 'server; public $dbusername = 'user'; public $dbpassword = 'pass'; public $dbname = 'db'; function openDb() { $dbhandle = mysql_connect($this->dbserver, $this->dbusername, $this->dbpassword); if (!$dbhandle) { die('Could not connect: ' . mysql_error()); } $selected = mysql_select_db($this->dbname, $dbhandle) or die("Could not select the database"); return $dbhandle; } function closeDb($con) { mysql_close($con); } } On my regular page, I do this: <?php require 'classes_db.php'; session_start(); //create instance of the DB class $db = new classes_db(); //get dbhandle $dbhandle = $db->openDb(); //process query $result = mysql_query("update user set username = '" . $usernameFromForm . "' where iduser= " . $_SESSION['user']->iduser); //close the connection if (isset($dbhandle)) { $db->closeDb($dbhandle); } ?> My questions is: how to do it right and make it OO and secure? I know that I need incorporate prepared queries- how to do it the best way? Please provide some code

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  • Table cellspacing with CSS

    - by antpaw
    hey, is it possible to add spacing between the cells? <!DOCTYPE html> <html> <head> <meta charset="utf-8" /> <title>test</title> <style> #wrap { display: table-row; /* there is no also cellspacing in CSS afaik :( */ } #wrap div { display: table-cell; /* margin: 40px; doesn't work for table-cell :( */ /* border: 30px solid transparent; works but it's as good as adding padding */ /* border: 30px solid white; works but the page background is invisible */ /* decoration: */ background-color: green; padding: 20px; } </style> </head> <body> <!-- don't touch the markup --> <div id="wrap"> <div>text1</div> <div>text2</div> </div> </body> </html>

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  • how to print data in a table from mysql

    - by robertdd
    hello, i want to extract the last eight entries from my database and print them into a 2 columns table!like this: |1|2| |3|4| |5|6| |7|8| is that possible? this is my code: $db = new Database(DB_SERVER, DB_USER, DB_PASS, DB_DATABASE); $db->connect(); $sql = "SELECT ID, movieno FROM movies ORDER BY ID DESC LIMIT 8 "; $rows = $db->query($sql); print '<table width="307" border="0" cellspacing="5" cellpadding="4">'; while ($record = $db->fetch_array($rows)) { $vidaidi = $record['movieno']; print <<<END <tr> <td> <a href="http://www.youtube.com/watch?v=$vidaidi" target="_blank"> <img src="http://img.youtube.com/vi/$vidaidi/1.jpg" width="123" height="80"></a> </td> </tr> END; } print '</table>';

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  • Querying using table-valued parameter

    - by antmx
    I need help please with writing a sproc, it takes a table-valued parameter @Locations, whose Type is defined as follows: CREATE TYPE [dbo].[tvpLocation] AS TABLE( [CountryId] [int] NULL, [ResortName] [nvarchar](100) NULL, [Ordinal] [int] NOT NULL, PRIMARY KEY CLUSTERED ( [Ordinal] ASC )WITH (IGNORE_DUP_KEY = OFF) ) @Locations will contain at least 1 row. Each row WILL have a non-null CountryId, and MAY have a non-null ResortName. Each row will have a unique Ordinal, the first being 0. The combinations of CountryId and ResortName in @Locations will be unique. The sproc needs to search against the following table structure. The image can be seen better by right-clicking it and View Image, or similar depending on your browser. Now this is where I'm stuck, the sproc should be able to find Tours where: The Tour's 1st TourHotel (Ordinal 0) has the same CountryId (and ResortName if specified) of the 1st row of @Locations (Ordinal 0). And also if @Locations has 1 row, the Tour must have additional TourHotels, ALL of which must be in the remaining CountryIds (and ResortNames if specified) of these remaining @Locations rows. Edit This is the code I finally used, based on Anthony Faull's suggestion. Thank you so much Anthony: select distinct T.Id from tblTour T join tblTourHotel TH on TH.TourId = T.Id join tblHotel H ON H.Id = TH.HotelId JOIN @Locations L ON ( ( L.Ordinal = 0 AND TH.Ordinal = 0 ) OR ( L.Ordinal > 0 AND TH.Ordinal > 0 ) ) AND L.CountryId = H.CountryId AND ( L.ResortName = H.ResortName OR L.ResortName IS NULL ) cross apply( select COUNT(TH2.Id) AS [Count] FROM tblTourHotel TH2 where TH2.TourId = TH.TourId ) TourHotelCount where TourHotelCount.[Count] = @LocationCount group by T.Id, T.TourRef, T.Description, T.DepartureDate, T.NumNights, T.DepartureAirportId, T.DestinationAirportId, T.AirlineId, T.FEPrice having COUNT(distinct TH.Id) = @LocationCount

