Jump to content

Brian W

Members
 View Profile  See their activity

  • Posts

    867

  • Joined

  • Last visited

 Content Type 

Posts posted by Brian W

    Insert Script Error

    in PHP Coding Help

    Posted

    try

    "INSERT INTO `users` (`username`,`password`,`email`) VALUES (%s,%s,%s)"

    for testing reasons. also use

    mysql_query($insert) or die("Could not insert comment" . mysql_error().'<br>'.$insert);

    that way you can see the query.

    if that doesn't work, let us know what the query is (copy and past the output of the error report above)

    Occasional problems

    in PHP Coding Help

    Posted

    in code:

    <?php //Find Rows
$Result2= mysql_query("SELECT * FROM users WHERE ID='$ID'");
$Rows2= mysql_fetch_array($Result2);

//Define User Variables
$Pokémon= $Rows2['VolcanoValley'];
$Pokémon2= $Rows2['VolcanoValley2'];
//If Action Is Not Null
if($Action=="catch")
{
  //If Form Is Not Submitted
if(!$Submit)
{

    //Define Random Variables
    $PokémonRand= rand(1, 100);

    //If Pokémon Is Charmander
    if($PokémonRand >= 1 && $PokémonRand <= 25)
    {
      //Define Pokémon Variables
      $PokémonNumber= 1;
      $PokémonName="Charmander";
      $PokémonImage="http://i13.tinypic.com/6xvsvw7.png";
    }

    //If Pokémon Is Magby
    elseif($PokémonRand >= 26 && $PokémonRand <= 50)
    {
      //Define Pokémon Variables
      $PokémonNumber= 2;
      $PokémonName="Magby";
      $PokémonImage="http://i10.tinypic.com/8also6f.png";
    }

    //If Pokémon Is Numel
    elseif($PokémonRand >= 51 && $PokémonRand <= 75)
    {
      //Define Pokémon Variables
      $PokémonNumber= 3;
      $PokémonName="Numel";
      $PokémonImage="http://i5.tinypic.com/73e7me8.png";
    }

    //If Pokémon Is Nothing
    elseif($PokémonRand >= 76 && $PokémonRand <= 99)
    {
      //Define Pokémon Variables
      $PokémonNumber= 0;
      $PokémonName="Nothing";
      $PokémonImage="";
    }

    //If Pokémon Is Entei
    else
    {
      //Define Pokémon Variables
      $PokémonNumber= 4;
      $PokémonName="Entei";
      $PokémonImage="http://i8.tinypic.com/7xxnyud.png";
    }

    //If Pokémon Is Nothing
    if($PokémonName=="Nothing")
    {
      //Display Message
      echo "You have found nothing when searching for Pokémon.<br /><br />

      <a href='map1.php?action=catch'><img border='0' src='http://i13.tinypic.com/8104aqe.jpg'></a>";
    }

    //If Pokémon Is Not Nothing
    else
    {
      //Define Random Variables
      $PokémonRand2= rand(1, 49);

      //If Gender Is Male
      if($PokémonRand2 >= 1 && $PokémonRand2  <= 24)
      {
        //Define Gender Variables
        $PokémonGender="Male";
        $PokémonGSign="(M)";
      }

      //If Gender Is Female
      elseif($PokémonRand2 >= 25 && $PokémonRand2  <= 48)
      {
        //Define Gender Variables
        $PokémonGender="Female";
        $PokémonGSign="(F)";
      }

      //If Gender Is Ungendered
      elseif($PokémonRand2== 49)
      {
        //Define Gender Variables
        $PokémonGender="Ungendered";
        $PokémonGSign="(U)";
      }

      //If Gender Is Unknown
      else
      {
        //Define Gender Variables
        $PokémonGender="Male";
        $PokémonGSign="(M)";
      }

      //Update Information
      mysql_query("UPDATE users SET VolcanoValley='$PokémonNumber', VolcanoValley2='$PokémonGSign' WHERE ID='$ID'");
      //Define User Variables

      //Display Form
      echo "

      <form action='?action=catch' method='POST'>
      <input type='hidden' name='Submit' value=1'>
      <img src='$PokémonImage' alt='$PokémonName' title='$PokémonName'><br />
      <b>$PokémonName</b><br /><br />

      You have encountered a wild $PokémonName <b>$PokémonGSign</b>!<br /><br />

      <input type='submit' value='Catch $PokémonName' onclick=\"this.disabled='true'; this.value='Please Wait...'; this.form.submit();\"><br /><br />
      </form>";

      echo "<a href='map1.php?action=catch'><img border='0' src='http://i13.tinypic.com/8104aqe.jpg'></a>";
    }
  }
?>

    just to help me...

