jeff5656

Actually that doesn't work. The $_SERVER['DOCUMENT_ROOT'] that I get back is not the actual location where my index file is. I think this link explains why: http://www.helicron.net/php/ Perhaps my host is not using apache (since above only works in apache). In any case, is there a way to solve this with relative paths? I thought that if I was in a subdirectory and I want to point to another subdirectory I would point to include "../anotherdirectory/filename.php"; but that doesn't work. Obviously the best solutions would be some way to automatically determine the root directory (regardless of whether the host has apache). Any ideas?

Having problems pointing to where files are. Is this the root?: include "../connectdb.php"; That file is locate din the root. If I am already in the root when I include that statment, the file isn't found. If I use the above code while I am in the subdirectory hello, (root/hello), it works. I use this in every page, so I think I need a way to point to the absolute root, not relative.

Oh I wasn't familiar with mysql SUM(). I looked it up and the following code works perfectly thanks! $query = "SELECT user_id, SUM(num_hours) FROM timesheet GROUP BY user_id"; while($row = mysql_fetch_array($result)){ echo "Total ". $row['user_id']. " = ". $row['SUM(num_hours)']; echo "<br />"; }

I want to add up all the hours in a field called "num_hours" When I run this code I get a result of 8, even though if you add up the hours it comes to about 20. Interestingly, there are 9 records - so it seems like my code is acting like mysql_num_rows()! $q = "select num_hours from timesheet "; $results = mysql_query ($q) or die (mysql_error()); $invarray = mysql_fetch_array ($results); echo "tot hours =".array_sum($invarray);

Wow you are smart. The error was stuck inside the dropdown in HTML so I never saw it untik I viewed the source. Very clever of you :-) <select name="state">Unknown column 'states' in 'order clause' I turns out the fieldname was "name" not state. Thanks!!

I have a table called states, but when I do a while loop to populated a dropdown, the dropdown has no content. I use this code for other dropdowns with no problems. Any ideas? <select name="state"><?php $q = "SELECT * FROM states ORDER BY states"; $result=mysql_query($q) or die (mysql_error()); $num_svcs=mysql_num_rows($result); while ($row = mysql_fetch_assoc ($result)) { ?><option value="<?php echo $row['abbrev'];?>"><?php echo $row['name'];?></option> <?php }?> </select>

Aha...I see thank you!

I get Parse error: syntax error, unexpected T_STRING in with this code. (Line 20 refers to the create table line) I copied this right from http://snipplr.com/view.php?codeview&id=6637 and there are no comments pointing out any errors, so I am at a loss. here is the code (I have always created tables right in phpmyadmin, so it would be great if I could do it this way because then I can modify the code as I create new tables): CREATE TABLE IF NOT EXISTS states ( id INT NOT NULL auto_increment, name CHAR(40) NOT NULL, abbrev CHAR(2) NOT NULL, PRIMARY KEY (id) ); INSERT INTO states VALUES (NULL, 'Alaska', 'AK'); INSERT INTO states VALUES (NULL, 'Alabama', 'AL'); and at the end of all those states: INSERT INTO states VALUES (NULL, 'Armed Forces Middle East', 'AE'); INSERT INTO states VALUES (NULL, 'Armed Forces Pacific', 'AP'); ?>

I get the above error with my code. The error says line 147m which is the 1st line shown below: $qq "select service from services where id = '". $userarray['id'] ."' LIMIT 1"; $results2 = mysql_query ($qq) or die (mysql_error()); $qqr = mysql_fetch_assoc ($results2);

I get the above error (Column count doesn't match value count at row 1) with this code: $query = "INSERT INTO shop (id,store, item, priceunit, units, unitname, dateadded) VALUES ('','$store', '$item', '$priceunit','$units', '$unitname' '$dateadded')"; echo $query; mysql_query($query) or die("<br>error. ".mysql_error()); here's my table: CREATE TABLE `shop` ( `id` int(10) NOT NULL auto_increment, `store` varchar(40) NOT NULL default '', `item` varchar(60) NOT NULL default '', `units` varchar(20) NOT NULL default '', `priceunit` varchar(20) NOT NULL default '', `dateadded` date NOT NULL default '0000-00-00', `unitname` varchar(40) NOT NULL default '', PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=1 ;

Trying to learn about functions. Before I call a user-created function, I have to read that function correct? So if I have a file called "functions.php" that lists a bunch of functions, do I have to put include ../functions.php at the beginning of *every* page that I will be calling a function? Or can I include the functions.php in just the index.php page? That seems a little tedious I guess, but I want to do what the standard is for reading-in the user-created functions. Because php's built in functions are somehow read without having an include file, so is there a way to treat my created functions the same way? Thanks!

