fugix

May 9, 2011

i found this code from a website...looks to be what you need...will need to change the query to NOT LIKE and the var values

$keywords= array("big","blue","sky");
$keycount= count($keywords);
for ($i=0; $i<$keycount; $i++)
{
$sql = "select * from ` domains ` where ` keyword ` like ' % " . $keywords [ $i ] . " % ' ";
$list = mysql_query($sql);
while ($lst = mysql_fetch_array($list))
{ ... }
mysql_free_result($list);
}

May 9, 2011

you might be able to call the plain array

if (preg_match_all($regexp, $html, $matches)) {
echo $matches;
}

but if not, then you might have to draw them out manually

if (preg_match_all($regexp, $html, $matches)) {
echo $matches[0][0];
        echo $matches[][1];
        echo $matches[1][0];
}

etc...

May 9, 2011

okay well thats good. Basically what you need is however you got the photoid for the pictures...i think you got it from you database...you need to use that same id as your action in your comment form....thats what im trying to do here....if you know what im talking about you will have an easier time doing it since you coded it

May 9, 2011

try changing your if statement to this

if (preg_match_all($regexp, $html, $matches)) {
echo $matches[0][0];
}

May 9, 2011

this article talks about using salt...hope its helpful...http://pbeblog.wordpress.com/2008/02/12/secure-hashes-in-php-using-salt/

May 9, 2011

yeah you could use a foreach() loop to pull out each key and trim each one...not sure if thats what you're looking for

May 9, 2011

try

$strQuery = "select * from Comments where Message LIKE '%$Bad_Words%' order by dtAdded desc";

May 9, 2011

im guessing that your message has multiple words..in this case it will still show everything....lets say your message is something like. "Test Message, crap".....that entire message does not equal "crap" even though it has the word in it...make sense?

May 9, 2011

Notice: Undefined index: submit in /home/ebermy5/public_html/form.php on line 9\

that error means that either $_POST['submit'] hasnt been submitted yet or doesnt exist...do what PFMaBiSmAd said and use isset()

May 9, 2011

i dont see why it wouldn't work, would add extra protection for sure

May 9, 2011

yeah, the query you are using now is efficient for you needs

May 9, 2011

sweet, welcome to phpfreaks!

May 9, 2011

firstly you need to wrap your names in quotes this

<select name=provincias>

should be

<select name="provincias">

May 9, 2011

hey im back...fell asleep last night. i wanted you to change the query code that i posted, for you image_show.php so you would have this

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>What do you know about.......</title>
<style type="text/css">
#What {
text-align: center;
}
</style>
</head>

<body>

<?php
$hostname='yoddce.com';
$username='yd29';
$password='Ad1!';
$dbname='d9';
$usertable='do';

mysql_connect('yourdddd!') or die(mysql_error());
mysql_select_db("d") or die(mysql_error());

$id = (int)$_GET['id'];
if( $id > 0 ) {
  $result = mysql_query("SELECT `name`, `photopath`, `state` FROM `photo` WHERE `id` = $id") or die(mysql_error());
  
  $row = mysql_fetch_array( $result );
  echo " <img src=\"/{$row['photopath']}\" height=\"400px\" width=\"350px\" />". "<br />
";
  echo $row['name']. "\n". "<br/>";
  echo $row['state']. "\n";
} else {
  echo 'BAD PHOTO ID';
}
?>
<?php
$hostname='dm';
$username='yd';
$password='d1!';
$dbname='d9';
$usertable='dto';

$myconn=mysql_connect('yddd!') or die(mysql_error());
mysql_select_db("dd") or die(mysql_error());

$name=$_POST['name'];
$comment=$_POST['comment'];
$submit=$_POST['submit'];
if($submit)
{
    if($name&&$comment)
    {
    $query=mysql_query("INSERT INTO comment (id,name,comment) VALUES ('','$name','$comment')");
    }
    else
    {
        echo "Please fill out all the fields.";
    }
}
$query_Recordset1 = "SELECT * FROM photo";
$Recordset1 = mysql_query($query_Recordset1) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
?>
<body>

<div id="comments">
<form action="image_show.php?id=<?php echo $row_Recordset1['id'];?>" method="POST">
<label>Name:  </label><br /><input type="text" name="name" input id="name"value="<?php echo "$name" ?>" /><br /><br />
<label>Comment:  </label><br /><input type="text" input id="comment"/><textarea name="comment" "cols="25" rows="7"></textarea><br /><br /><br />
<input type="submit" name="submit" value="Comment" /><br />

</form></div>
<hr width="1100px" size="5px" />
</body>
<?php

$query=mysql_query("SELECT * FROM comment ORDER BY id DESC");
while($rows=mysql_fetch_assoc($query))
{
     $id=$rows['id'];
     $name=$rows['name'];
     $comment=$rows['comment'];
     $linkdel="<a href=\"delete.php?id=" . $rows['id'] . "\">Delete User</a>";
     echo '<font color="red">Name:</font>  ' . $name . '<br />' . '<br />' . '<font color="red">Comments:</font>  ' . '<br />' . $comment . '&nbsp' . '&nbsp' .
      '&nbsp' . '&nbsp' .  $linkdel . '<br />' . '<br />' . 
     '<hr size="5px" width="500px" color="blue" />'  ;    
}
?>
</html>
</body>
</html>

May 9, 2011

yes that will need to be changed to, your form and header need to go to the dynamically created page, not the base page okay so change this

$query_Recordset1 = "SELECT `id`, `name`, `photopath`, `state` FROM photo ORDER BY `name` ASC";
$query_limit_Recordset1 = sprintf("%s LIMIT %d, %d", $query_Recordset1, $startRow_Recordset1, $maxRows_Recordset1);
$Recordset1 = mysql_query($query_limit_Recordset1, $myconn) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
?>

to

$query_Recordset1 = "SELECT * FROM photo";
$Recordset1 = mysql_query($query_Recordset1) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
?>

May 9, 2011

can i see the form that sends to this page too please

May 9, 2011

can i see your delete script please

May 9, 2011

it is now since i told you to change it, im perplexed as to why it is still not showing an id...what does you "view source" on your browser show the photo id being now?

