[Obtaining Values Between Characters]

need to add an underscore to your char class [a-zA-Z0-9_]

Thanks a lot, that worked . (off topic, but I remember you from when I frequented phpfreaks ^^)

[Obtaining Values Between Characters]

As I am not skilled in Regex, I am having trouble with this line of code. I use this:

preg_match_all('/f_[a-zA-Z0-9]*_/', $string, $match);

to locate the string (including f_ and _). For example, I have a function named f_get_value_ and f_set_value_ and I would like to locate these function names. However, when I do print_r($match[0]), it returns with:

Array
(
    [0] => f_set_
    [1] => f_get_
)

Obviously it is cutting it off at the first underscore it comes across (after f_) rather than what I want it to do (cut it off at the final underscore). I cannot guarantee that there will be only 1 underscore between the words (get/set_value). There could be 2, 3, 4, or more. I am looking for a regular expression that will accomplish this goal and I thank anyone who attempts to help.

[help in this line ....]

You didn't use $ before the variable names... and I recommend using brackets when directly using variables within double quotes:

for ($i=1; $i<11; $i++)
  {
     echo "<tr>";
    echo"<td><p><strong>[{$i}]</strong></p></td> ";
    echo"<td><textarea name='causes[{$i}]' cols='30' id='causes' value='{$causes[$i]}'></textarea></td>";
    echo"<td><textarea name='corrective_action[{$i}]' cols='50' id='corrective_action' value='{$corrective_action[$i]}'></textarea></td>";
    echo"<td><input name='name[{$i}]' type='text' id='name' value='{$name[$i]}' /></td>";
    echo "<td><input name='date[{$i}]' type='text' id='date' value='{$date[$i]}' /></td>";
    echo "<td><textarea name='remarks[{$i}]' cols='50' id='remarks' value='{$remarks[$i]}'></textarea></td>";
   echo"</tr>";
   
   }

 

I guess you already solved this though? o.O

[New conversion help please]

You're missing a double quote in your array:

15 => "fiften):

to

15 => "fiften"

[date() question]

Something like:

<?php
$thisYear = date("Y", time());
$i = 0;
do
{
     echo intval($thisYear)+$i;
     $i++;
}
while(true);
?>

[help with something i cant figure out]

I'm thinking it's this line:

$resultv = mysqli_query($cxn,$sqv)
             or die (mysqli_error($result)); 

 

Try changing it to:

$resultv = mysqli_query($cxn,$sqv)
             or die (mysqli_error($cxn)); 

[Help with invalid argument with foreach]

See what happens when you do (instead of the foreach):

print_r($_POST['manager']);

[Sorting Order Drop Down Menu]

You could try something like:

Sorting Order</span><br>
                <select name="sort" class="textBox_center" id="sort" style="width:150px;">
                <option value="1" onclick="newestentries.display='block';youngesttrees.display='none'">Newest Entries</option>
                <option value="2" onclick="newestentries.display='none';youngesttrees.display='block'">Youngest Trees</option>
              </select>

 

And have two sections that would correspond to this:

<div id="newestentries" style="display: none">
<?php
      $result = mysql_query("SELECT * FROM album ORDER BY dtime DESC",$connect);
      while ($row = mysql_fetch_assoc($result))
      {
            echo(" <a href=\"image_script.php?albumid=" . $row["albumid"] . "\">" . $row["title"] . "</a><br />");
      }
?>
</div>
<div id="youngesttrees" style="display: none">
<?php
      $result = mysql_query("SELECT * FROM album ORDER BY age DESC",$connect);
      while ($row = mysql_fetch_assoc($result))
      {
            echo(" <a href=\"image_script.php?albumid=" . $row["albumid"] . "\">" . $row["title"] . "</a><br />");
      }
?>
</div>

 

Not sure if that's what you're looking for though.

[POLL: Pick which php bbcode button looks better]

I chose right, but I don't think a button is needed - it's just as easy, if not easier, to type in the bbcode manually.

[PHP Upload Image Turns Into Gibberish]

Can't think of anything else, but I recommend you check your personal messages. Also, there is a "Solved" button somewhere around your topic you can press xD. I think it's at the very bottom, but I haven't posted any topics too recently so I don't remember.

[PHP Upload Image Turns Into Gibberish]

Ah, I meant the actual data itself (like the base64_encrypted data). Right now you could try using this code instead of the old one:

(below)

 

Try creating another album and see if you get the same little icon (and check if the information is different in the table for the two different images).

 

EDIT:

Actually, after looking at it, this ($result = mysql_query("select * from album where albumid='".addslashes($image).".jpg'") would not work if your "albumid" is an auto_incrementing integer. Try using these:

<?php
include("config.php");

$image = stripslashes($_GET['image']);
$result = mysql_query("select * from album where albumid='".
addslashes($image)."'");
$myrow = mysql_fetch_assoc($result);
$imagebytes = $myrow['imgdata'];
header("Content-type: image/jpeg");
print base64_decode($imagebytes);
?>

(the display script):

<?php

include("config.php");

        $albumid = $_GET['albumid'];
        
        $result = mysql_query("SELECT * FROM album WHERE albumid='$albumid' ",$connect);
        while($myrow = mysql_fetch_assoc($result))
             {
               echo "<b>Title: </b>";
               echo $myrow['title'];
               echo "<b><br>Posted: </b><i>";
               echo $myrow['dtime'];
            echo "</i><b><br>Age (Years):</b>: ";
               echo $myrow['age'];
               echo "<br><br><a href=\"javascript:self.history.back();\"><-- Go Back</a>";
            echo "<b><br><br><br><br>Image: <br><br></b>";
            echo "<img src=\"get_image.php?image={$myrow['albumid']}\">";
             }
          
?>

[PHP Upload Image Turns Into Gibberish]

I was getting confused on your display image script (because I thought it actually took image data through the url, which couldn't have worked), but basically what premiso said is correct. Not sure why it outputs a small icon :\ Can you PM or post some of your data in the album table?