    PHP post varieable problem - please help

    in PHP Coding Help

    Posted

    with my version

    <input type="submit" value="Submit" style="border:0px solid; width:150px; height:35px; background:url(Images/bg_button.gif);"/>

    you could add some javascript to ask them to confirm which is nice and if they don't have javascript enabled, the button simply just works without confirmation. unnecessary but cute...lol    going above and beyond!

    Select first letter

    in PHP Coding Help

    Posted

    well, think this may be what you are looking for...

    <?php

$letter = $_GET['letter'];
$system = $_GET['system'];

$sql = "SELECT * FROM OSG_Games WHERE system_id = $system AND title LIKE '$letter%' ORDER BY title ASC";

$query = mysql_query($sql);

while($row = mysql_fetch_array($query)) {
   echo $row['title'];
   echo "<br>";
   
}

    PHP post varieable problem - please help

    in PHP Coding Help

    Posted

    I think this would also work... but am not 100%

    <form method="post" action="unsub.php">
    <div style="text-align: center;">
        <span style="font-weight: bold;">Email Address </span>
        <input size="35" name="EmailAddress">     <br>
    </div>
<input type="submit" value="Submit" style="border:0px solid; width:150px; height:35px; background:url(Images/bg_button.gif);"/>
</form>

     

    Reorder MySQL table

    in PHP Coding Help

    Posted

    I've thought about this before... there isn't any easy way to do so as there isn't a SQL command to do so and when you start looking at even as little as 100+ records even a server side process of some sort gets ridiculous.

    "why" is funny to me because whether they be crazy, dumb, or just OCD... it doesn't matter what they will do with it, the challenge of "how" is more important.

    wtf is system.byte[]

    in Other RDBMS and SQL dialects

    Posted

    Okay, my co-worker who is an ASP developer has the entire office shitting bricks cuz the boss is shitting bricks. We moved a website onto our server yesterday seemingly without issues but not entirely without (of course)...

    Though he is using ASP, this sounds like a mysql problem sorta.

     

    What used to display First, middle, and last name and a float delimited by commas with something like CONCAT_WD(' ', Firstname, Middle, Last, Hours)  now errors saying "system.byte[]".

     

    What makes me think this is a mysql issue is that it worked on the old server which was running an older version of MySql, ODBC, and IIS6. Now it is on a system with the latest a greatest versions of mysql, ODBC and even IIS7. Huge environment changes like this make me paranoid... god bless testing the water and such but we didn't catch this bug till it hit production.

     

    Any ideas? Thanks

    [SOLVED] Strange brew (symbols)

    in PHP Coding Help

    Posted

    Where I copied the headers from already had that. I changed it to ISO-8859-1 while using the UTF8_decode() but it still feed me the ?'s instead of symbols.

    With UTF8_encode() and utf-8 as my charset i get:

    â¦letâs

    and still if I have UTF8_decode() and utf-8 as my charset i get it with the ?'s

     

    BUT! there is a BUT! I figured it out...

    I used htmlspecialchars() with charset=UTF-8 and it worked!

    I realized that what I needed was to not make it HTML safe but to make it HTML... lol

     

    Thanks for the help guys.

    [SOLVED] Strange brew (symbols)

    in PHP Coding Help

    Posted

    I was just finished my test of utf8_decode() when I read your post, premiso...  :D

    my db handles the the symbols fine, just not the emails.

    Little better now. This is what gets emailed now...

    ?let?s

    This detail may or may not be effecting things. here is some of my code for the email:

    $body = '<b>Summary: </b>"'.utf8_decode($_POST['Summary']).'"<br>';
$headers = "MIME-Version: 1.0\n";
$headers .= "Content-type: text/html; charset=us-ascii\n";

    [SOLVED] Strange brew (symbols)

    in PHP Coding Help

    Posted

    Need some help here with weird symbols. First off, here is the situation:

    I have a form that posts text to a page to be processed. The input is a normal HTML text field.