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  • Print table data mysql php

    - by Marcelo
    Hi people, i'm having a problem trying to print some data of a table. I'm new at this php mysql stuff but i think my code is right. Here it is: <html> <body> <h1>Lista de usuários</h1> <?php $host="localhost"; // Host name $username="root"; // Mysql username $password=""; // Mysql password $db_name="sabs"; // Database name $tbl_name="doador"; // Table name // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $sql="SELECT * FROM $tbl_name"; $result=mysql_query($sql); while($rows = mysql_fetch_array($result)){ echo $row['id'] . " " .$row['nome'] . " " . $row['sobrenome'] . " " . $row['email'] . " " . $row['login'] . " " . $row['senha'] . " " . $row['idade'] . " ". $row['peso'] . " " . $row['fuma'] . " " . $row['sexo'] . " " . $row['doencas']; echo "<BR/>"; } mysql_close(); ?> </body> </html> All columns of the echo command exist in my table in the database. Don't get why it's not printing those values. Thanks for the attention.

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  • Add to a table with javascript

    - by incrediman
    I have a table which looks like this: <table id='t'> <thead> .... </thead> <tbody id='t_tbody'> </tbody> </table> thead is filled with content tbody is empty I want to use javascript to add this (for example) to t_tbody: <tr> <td>Name</td> <td>Points</td> <td>Information</td> </tr> How can I do this? I need a function which adds the above to t_tbody. Note that simply using document.getElementById('t_tbody').innerHtml+="<tr>...</tr>" works fine in FF, but not in IE. Also note that I need to use raw javascript (ie. no frameworks) for this project, as I am finishing a project which has already been mostly completed using raw javascript.

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  • related categories - database design

    - by mike
    Hello! I'm looking for a little database design advice... I have a spreadsheet with a few columns in it. Column 1 being a list of categories and the rest being related categories(to the category in column 1). I'm trying to figure out what the best way to setup the tables would be... My thought so far is to have a table that just lists the categories then have a table with 2 columns that holds the id of the category and the id of a related category.... Would this be the best way to do this? Any better ideas?

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  • Is there anyway to show a hidden div in the last row of a html table

    - by oo
    i have an html table. here is a simplified version: <table> <tr> <td><div style="display: none;" class="remove0">Remove Me</div></td> </tr> <tr> <td><div style="display: none;" class="remove1">Remove Me</div></td> </tr> <tr> <td><div class="remove2">Remove Me</div></td> </tr> </table> i have javascript that clicks on Remove Me in the last row and it deletes the html row using: $(this).parents("tr:first").remove(); the issue is that when i remove this last row, i also want the "Remove Me" text to now show up on the second row (which is now the new last row). how would i show this div so that it would dynamically show the "remove me" from the new last row?

