Unseeeen
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Posts posted by Unseeeen
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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.
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There can only be one match per person, but there will be multiple people.
My code works exactly how I'd like it. The result just annoys me though.
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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?
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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.
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read up on mod_rewrite in .htaccess files.
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Haha wow, I would of never saw that.
Thanks! That's what fixed it. :) -
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. -
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. -
Bingo!
Thanks a lot.
Also, how common is it for php short tags to NOT be useable? Just seems like all the hosts I've been on have had short tags enabled. -
Can anyone see what's possibly wrong? I've been stuck for a while now.
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Yeah I can show you, but only after I finish it up.
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Yes and yes. The part where the query result should be showing up is empty.
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Maybe try putting the
[code] $a = '1234.txt';[/code]
before the first if?
I think this is what you're trying to do...if id == 1, then display a link with a in it. -
Thanks, but I don't think I quite understand. I tried:
[code]$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? -
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] -
Hey thanks man, I appreciate it.
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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.
[SOLVED] How would I do this?
in PHP Coding Help
Posted
That's awesome. Thanks for all the help