May 9, 2011

well i know why your photo id is dissapearing, when you click on the photo link

<a href="image_show.php?id=<?php echo $row_Recordset1['id'];?>"><img src='<?php echo $row_Recordset1['photopath'] ;?>' height="100" width="100"/></a>

it is setting the url to image_show.php?id=<?php echo $row_Recordset1['id'];?/>

however, when you post a comment, you have the action set to just image_show.php

therefore when the form is submitted, you are going from a dynamic url to plain image_show.php, thus you lose your id

May 9, 2011

okay so you're going to try this

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>What do you know about.......</title>
<style type="text/css">
#What {
text-align: center;
}
</style>
</head>

<body>
<div id="What"><img src="images/what.png" width="477" height="36"  alt="do you know something i don't" /></div>
<?php
$hostname='fd.db.fdffd.d.com';
$username='d';
$password='d!';
$dbname='d';
$usertable='d';

mysql_connect('yd.com', 'yd9', 'Ad!') or die(mysql_error());
mysql_select_db("yd9") or die(mysql_error());

$id = (int)$_GET['id'];
if( $id > 0 ) {
  $result = mysql_query("SELECT `name`, `photopath`, `state` FROM `photo` WHERE `id` = $id") or die(mysql_error());
  
  $row = mysql_fetch_array( $result );
  echo " <img src=\"/{$row['photopath']}\" height=\"400px\" width=\"400px\" />". "<br />
";
  echo $row['name']. "\n". "<br/>";
  echo $row['state']. "\n";
} else {
  echo 'BAD PHOTO ID';
}
?>
<?php
$hostname='durce.com';
$username='yod9';
$password='Ad1!';
$dbname='yd';
$usertable='d';

mysql_connect('yd, 'yd9', 'Ad!') or die(mysql_error());
mysql_select_db("yd") or die(mysql_error());

$name=$_POST['name'];
$comment=$_POST['comment'];
$submit=$_POST['submit'];
if($submit)
{
    if($name&&$comment)
    {
    $query=mysql_query("INSERT INTO comment (id,name,comment) VALUES ('','$name','$comment')");
    }
    else
    {
        echo "Please fill out all the fields.";
    }
}
$query_Recordset1 = "SELECT `id`, `name`, `photopath`, `state` FROM photo ORDER BY `name` ASC";
$query_limit_Recordset1 = sprintf("%s LIMIT %d, %d", $query_Recordset1, $startRow_Recordset1, $maxRows_Recordset1);
$Recordset1 = mysql_query($query_limit_Recordset1, $myconn) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
?>
<body>

<div id="comments">
<form action="image_show.php?id=<?php echo $row_Recordset1['id'];?>" method="POST">
<label>Name:  </label><br /><input type="text" name="name" input id="name"value="<?php echo "$name" ?>" /><br /><br />
<label>Comment:  </label><br /><input type="text" input id="comment"/><textarea name="comment" "cols="25" rows="7"></textarea><br /><br /><br />
<input type="submit" name="submit" value="Comment" /><br />

</form></div>
<hr width="1100px" size="5px" />
</body>
<?php

$query=mysql_query("SELECT * FROM comment ORDER BY id DESC");
while($rows=mysql_fetch_assoc($query))
{
     $id=$rows['id'];
     $name=$rows['name'];
     $comment=$rows['comment'];
     $linkdel="<a href=\"delete.php?id=" . $rows['id'] . "\">Delete User</a>";
     echo '<font color="red">Name:</font>  ' . $name . '<br />' . '<br />' . '<font color="red">Comments:</font>  ' . '<br />' . $comment . '&nbsp' . '&nbsp' .
      '&nbsp' . '&nbsp' .  $linkdel . '<br />' . '<br />' . 
     '<hr size="5px" width="500px" color="blue" />'  ;    
}
?>
</html>
</body>
</html>

and let me know what you get

May 9, 2011

okay so let me get organized here when you click this link

<a href="image_show.php?id=<?php echo $row_Recordset1['id'];?>"><img src='<?php echo $row_Recordset1['photopath'] ;?>' height="100" width="100"/></a>

everything works as it should....the picture shows and everything, however when someone writes and comment and hits the submit button...the image does not show anymore?

May 9, 2011

and from where in your code is the

$id = (int)$_GET['id'];

coming from?

this line

$id = (int)$_GET['id'];

is causing all of your problems, if we can just get this sorted out you'll be good

May 9, 2011

maybe, found this tough.

May 9, 2011

il check my cPanel for it

May 9, 2011

sweet, what you were doing by have (!result) was telling it that if the query failed, to echo your 'Never achieved'. But even if the query returned 0 rows it still wasnt failing, so you would never see your empty query message..glad i could help

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fugix

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Posts posted by fugix

Query to Exclude Certain Results

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sha1(md5())

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sha1(md5())

Better type of query to use

VZdemon

what I´m doing wrong... It doesn´t work and don´t know why...

linking comment systen to template page

linking comment systen to template page

linking comment systen to template page

linking comment systen to template page

linking comment systen to template page

linking comment systen to template page

linking comment systen to template page

linking comment systen to template page

linking comment systen to template page

PHP copy

PHP copy

problem with data_seek

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