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  • Find class in table row

    - by meep
    Hello. Take a look at this table: <table cellpadding="0" cellspacing="0" class="order_form"> <tr> <th>Amount</th> <th>Desc</th> <th>Price</th> <th>Total</th> </tr> <tr> <td><input type="text" class="order_count" /></td> <td> <span class="order_desc">Middagstallerken</span> </td> <td> <span class="order_price">1,15</span> </td> <td> <span class="order_each_total"></span> </td> </tr> [...] </table> Upon entering amount I need to select the class "order_price" and multiply it with the value of the input "order_count" and place it in "order_each_count". I have millions of these rows so I need to find the next class in the row. I have tried using some function like this but without result: <script type="text/javascript"> $(document).ready(function(){ $('.order_count').keyup(function() { var each_price = $(this).prevUntil("tr").find("span.order_price").text(); }); }); </script> I hope someone have a good solution :-)

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  • BNF – how to read syntax?

    - by Piotr Rodak
    A few days ago I read post of Jen McCown (blog) about her idea of blogging about random articles from Books Online. I think this is a great idea, even if Jen says that it’s not exciting or sexy. I noticed that many of the questions that appear on forums and other media arise from pure fact that people asking questions didn’t bother to read and understand the manual – Books Online. Jen came up with a brilliant, concise acronym that describes very well the category of posts about Books Online – RTFM365. I take liberty of tagging this post with the same acronym. I often come across questions of type – ‘Hey, i am trying to create a table, but I am getting an error’. The error often says that the syntax is invalid. 1 CREATE TABLE dbo.Employees 2 (guid uniqueidentifier CONSTRAINT DEFAULT Guid_Default NEWSEQUENTIALID() ROWGUIDCOL, 3 Employee_Name varchar(60) 4 CONSTRAINT Guid_PK PRIMARY KEY (guid) ); 5 The answer is usually(1), ‘Ok, let me check it out.. Ah yes – you have to put name of the DEFAULT constraint before the type of constraint: 1 CREATE TABLE dbo.Employees 2 (guid uniqueidentifier CONSTRAINT Guid_Default DEFAULT NEWSEQUENTIALID() ROWGUIDCOL, 3 Employee_Name varchar(60) 4 CONSTRAINT Guid_PK PRIMARY KEY (guid) ); Why many people stumble on syntax errors? Is the syntax poorly documented? No, the issue is, that correct syntax of the CREATE TABLE statement is documented very well in Books Online and is.. intimidating. Many people can be taken aback by the rather complex block of code that describes all intricacies of the statement. However, I don’t know better way of defining syntax of the statement or command. The notation that is used to describe syntax in Books Online is a form of Backus-Naur notatiion, called BNF for short sometimes. This is a notation that was invented around 50 years ago, and some say that even earlier, around 400 BC – would you believe? Originally it was used to define syntax of, rather ancient now, ALGOL programming language (in 1950’s, not in ancient India). If you look closer at the definition of the BNF, it turns out that the principles of this syntax are pretty simple. Here are a few bullet points: italic_text is a placeholder for your identifier <italic_text_in_angle_brackets> is a definition which is described further. [everything in square brackets] is optional {everything in curly brackets} is obligatory everything | separated | by | operator is an alternative ::= “assigns” definition to an identifier Yes, it looks like these six simple points give you the key to understand even the most complicated syntax definitions in Books Online. Books Online contain an article about syntax conventions – have you ever read it? Let’s have a look at fragment of the CREATE TABLE statement: 1 CREATE TABLE 2 [ database_name . [ schema_name ] . | schema_name . ] table_name 3 ( { <column_definition> | <computed_column_definition> 4 | <column_set_definition> } 5 [ <table_constraint> ] [ ,...n ] ) 6 [ ON { partition_scheme_name ( partition_column_name ) | filegroup 7 | "default" } ] 8 [ { TEXTIMAGE_ON { filegroup | "default" } ] 9 [ FILESTREAM_ON { partition_scheme_name | filegroup 10 | "default" } ] 11 [ WITH ( <table_option> [ ,...n ] ) ] 12 [ ; ] Let’s look at line 2 of the above snippet: This line uses rules 3 and 5 from the list. So you know that you can create table which has specified one of the following. just name – table will be created in default user schema schema name and table name – table will be created in specified schema database name, schema name and table name – table will be created in specified database, in specified schema database name, .., table name – table will be created in specified database, in default schema of the user. Note that this single line of the notation describes each of the naming schemes in deterministic way. The ‘optionality’ of the schema_name element is nested within database_name.. section. You can use either database_name and optional schema name, or just schema name – this is specified by the pipe character ‘|’. The error that user gets with execution of the first script fragment in this post is as follows: Msg 156, Level 15, State 1, Line 2 Incorrect syntax near the keyword 'DEFAULT'. Ok, let’s have a look how to find out the correct syntax. Line number 3 of the BNF fragment above contains reference to <column_definition>. Since column_definition is in angle brackets, we know that this is a reference to notion described further in the code. And indeed, the very next fragment of BNF contains syntax of the column definition. 1 <column_definition> ::= 2 column_name <data_type> 3 [ FILESTREAM ] 4 [ COLLATE collation_name ] 5 [ NULL | NOT NULL ] 6 [ 7 [ CONSTRAINT constraint_name ] DEFAULT constant_expression ] 8 | [ IDENTITY [ ( seed ,increment ) ] [ NOT FOR REPLICATION ] 9 ] 10 [ ROWGUIDCOL ] [ <column_constraint> [ ...n ] ] 11 [ SPARSE ] Look at line 7 in the above fragment. It says, that the column can have a DEFAULT constraint which, if you want to name it, has to be prepended with [CONSTRAINT constraint_name] sequence. The name of the constraint is optional, but I strongly recommend you to make the effort of coming up with some meaningful name yourself. So the correct syntax of the CREATE TABLE statement from the beginning of the article is like this: 1 CREATE TABLE dbo.Employees 2 (guid uniqueidentifier CONSTRAINT Guid_Default DEFAULT NEWSEQUENTIALID() ROWGUIDCOL, 3 Employee_Name varchar(60) 4 CONSTRAINT Guid_PK PRIMARY KEY (guid) ); That is practically everything you should know about BNF. I encourage you to study the syntax definitions for various statements and commands in Books Online, you can find really interesting things hidden there. Technorati Tags: SQL Server,t-sql,BNF,syntax   (1) No, my answer usually is a question – ‘What error message? What does it say?’. You’d be surprised to know how many people think I can go through time and space and look at their screen at the moment they received the error.