[PHP Upload Image Turns Into Gibberish]

For uploading image data, try using base64_encode (on the image data before you insert it into the database) and to display it use base64_decode.

[PHP Created Forums Question]

I agree with thorpe...

 

Have you already implemented a commenting system and want to be able to somehow tell the forum that a new reply has been posted, or are you having trouble creating a script that allows users to reply to threads?

[Prevent SQL Injection attacks!]

You could replace:

$username=$_POST['username'];
$password=$_POST['password'];

 

With:

$username=get_magic_quotes_gpc() ? stripslashes($_POST['username']) : $_POST['username'];
$password=get_magic_quotes_gpc() ? stripslashes($_POST['password']) : $_POST['password'];

To make sure slashes have not already been added. Also, you forgot to put $ before your variable name for the "query" variable.

[Array (Duplicate value from other value)]

Place all of the longer strings before the shorter ones in the array. Or:

<?php
function cmp($a, $b)
{
    return strlen($a)-strlen($b);
}

$array = array(""); //original array
usort($array, "cmp");
$array = array_reverse($array); //new organized array
?>

[[SOLVED] password is not being accepted using SHA]

It would help with the actual code, seeing how "SHA" isn't a function and, assuming you meant sha1, executing sha1('$p'); would return "fc23764ac5b792f40bb1a00c0e3284e45f3f49c0".

 

 

[Clueless]

You would add a closing brace right before you close the mysql connection (after you use "include("registered.php")").

[Display a default image]

What do variables $myrow and $field1_name contain?

 

<?php
    if( file_exists( "thumbs/" . $myrow[$field1_name] ) )
        {
                echo( "<img src=\"./thumbs/" . $myrow[$field1_name] . "\" />" );
        }
    else
        {
                echo( "<img src=\"./thumbs/nophoto.jpg\" />" );
        }
?>

 

This should work if the file doesn't exist and the "./thumbs/nophoto.jpg" exists, although this code is basically the same as what mrMarcus suggested (except I use ./ before the location).

[Read from forum]

Without direct access (like, if the script was on their server), then you would have to first access the page (via file_get_contents, fopen/fread, cURL, etc.) and then parse the page to get all of the links to the threads; afterwards, you would have to access all of those links (through a loop with your preferred method) and parse the information from that.

[Help with lottery style system?]

Why in the world does this have to do with cron jobs?

 

You mean besides him mentioning how he is "...still new to cron jobs but will have to figure out how to set them up to draw the numbers every hour" in his original post? I have no clue.

[How to file_get_contents when login required?]

Make sure the cookie file is being created - and I believe you need to "restart" cURL (initiate it again) after you execute it.

[How do I get the row number from from an sql table query]

Well you could use LIMIT. For example, if you started on page one, a previous button would be disabled and the next button would go to page two (assuming there is more than one "page"). On page two, the previous button would be enabled. Obviously, next would add one to the page number and previous would subtract one. Lets say you want to show 3 results per page:

 

<?php
//mysql connect, select db, queries done, basically the script you had before excluding outputting the information

$page = intval($_GET["page"]);
$total = mysql_num_rows($query);
$increment = 3;
$current = ($page-1)*$increment;
$previous = false;
$next = false;

if ($current > 0)
{
	$previous = true;
}

if ($current < $total)
{
	$next = true;
}

if ($previous)
{
	//display previous, $page -1
}

if ($next)
{
	//display next, $page +1
}

$sql = "SELECT * FROM table LIMIT {$current},{$increment}";
$query = mysql_query($sql);
$numrows = mysql_num_rows($query);
if ($numrows > 0)
{
	while ($rows = mysql_fetch_array($query))
	{
		//output the information or something
	}
}
else
{
	die("Page information not located");
}
?>

 

I did that pretty fast, so replace the comments with whatever is needed. The main thing you need to look at is SQL and the variables at the top. If, after the comments are replaced, the script doesn't work (through an error of my own), please note that I did not test it, but I'm pretty sure there shouldn't be any errors.

[Prevent SQL Injection by escaping?]

I the injection would have not been "1' OR '1=1", but rather:

1' OR '1'='1

Resulting in an unescaped result of:

SELECT * FROM login WHERE username = '1' AND password = '1' OR '1'='1'

 

Escaped:

SELECT * FROM login WHERE username = '1' AND password = '1\' OR \'1\'=\'1'

 

Try your example again with:

username: a username that works (I guess you used "1" before)

password: 1' OR '1'='1

[How do I get the row number from from an sql table query]

That looks like what I need

 

Is the $row['id'] generated automaticaly by the query

 

You would have to create your own id row, but you could make it auto-increment.