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Unseeeen

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Posts posted by Unseeeen

  1. I get an error with that...

     

    Parse error: syntax error, unexpected T_LNUMBER, expecting T_VARIABLE or '$'

     

    I don't really understand what that's trying to do anyways.

     

    But what I mean by it annoys me, the result is in an array. I don't want it in an array.

  2. Thanks guys, that helps a lot. Here's my code...

     

    <form action="" method="post">
    <textarea name="first"></textarea><br>
    <textarea name="second"></textarea>
    <input value="Submit" type="submit">
    
    </form>
    <?
    $first = $_POST['first'];
    $second = $_POST['second'];
    
    
    $blah = explode(" ", $first);
    $blahtwo = explode(" ", $second);
    
    $result = array_intersect($blah, $blahtwo);
    print_r($result);
    ?> 

     

    It's simple, the way I like it. Anyways, how would I echo the info without having that "Array ( [0] => joe )" thing at the end?

  3. So basically I have 2 lists of information.

     

    ex:

    list 1- bob, joe, dan, billy

    list 2- paul, joe, jim, mike

     

    After putting both of those lists into 2 text boxes, how would I make it so only the common ones show up? In this case joe would be the only name that should show up.

     

    Thanks.

  4. v3x: That didn't fix the error.

    phpstuck: the reason why I have that if there is because that's not all of it...I made it so if there's a post, basically send it to the database, else post the form. I didn't bother posting the form because I didn't feel it was needed and would just make my post longer.
  5. I honestly cant find anything wrong with this script, other than it's extremely insecure (and I'll fix that later [img src=\"style_emoticons/[#EMO_DIR#]/unsure.gif\" style=\"vertical-align:middle\" emoid=\":unsure:\" border=\"0\" alt=\"unsure.gif\" /])

    [code]<?php
    $dbconnect = mysql_connect ("localhost", "***", "***") or die ('I cannot connect to the database because: ' . mysql_error());
    mysql_select_db ("***");
    $title = $_POST['title'];
    $author = $_POST['author'];
    $news = $_POST['news'];
    $submit = $_POST['submit'];
    $time = date('D M j');

    if(isset($submit))
    $sql = "INSERT INTO news ( id, title, newsbody, username, time ) values ( 'NULL', '$title', '$news', '$author', '$time' )";
    mysql_query ($sql) or die(mysql_error();
    echo "News successfully submitted!";
    } else {
    ?>[/code]

    I get a unexpected ';' error on line 12, and thats

    [code]if(isset($submit))[/code]

    I'm pretty sure I used isset() correctly?

    I was thinking there could be an error on the next line...but I've been trying everything I can think of and it just won't work

    Can anyone see what's wrong with it? Thanks.
  6. Thanks, but I don't think I quite understand. I tried:

    [code]$query = "SELECT * FROM news ORDER BY id DESC LIMIT $frontpagelimit";
    $result = mysql_query($query);
    $row = mysql_fetch_object($result);
    while ($row = mysql_fetch_assoc($result))
    {
    ?>
    <center><table class="news">
    <tr><td><b>Title</b>:<i><? echo $row->title; ?></i></td></tr>
    <tr><td><? echo $row->newsbody;?></td></tr>
    <tr><td><b>Posted by:</b> <i><? echo $row->username; ?></i> <b>on</b> <i>Jan 1st, 2006</i> <b>at</b> <i><? echo $row->time; ?></i></td></tr>
    </table>
    <?
    }
    ?>[/code]

    But nothing shows up? What's wrong?
  7. I have another question...

    My code, will only display one result from the database, but I'm trying to get it to display 5. Here's my code...once again, I dont see what's wrong with it.

    [code]<?
    include 'functions.php';
    $query = "SELECT * FROM news ORDER BY id DESC LIMIT $frontpagelimit";
    $result = mysql_query($query);
    $row = mysql_fetch_object($result);
    ?>
    <center><table class="news">
    <tr><td><b>Title</b>:<i><? echo $row->title; ?></i></td></tr>
    <tr><td><? echo $row->newsbody;?></td></tr>
    <tr><td><b>Posted by:</b> <i><? echo $row->username; ?></i> <b>on</b> <i><? echo $row->date; ?></i> <b>at</b> <i><? echo $row->time; ?></i></td></tr>
    </table>[/code]
  8. Well, I haven't done anything with php or sql in a while and I just had a simple question...

    I'm making a very simple news script and right now all I want to do is insert & display results to the database.

    Here's my code so far.
    [code]
    <?
    $frontpagelimit = "5";
    $dbconnect = mysql_connect ("localhost", "user", "pass") or die ('I cannot connect to the database because: ' . mysql_error());
    mysql_select_db ("db");
    $query = "SELECT * FROM news ORDER BY id DESC LIMIT $frontpagelimit";
    mysql_query($query);
    $row = mysql_fetch_object($query);
    echo $row->newsbody;
    ?>[/code]

    What I'm trying to do at this stage is get the newsbody that has already been inserted.

    I get a
    [!--quoteo--][div class=\'quotetop\']QUOTE[/div][div class=\'quotemain\'][!--quotec--]Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource[/quote]
    on line 7 error.

    From what I remember...this should be right. What's my problem? And how, once I've grabbed everything, display more than one news insert at a time? I'm thinking with maybe an array, but how would I do that?


    Thanks.
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