    The php page that processes the data first inputs the text into my mysql database and if successful, sends an email to an email address I've specified.

    Sometimes I get weird things like "…let’s" (which is supposed to be ...let's). I've narrowed it down to the fact that some people like to copy text from things like MS WORD into my text field which doesn't email pretty when they use symbols that are not HTML friendly (Examples: "…" instead of "..."{there is a difference} and " ’ " instead of " ' ").

    How do I convert their input into HTML friendly text?

     

    Thanks in advance.

    Dynamic droplists off of a dropdown lists selection

    in PHP Coding Help

    Posted

    Your looking at needing Ajax my friend. You could find out how to do that yourself or you could pay one of the code monkeys around here to do it but It is in your favor to learn how it is done.

    Here is a brief overview of how I would do it (which may not be the best way...)

    Create three divs with id's "Cabinet", "Folder", "Item".

    Create two other php pages that only contain a dynamically created menu based on a $_GET value.

    :::This would likely be a unique ID passed from the above level:::

    Use java script to trigger pulling in your "Folder" page into the div "Folder" with a URL variable to pass the "Cabinet" ID or whatnot.

    Same thing for the "Folder" level to the "Item" level.

     

     

    [SOLVED] Create dropdown list by querying database?

    in PHP Coding Help

    Posted

    THANK YOU FOR CLARIFYING!!!  lol  jk, but that does help.

    on this line i gave you:

    echo "<a href=\"paystubs.php?Date=".$Date."\">";

    instead say

    echo "<a href=\"paystubs.php?Date=".$endingdate."\">";

    Then, in your query, say

    ...WHERE `date_column`=".$_GET['Date']."...

    that should pass your block date (12152008) to the next page specified by the link above and make it usable for the query.

    Additional comment: use

    if(!is_numeric($_GET['Date'])){ die('Bad Date Input'); }

    That will prevent SQL injection (type of hacking if you didn't know)...

    [SOLVED] Create dropdown list by querying database?

    in PHP Coding Help

    Posted

    That is not a form dude. Your form won't use links like that. So, forget the form if you wan't, but you'll need to stop trying to use the like they are Inputs. Lets do this with $_GET then.

    <?php

$RESULTDS=mssql_query("SELECT DISTINCT M2.[EMPNO], PS.[PSTUB5] 

FROM MASTERL2 M2 

LEFT JOIN PAYSTUBS PS
ON PS.[PSTUB2]=M2.EMPNO 

WHERE M2.[EMPNO] = '".$_SESSION['empcode']."' ORDER BY PS.[PSTUB5]");
   
$RESULT=mssql_fetch_assoc($RESULTDS);

mssql_data_seek($RESULTDS,0);

echo "<CENTER>";
echo "<font size=2 color=#000000 face=arial>";
echo "Please select a stub to view:";
echo "</font>";
echo "</CENTER>";
echo "<BR>";

echo "<CENTER><font size=2 color=#0000ff face=arial>";

while ($RESULT = mssql_fetch_assoc($RESULTDS))   {

$endingdate = $RESULT['PSTUB5'];
$month = substr("$endingdate", -8, 2);
$day = substr("$endingdate", -6, 2);
$year = substr("$endingdate", -4, 4);




$Date = $month."-".$day."-".$year; 

echo "<a href=\"paystubs.php?Date=".$Date."\">";

echo $Date;

echo "</a>";

}

echo "</font> </CENTER>";

echo "<BR>";
?>

    Sorry, changed your date format to use "-" instead of "/" because you shouldn't use / in the URL unless you are using it correctly (for directories). To use the variable as a $_GET variable: once its been clicked on and the variable is added to the URL query string, use echo $_GET['Date'].

    I'm not sure how you are using advanced SQL and even some pretty non novice PHP functions and not be using forms correctly.

    If you want to still use a drop down menu, you (or I) can do so by using some javascript.

     

    BTW, why are you even bothering breaking the date up if your just going to put it back together?

     

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.