Pass $row['date'] as a parameter to this date function and return the changed value.
Eg.

function changeDate($date) {
$newdate = date('d/n/y',strtotime($date));
return $newdate;
}

Call...

Great :) Cheers!

This is wrong. Try,

$query = "SELECT * FROM Employee WHERE $method = '$search'";
$result = mysql_query($query);

The only problem with this is , if a user loses his interest midway [ie., after submitting 1st form], there will be a obsolete record in the table. Or else, this is good.

You can do it in 2 simple ways (maybe there are even more, but I am not aware of it).
You can either save the data of the first form in session after it is posted, and then, when the second form is...

The echo statement is incorrect.

echo "<a href=\"inventorid.php?inventorid=".$data2['inventorid']."\">".$data2['inventorid']." ".$data2['firstname']." ".$data2['lastname']."</a>\n";

is the...

:) You forgot the column name as the 2nd parameter for mysql_result.
http://php.net/manual/en/function.mysql-result.php

Condition is a mysql reserved keyword. In order to use it, you have to use ` [don't confuse it for single quote]. ie., `condition`. In my opinion, use some other word instead of condition to avoid...

Yeah, ofcourse. :) $string is just a variable name. You can give any name to a variable.
See Php naming convention here (http://www.tizag.com/phpT/variable.php).

Clear @ before mysql_query and see if it throws any error. Also, you can have $result = mysql_query($query) or die(mysql_error()); to print the error when the query fails.

P.S. @ suppresses the...

Or simply, $string = implode(",",$_POST['Prefered_Engineers']);

Hmm.. You are right! That makes sense..

@cwarn23, I think the OP was saying, that it shows only 1 record instead of 2 !
@levsha,
Are you sure you have 2 records for that query ? If you have 2 records, it *should* display them. There is...

You are supposed to put mysql_fetch_array in a while loop.

while( $data2 = mysql_fetch_array($query)) {
echo $data2['firstname'];
echo $data2['lastname'];
echo $data2['task'];
}

I don't recommend using mysql_fetch_row though. This will return an array with numeric indices and the script will fail if you add a new column to your table.
I personally prefer using...

Probably you didn't get my point. I was trying to demonstrate the use of exit after header function. Even though it will throw an error, it will still print This data is supposed to be secure... ....

The exit function is used as a safety measure just in case the end user gets to echo something before header function (that will cause an error and header will not redirect).
Here is an example,
...

There are still some typos. $GET_['formfield'] instead of $_GET['formfield']. One simple suggestion though. You don't need a switch/if-elseif-else condition to find out the day. A simple array will...

Pretty simple. Fetch the data from the database and place it between <textarea> tags. Here is an example.

<?php
if(isset($_POST['save'])) {
print "<pre>";
print_r($_POST);
print "</pre>";...

Instead of storing the password just like a normal string, save it as an encrypted string, using SHA1 or something. This will provide you more security as well as solves your issue.
Here is an...

What is the problem ? You would be selecting CourseName and courseID. So, when you post the form, You will have "MATHS" in $_POST['CourseName'] and "101" in $_POST['courseID']. Then it is all about...

:) Thanks for waking me up. That is absolutely right.

Can you post your code ? (and print the variable $sql and post it here ?)
I guess the variable $table is empty or something.

:) Wohoo! Good luck..

Are you using Linux server ? If yes, then make sure to check the case (lowercase/uppercase) of table and column names. Linux is case-sensitive. :)
Did you try printing out the query and executing...

Hmm.. I tried your code snippet and it worked fine. There is one thing that I'd always do after having a header, an exit. This might or might not fix the problem, but try having an exit after...

Try <?php ob_start(); ?> to turn on output buffering (should be the 1st line of your code) and <?php ob_flush(); ?> as the last line to flush the buffered output.

1. Do you have mysql_connect ?
2. Are you selecting the database ?
3. Does that table exist ?
4. Are you sure the query is correct ? Ie., echo the query and test it in phpmyadmin/mysql console. ...

Ah! I see.. Happy learning :) Cheers!

<label>Content: <textarea name="Content" /><?php echo $row['Content'];?></textarea></label><br />

Pretty simple ? eh ? :)

http://www.tizag.com/mysqlTutorial/mysqlfetcharray.php
You should use the array returned by mysql_fetch_array and not the result resource itself. ie., $row['content'], $row['name'] etc and not...

Read this FAQ. (http://www.daniweb.com/forums/thread191031.html)

mysql_num_rows (http://in2.php.net/mysql_num_rows) will tell you how many records were returned by the query.

Which query are you exactly using ? Because, earlier you had posted

//wrong query, quotes are all mixed up
$query = 'SELECT * FROM t_master_reg dt_reg_reminder >= '$today' AND dt_reg_end...

Did you print the query and execute it in mysql console or phpmyadmin ? Does that return any records ?
If Yes, then it should work in this case too.
If No, then you don't have any records in the...

Print out the query, execute it in phpmyadmin, Or, simply add, or die(mysql_error());
ie.,

// sent from form
$textbox_value = ($_POST['code']);
$checkbox = ($_POST['checkbox']);...

I don't get it ! If you have the entire address in the table, why not simply use,
echo "<img src='".$row['image']."'>"; ?

What does $row['image'] print ? If you are in the subdirectory and if you want to print an image in the main directory, use .. .
ie., ../imagename.jpg

Instead of having different names for your checkboxes, use a checkbox array and on submit, loop through all the selected checkbox and insert a record to the table.
For eg.,

<form name="addcode"...

I checked your script.
is wrong. If you want to have a hidden element with php variable's value, you should do,

<input type="hidden" name="autoid" value="<?php echo $autoid; ?>">

Secondly,...