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Thread: Php problem

  1. #1
    Join Date
    Nov 2013
    Posts
    6

    Php problem

    Gooed evening;

    Im new with php and i got some application that i create as learn proces.

    But nog i have an problem. and i can't solve :s

    Im trying to edit one row on mij sql data. But if i do the data of all rows would be edited. And that not my intention i want to update that one specific row. Plz help me!

    http://st-joris-turnhout.be/man/log-ins-test.php a working example of the bug.

    code:

    my database:
    Code:
    CREATE TABLE IF NOT EXISTS `logins` (
      `ID` int(11) NOT NULL AUTO_INCREMENT,
      `Website_applicatie` varchar(150) NOT NULL,
      `Login` varchar(150) NOT NULL,
      `Pass` varchar(150) NOT NULL,
      `Notitie` varchar(255) NOT NULL,
      PRIMARY KEY (`ID`)
    ) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=35 ;
    function:
    PHP Code:
    $id =$_REQUEST['ID'];

    $result mysql_query("SELECT * FROM logins WHERE ID  = '$id'");
    $row mysql_fetch_array($result);
    if (!
    $result
            {
            die(
    "Error: Data is niet gevonden");
            }

                    
    // Laad de bestaande data in de form
                    
    $web_app=$row['Website_applicatie'] ;
                    
    $login=$row['Login'] ;                    
                    
    $pass=$row['Pass'] ;
                    
    $notitie=$row['Notitie'] ;
                    
    $weblink=$row['Hyperlink'];

    if(isset(
    $_POST['save']))
    {    
        
    // Wijzigd de data
        
    $web_app_save $_POST['webapp'];
        
    $login_save $_POST['login'];
        
    $pass_save $_POST['pass'];
        
    $notitie_save $_POST['notitie'];
        
    $hyperlink_save $_POST['hyperlink'];

        
    mysql_query("UPDATE logins SET Website_applicatie ='$web_app_save', Login ='$login_save', Pass ='$pass_save', Notitie ='$notitie_save', Hyperlink ='$hyperlink_save'")
                    or die(
    mysql_error()); 
        echo 
    "Uw wijziging is doorgevoerd!";
        
        
    header("Location: ../log-ins-test.php");            
    }
    mysql_close();
    ?> 
    trigger:
    HTML Code:
    <a class="label label-default" href="includes/edit.php?ID=129"><span class="glyphicon glyphicon-pencil"></span> Edit</a>
    .

    I hiope u can help me guys

  2. #2
    Join Date
    Aug 2004
    Location
    Ankh-Morpork
    Posts
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    Looks like you need a WHERE clause in your UPDATE query to tell it which row to update, presumably on the ID field, which it appears will be available in $_GET['ID'].
    "Please give us a simple answer, so that we don't have to think, because if we think, we might find answers that don't fit the way we want the world to be."
    ~ Terry Pratchett in Nation

    eBookworm.us

  3. #3
    Join Date
    Nov 2013
    Posts
    6
    Quote Originally Posted by NogDog View Post
    Looks like you need a WHERE clause in your UPDATE query to tell it which row to update, presumably on the ID field, which it appears will be available in $_GET['ID'].
    can u give me some more info plz i don't understand it well. or can u give me a patch

  4. #4
    Join Date
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    Location
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    PHP Code:
    if(empty($_GET['id']) or !ctype_digit($_GET['ID'])) {
      die(
    "ID not received or it is not an integer.");
    }
    $id = (int) $_GET['ID'];
    $sql "
      UPDATE logins SET 
        Website_applicatie ='
    $web_app_save',
        Login ='
    $login_save',
        Pass ='
    $pass_save',
        Notitie ='
    $notitie_save',
        Hyperlink ='
    $hyperlink_save'
      WHERE ID = 
    $id
    "
    ;
    mysql_query($sql); 
    "Please give us a simple answer, so that we don't have to think, because if we think, we might find answers that don't fit the way we want the world to be."
    ~ Terry Pratchett in Nation

    eBookworm.us

  5. #5
    Join Date
    Nov 2013
    Posts
    6
    Quote Originally Posted by NogDog View Post
    PHP Code:
    if(empty($_GET['id']) or !ctype_digit($_GET['ID'])) {
      die(
    "ID not received or it is not an integer.");
    }
    $id = (int) $_GET['ID'];
    $sql "
      UPDATE logins SET 
        Website_applicatie ='
    $web_app_save',
        Login ='
    $login_save',
        Pass ='
    $pass_save',
        Notitie ='
    $notitie_save',
        Hyperlink ='
    $hyperlink_save'
      WHERE ID = 
    $id
    "
    ;
    mysql_query($sql); 
    and how i implement this code?

  6. #6
    Join Date
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    Location
    Ankh-Morpork
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    It would replace this line:
    PHP Code:
        mysql_query("UPDATE logins SET Website_applicatie ='$web_app_save', Login ='$login_save', Pass ='$pass_save', Notitie ='$notitie_save', Hyperlink ='$hyperlink_save'"
    "Please give us a simple answer, so that we don't have to think, because if we think, we might find answers that don't fit the way we want the world to be."
    ~ Terry Pratchett in Nation

    eBookworm.us

  7. #7
    Join Date
    Nov 2013
    Location
    India
    Posts
    1
    Thanks For your explanation, but i need some more explanation. Because now only i am learning PHP.

  8. #8
    Join Date
    Nov 2013
    Posts
    6
    Quote Originally Posted by NogDog View Post
    It would replace this line:
    PHP Code:
        mysql_query("UPDATE logins SET Website_applicatie ='$web_app_save', Login ='$login_save', Pass ='$pass_save', Notitie ='$notitie_save', Hyperlink ='$hyperlink_save'"
    but then i recive this error"ID not received or it is not an integer."

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