[Problem with str_word_count()]

I would do:

[code]<?php

$string = "This is sample string and th--is is another";
$string_formatted = str_replace("--", "", $string);

print_r(str_word_count($string_formatted));
?>[/code]

That's if I understood your question and needs properly.

[Need to add a field to PHP/HTML Form]

I am doing this right now, but I need to know a few things. What is the maximum length for the field and what type of field is it, (textarea, text, password, etc.)?

[** solved ** user profile]

What do you mean by yeah it would've? I have fixed the WHERE misspelling, and comment the code so you know what's going on. I have also changed the code a bit, because on a single profile page only one row would have to be returned.

[code]
<?php
// Include connect file
include("db_connect.php");

// Setup query and do query
$query = "SELECT * FROM users WHERE id = '{$id}' LIMIT 1"; // Putting your variables in { and } makes them better somehow, it's slipped my mind ATM, but I know it's like a safety net.
$result = mysql_query($query) or die(mysql_error());

// Count the number of rows returned
$count = mysql_num_rows($result);

// If the amount of rows returned is more than 0, continue.
if ($count > 0){
    // Put user info into array.
    $user = mysql_fetch_assoc($result);
    // Do stuff here....
} else {
    // Show error message
    echo "This user does not exist.";
    // Template stuff, if any.
   
    // Exit the script
    exit();
}

?>[/code]

[** solved ** user profile]

For your original example, you wouldn't put mysql_num_rows() in the while() part because it doesn't return an array or anything like that, just a number. For the second example by desithugg, that wouldn't work if my table key was p_id or something like that. Here is what I would use:

[code]
<?php
include("db_connect.php");

$query = "SELECT * FROM users WHRE id = '{$id}'";
$result = mysql_query($query) or die(mysql_error());

$count = mysql_num_rows($result);

if ($count > 0){
    while ($user = mysql_fetch_assoc($result)){
        // Do stuff here....
    }
} else {
    echo "This user does not exist.";
    // Template stuff, if any.

    exit();
}

?>[/code]

[[SOLVED] Show first few character of a string]

To return the first 9 characters:

[code]
$test = "This is the string";

$output = substr($test, 0, 8);

echo $output;
// This is t
?>
[/code]

It doesn't return "he" because PHP counts whitespace as a character.

EDIT: I accidentally pressed Preview so the reply never got sent. Also, you would use 0 and 8, not 9, because 0 - 8 is 9, 0 - 9 is 10.

[XML in PHP4]

Hello,

I am trying to parse an XML file for 3 of it's fields, but I have no idea how to. I know in PHP5 we have SimpleXML, but what can I use in PHP4? Below is an example of one of the items in the XML file. The fields in bold are the ones I need.

[QUOTE]
<item>
[b]<title>Gemineye - Break Neck Radio (The Afro-Takeover)</title>[/b]
−
<link>
http://podcast.digital-djs.com/imix/default.asp?id=11070
</link>
−
[b]<description>
01. (Intro) Spencer Doran - First of All 02. J. Dilla - Thunder 03. Blu - Soul Amazing (Soul Provider Remix) 04. Session - Freestyle 05. Jahi feat. Dwele - On My Own 06. Q-Tip feat. Andre 3000 - That's Sexy 07. (Break #1) Cyrus Tha Great - Island Music 08. Jay-Z - 99 Problems (DL's Firefly Remix) 09...
</description>[/b]
<pubDate>Mon, 31 Jul 2006 10:27:33 EDT</pubDate>
<source url="http://feeds.feedburner.com/DDJiMix">Digital-DJs.com (Podcasts)</source>
<bd:channelLink>http://www.digital-djs.com</bd:channelLink>
<enclosure [b]url="http://broadcast.underground-fusion.com:8080/iMixes/Gemineye/Break%20Neck%20Radio%20(The%20Afro-Takeover).mp3"[/b] type="audio/mpeg"/>
<media:content url="http://broadcast.underground-fusion.com:8080/iMixes/Gemineye/Break%20Neck%20Radio%20(The%20Afro-Takeover).mp3" type="audio/mpeg"/>
</item>
−
<item>
[/quote]


Thanks,
Michael Boutros

[Image Resizing Problem]

Anyone at all? Please?

[Image Resizing Problem]

It's because I am creating a script for users to upload and change their images, and yes, it is possible.

[Image Resizing Problem]

Hello everyone. I am writing a script to resize images, and am stuck with this problem. I know for a fact that the problem is NOT that any of the variables are unset, because I have checked and they are all set. If anyone could help I would be very appreciative :)

[code]// Make new image
$new = imagecreatetruecolor($new_width, $new_height);

// Create source image
if ($file_type_ns == "jpg" OR $file_type_ns == "jpeg"){
   $source = imagecreatefromjpeg("uploads/$file_name");
}
    if ($file_type_ns == "gif"){
   $source = imagecreatefromgif("uploads/$file_name");
}
    if ($file_type_ns == "png"){
   $source = imagecreatefrompng("uploads/$file_name");
}

// Build new image
imagecopyresampled($new, $source, 0, 0, 0, 0, $new_width, $new_height, $width, $height);

// Output new image
if ($info_type == "jpeg" OR $info_type == "jpg"){
  imagejpeg($new, "images/$file_name_ns.jpeg", $quality);
}

if ($info_type == "gif"){
  imagegif($new, "images/$file_name_ns.gif");
}

if ($info_type == "png"){
  imagepng($new, "images/$file_name_ns.png");
}[/code]

$file_type_ns: the file type from $_FILES with the suffix extension
$info_type: the new file type that a user has already chosen from a form

Once again, all these variables have already been set. I get no errors or anything, but when I go to look at the new image, it is the right size and type, but it is completely black. Thanks to anyone who can help :)