I am about to take the plunge and learn to write some functions to make my coding easier. However, what I can't seem to find in any of the tutorials is *where* do I store these functions and when do I run them? DO I have a file called functions.php and then put an include line in every single php page I have? I that was the case, it would kind of negate the advantgae of having a function - so I think I am not "getting" something. I figured I would just ask the experts :-)

ok thanks. I use the same process page for different forms (just different if statements to direct to correct section), so I may need to use sessions.

If I have a form, and I have aform processing page like process.php and let's say I want to redirect back to the form if there is an error (i.e. passwords don't match). How can I redirect back to the form and retain the contents of the other fields, so the user doesn't have to type it all over again? BTW I know I can do this with java on the actual form.php page, but was wondering if there was a PHP solution)

"SELECT table1.field as field1, table2.field as field2 FROM ........" Is there any way to do that and still use the wildcard *? I have many fieldnames to selct from both tables and don't want to have to write them all out...

if I join two tables and one of the fieldnames is the same in both tables, how do I echo out the correct one? as an example, $query = "SELECT * FROM food, more_food "; $results = mysql_query ($query) or die (mysql_error()); while ($row = mysql_fetch_assoc ($results)) { Now the following "made sense", but it doesn't work. How do i speciy the variable from the correct table if the ieldnames are the same in both tables? echo "the type of apple is $row['food.apple'];

Great, thanks guys.

Hey now it says equal! Thanks. But why does that work. Was there an invisible carriage return in one?

I want to compare something in the database with a POSTed string to see if they are equal: $query = "SELECT * FROM `testbase` WHERE `id` = '".$id."' "; $result = mysql_fetch_array(mysql_query($query)) or die(mysql_error()); echo $result['t3']; echo "<br>"; echo $_POST['t3']; When I do that I get this: WBC 7 Hg 7 INR: 1.4 Cx: WBC 7 Hg 7 INR: 1.4 Cx: They look the same right? But when I do this, if($result['t3'] == $_POST['t3']) { echo "they are equal" } else { echo "Not equal" } I get Not equal. But they sure look equal!! Any ideas??

My server is 3 hours behind in another time zone. Does anyone know how I can adjust this code so that $d is +3 hours ahead? $d= date("Y-m-d H:i:s"); Echo $d;

I apologize I could not undertand how to use the first link: for($count=0;$count<$fields;$count++) { $field = mysql_fetch_field($result,$count); echo "<p>$field->name $field->type ($field->max_length)</p>"; and apply that to my code. what is "field ->type"?? If I have $row['f13'] how do I see what data type that is? Yoir send link: mysql_field_type(data,field_offset) What is a field offset? I do not know what number to start at.

$query = "SELECT * FROM `people` WHERE `s_status` = 'a' order by $srt"; $results = mysql_query ($query) or die (mysql_error()); while ($row = mysql_fetch_assoc ($results)) { echo $row['f13'] . "this field type is"."?????"; Where the ???? are in the code above, what do I put there to echo the filed type (i.e. "varchar" or "Date", etc?

Oh is that all? That's easy. Thanks!

I have names called f1_name, f2_name all the way to 17 (I get these from a queried database). I want to write a for loop so I don't have to write them all out in a form. Here is my code. Look at this array: $row['f1_name'] <?php for ($f=1;$f<18;$f++){ ?> <td bgcolor="#FFFFFF">F<?php echo $f;?></td> <td > <input name="f<?php echo $f.'_name';?>" type="text" size="20" value="<?php echo $row['f1_name']; ?>" /> </td> </tr> <?php } ?> How do I fix that so that the For-loop inserts the correct number for $row['fx_name'] (where x is $ff)? Hope that's clear! :-)

ahhh. Thanks!