PHP - Moved: Find The Second Hieghest Value In A Table

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Hello,
I want to know which is the table in the sql SELECT query.
What i have done to take fields :
Code: [Select]
$temp_sql = strtolower($sql);
$pos = stripos($temp_sql, 'select');
$str = substr($temp_sql, $pos);
$str_two = substr($str, strlen($start));
$second_pos = stripos($str_two, 'where');
$temp_fields = substr($str_two, 0, $second_pos);
$fields = trim($temp_fields);
But for table i don't know what to do. After the table can end the string or can be a WHERE or ORDER .
Is there any way to take it.
Until now i use
$temp = explode(' ',$sql);
$table= $temp[3];
Thank you

Ultimate goal is for the query to find the row in the database to avoid the validation and carry on with the rest of the script.

What actually happens is the script runs without any parse errors but it doesn't find the row within mysql.

$query = sprintf('SELECT
        name_id, title, description, price, width, height, depth, quantity
    FROM
        art
    WHERE
        product_code = "$u"',
mysql_real_escape_string($product_code, $db));
$result = mysql_query($query, $db)or die(mysql_error($db));

if (mysql_num_rows($result) != 1) {
    header('Location: home.php');
    mysql_free_result($result);
    mysql_close($db);
    exit();
}


I've echoed the where variable and the mysql_num_rows($result). the where variable is as expected but the result variable comes out as 0. I've checked my phpadmin and there is a entry in there. This worked fine before I sprinted the table. Any help appreciated.

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Hi guys I am new here

I have had a search around and can't really find out how to do this. Its quite difficult for me to explain and I am really new to PHP / mySQL - just starting out really. I hope to learn from doing small projects like this and seeing how it gets done.

So I have a mySQL DB with two different tables - one is called "notesnew" the other is called "users".

I have modified a common tutorial on how to make a "to do list" for what I am doing here.

Basically users can login and post messages on this list as them self. Each user's post appears as a list item, showing their username, the time they posted the message and the message content. The message content, time and owner of the message is stored in the "notesnew" table.

The list of user's, their passwords and their "avatar/profile" image URL location is stored in the "users" table.

Now I am having trouble getting each user's avatar image from being output next to each of their posts. The part that is confusing me is being able to match up the username's post with their content that is display - because I am posting the user's name along with their post, based on the logged in "session name" as their username. So basically, all the content being listed is being displayed from the "notesnew" table. I need to somehow figure out what each user's profile image is based on what is stored against their entry in the "users" table.

Here is the code I have so far. It is not working properly - it displays all content etc, but displays the avatar / image of the last user that has registered. Can anyone help me out here with some code that works, or pointers to try follow? I know the $finduserprofile part is incorrect in the way it goes through all user's and finds their profileimg, and that the image displayed next to each post is therefore incorrect (as it displays the last member to have registered's avatar, but its just that I don't know how else to do what I am trying to do.

Code: [Select]
<?php
//Connect to the database
$connection = mysql_connect('localhost', 'zzzz' , 'zzzzzzz');
$selection = mysql_select_db('zzzzz', $connection);

$username = $_COOKIE['ID_my_site'];

//Was the form submitted?
if($_POST['submit']){

//Map the content that was sent by the form a variable. Not necessary but it keeps things tidy.
$content = $_POST['content'];

//Insert the content into database
$ins = mysql_query("INSERT INTO `notesnew` (content, owner, dp_time) VALUES ('$content', '$username', NOW())");

//Redirect the user back to the members or index page
header("Location:index.php");
}

/*Doesn't matter if the form has been posted or not, show the latest posts*/

//Find all the notes in the database and order them in a descending order (latest post first).
$find = mysql_query("SELECT * FROM `notesnew` ORDER BY id DESC");
$finduserprofile = mysql_query("SELECT * FROM `users` ORDER BY id DESC");

//Setup the un-ordered list
echo '<ul>';

while($row = mysql_fetch_array($finduserprofile))
{
$imagelocation = $row['profileimg'];
//Continue looping through all of them
while($row = mysql_fetch_array($find))
        {

$owner = $row['owner'];
//For each one, echo a list item giving a link to the delete page with it's id.
echo '<li>' . '<img src ="' . $imagelocation . '">' . $row['owner'] . ' said at ' . $row['dp_time'] . ': ' . $row['content'] . ' <a id="' . $row['id'] . '" href="delete.php?id=' . $row['id'] . '"><img src="delete.png" alt="Delete?" /></a></li>';

}
}

//End the un-ordered list
echo '</ul>';

?>

Hi there

I'm trying to work out how I'd go about searching my database for URLs that have been submitted by users that are duplicates.
Now this isn't as simple as http://www.example.com/something.php and http://www.example.com/something.php oh no!

People have entered links such as http://www.example.com//something.php or http://www.example.com///something.php

The script (I didn't write it) currently see's these links as unique, but they're not. They function exactly the same way.

Anyone have any advice on how to weed these out?

I was thinking maybe a straight forward LIKE '%//%' in the mySQL syntax, but this will bring all URLs up due to the http:// aspect of the link

Any feedback would be appreciated!

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