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  • MVC - Cocoa interface - Cocoa Design pattern book

    - by Idan
    So I started reading this book: http://www.amazon.com/Cocoa-Design-Patterns-Erik-Buck/dp/0321535022 On chapter 2 it explains about the MVC design pattern and gives and example which I need some clarification to. The simple example shows a view with the following fields: hourlyRate, WorkHours, Standarthours , salary. The example is devided into 3 parts : View - contains some text fiels and a table (the table contains a list of employees' data). Controller - comprised of NSArrayController class (contains an array of MyEmployee) Model - MyEmployee class which describes an employee. MyEmployee class has one method which return the salary according to the calculation logic, and attributes in accordance with the view UI controls. MyEmployee inherits from NSManagedObject. Few things i'm not sure of : 1. Inside the MyEmplpyee class implemenation file, the calculation method gets the class attributes using sentence like " [[self valueForKey:@"hourlyRate"] floatValue];" Howevern, inside the header there is no data member named hourlyRate or any of the view fields. I'm not quite sure how does it work, and how it gets the value from the right view field. (does it have to be the same name as the field name in the view). maybe the conncetion is made somehow using the Interface builder and was not shown in the book ? and more important: 2. how does it seperate the view from the model ? let's say ,as the book implies might happen, I decide one day to remove one of the fields in the view. as far as I understand, that means changing the way the salary method works in MyEmplpyee (cause we have one field less) , and removing one attribute from the same calss. So how is that separate the View from the Model if changing one reflect on the other ? I guess I get something wrong... Any comments ? Thanks

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