Hello all,
I am looking for help on how to code this:
I would like to have a page where there is a box where registered users can type something then on pressing the button a new page is created displaying the text they entered and boxes for other people to type into.
Thanks in advance.
Jump to PostAre you looking for a commenting system?
Jump to PostPlease post some code so that users can help guide you. Don't expect anyone here to do your job for you =)
Jump to PostSet up the DB - for a simple system with tags for posts (pages):
posts (pages)
post_id (PK)
post_title
post_body
user_id (FK)
post_timestamptags
tag_id (PK)
tags_stringpost_tags (link table)
post_id (FK)
tag_id (FK)comments
comment_id (PK)
user_id (FK)
post_id (FK)
comment_timestampYou have just …
Jump to PostCan anyone provide some code for a basic version?
That's what Google is for. If you're stuck on a particular part of the code, paste what you have and we'll help. The purpose of this forum is not to be a 'rent-a-sap' site, rather to help users with their own …
Jump to PostI posted my code.
Yes you did, but you also asked for somebody to provide you with a basic system.
RE: your code, use mysql_insert_id() to get the last added id - no need to do a SELECT.
Also you don't clean your input, so any single quotes …
Are you looking for a commenting system?
Please post some code so that users can help guide you. Don't expect anyone here to do your job for you =)
Yeah, I suppose a comment system where a page is created with information already entered and boxes for people to comment/post in.
I want the new page to last forever, or until it is deleted from the database. It'd be nice if the url ended with an id number so the pages can be found.
I don't have any code to show before I don't know where to start or even what to do. I'm not asking for the full system, just code snippets or whatever.
I've attempted it but my code doesn't seem to work - throws back HTTP Error 500 (Internal Server Error): An unexpected condition was encountered while the server was attempting to fulfill the request. Can anyone take a look? Sorry for double posting.
<html>
<head>
</head>
<body>
<?php
if (isset($_POST['mytext']) {
$myt = $_POST['mytext'];
$mymysql = mysql_connect("host","user","pass") or die ("can't connect");
mysql_select_db("dbname", $mymysql );
$sql = "INSERT INTO content (content) VALUES ('$myt')";
mysql_query( $sql, $mymysql );
$sql = "SELECT * FROM content WHERE content = '$myt'";
$row = mysql_query( $sql, $mymysql );
$myid=$row['id'];
mysql_close ($mymysql);
?>
Your text has been added. Your link:<br>
<a href="http://website.com/link.php?id=<?php print ($myid) ?>"> Link</a>
<?php } else { ?>
<form action="blah.php" method="POST">
<input type="text" value="type your text here" name="mytext">
<input type="submit">
</form>
<?php } ?>
</body>
</html>blah.php = itself
I've coded link.php which seems to work.
Set up the DB - for a simple system with tags for posts (pages):
posts (pages)
post_id (PK)
post_title
post_body
user_id (FK)
post_timestamp
tags
tag_id (PK)
tags_string
post_tags (link table)
post_id (FK)
tag_id (FK)
comments
comment_id (PK)
user_id (FK)
post_id (FK)
comment_timestamp
You have just one page for displaying posts, lets call it posts.php
You then access the relevant post record with an 'id' parameter in the querystring:
<a href="posts.php?id=7">How to create a DB in mysql</a>Those are the bare bones of the system IMO.
I'm having trouble creating the custom link and displaying the information. Can anyone provide some code for a basic version?
Can anyone provide some code for a basic version?
That's what Google is for. If you're stuck on a particular part of the code, paste what you have and we'll help. The purpose of this forum is not to be a 'rent-a-sap' site, rather to help users with their own code. If it's a trivial thing, well fair enough, but a full commenting system is not a trivial thing.
I suppose a pro could knock one up in about 20 minutes or less, but again that's not the purpose of this forum.
I posted my code.
I posted my code.
Yes you did, but you also asked for somebody to provide you with a basic system.
RE: your code, use mysql_insert_id() to get the last added id - no need to do a SELECT.
Also you don't clean your input, so any single quotes in the $myt will kill the query.
It still doesn't seem to show new link, create a new page and therefore show the text. I posted my code below.
example.php
<?php
if (isset($_POST['mytext'])) {
$myt = $_POST['mytext'];
$mymysql = mysql_connect("","","") or die ("can't connect");
mysql_select_db("", $mymysql );
$sql = "INSERT INTO content (content) VALUES ('$myt')";
mysql_query( $sql, $mymysql );
$sql = mysql_insert_id();
$row = mysql_query($sql, $mymysql);
$myid=$row['id'];
mysql_close ($mymysql);
echo "<a href='http://site.com/link.php?id='.$myid>Link</a>";
}
?>link.php
<?php
if (isset($_GET['id'])) {
$myid = $_GET['id'];
$mymysql = mysql_connect("","","") or die ("can't connect");
mysql_select_db("", $mymysql );
$sql = "SELECT * FROM content WHERE id = '$myid'";
$row = mysql_query( $sql, $mymysql );
if ($row) {
print ($row['content']);
}
mysql_close ($mymysql);
}
?>Bump :o
mysql_select_db("", $mymysql );
$sql = "INSERT INTO content (content) VALUES ('$myt')";
mysql_query( $sql, $mymysql );
$id = mysql_insert_id();
$myid=$id;
mysql_close ($mymysql);Thanks, now it creates new url. However, the text doesn't show on the page. How would I go about showing it?
So you get the link: http://site.com/link.php?id=7 or similar.
Your code in the link.php is buggered. Is that right?
$sql = "SELECT * FROM content WHERE id = '$myid' LIMIT 1";
$row = mysql_query( $sql, $mymysql );
if (mysql_num_rows($row)>0) {
$d = mysql_fetch_array($row);
echo "Content exists:<br />";
echo $d['content']);
}else{
echo 'No content for this ID';
}Just tried that code for link.php and it displays an error.
<?php
if (isset($_GET['id'])) {
$myid = $_GET['id'];
$mymysql = mysql_connect("","","") or die ("can't connect");
mysql_select_db("", $mymysql );
$sql = "SELECT * FROM content WHERE id = '$myid' LIMIT 1";
$row = mysql_query( $sql, $mymysql );
if (mysql_num_rows($row)>0) {
$d = mysql_fetch_array($row);
echo "Content exists:<br />";
echo $d['content']);
}else{
echo 'No content for this ID';
}
}
?>Are you going to share the error message or am I supposed to guess...:)
Although it could be this:
echo $d['content']);change to this:
echo $d['content'];Solved! Thanks!
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