Congrats on solving your problem. I am glad you learnt something new :)

If you post your answer, even others may get to learn something from this thread.

Seems like you didn't read the member rules. (http://www.daniweb.com/forums/faq.php?faq=daniweb_policies)
Ch, ch. Read more on count and group by clauses. After reading them, you will be able to...

Dude, datatype as in int, varchar, date, datetime, etc.

ALTER TABLE clown_contest MODIFY COLUMN `location` varchar(40) AFTER `state`

Clear ? :)

You forgot the datatype! (doh, even I forgot it in the syntax).
Sorry, here is the right syntax.

ALTER TABLE tablename MODIFY COLUMN columnname datatype AFTER columnname

This works for sure!

Hey, sorry. I misunderstood the question.

/*
ALTER TABLE tablename MODIFY COLUMN columnname AFTER columnname
*/
Eg. ALTER TABLE members MODIFY COLUMN date_of_birth date AFTER username
...

Very simple.

ALTER TABLE tablename ADD columnname datatype AFTER columnname

Hi, can you elaborate your question ?
If you are inserting the value to the table from php, you can use substr (http://php.net/manual/en/function.substr.php) to insert only a part of a string. If...

You should escape ' before adding it to the database.
Use,

$link = mysql_real_escape_string($_POST['link']);
$sql="INSERT INTO catalog (link)
VALUES ('$link')";

You got that error...

All the directories will be in data folder under mysql directory. If you have installed WAMP, for example, the databases will be in c:\wamp\mysql\data :) Look for data directory.

1. I would never use form element's names like, Y M D, ie., with spaces. Try,

<input id="dmy" name="dmy" type="text" />

And in the php script,

$date =...

You can use @ before mysql statements to supress the warning and error messages.
@mysql_connect(...)

This is a comparison operator and you are comparing if the value of $conn_server is "true". So, you should use == instead of =.
Apart from that, I don't see anything wrong with your code which...

Please post your code and explain your problem in detail. Use [code] tags to wrap your code.

You are welcome! :)

escape single quotes and forward slashes by using mysql_real_escape_string. This will prevent sql injections as well.
Ie.,

$leavetype = mysql_real_escape_string($leavetype);
$fullday =...

Yeah.. 1 solution, good. 2 solutions, better :)

Get some sleep dude! Cheers!

@Dickersonka, I have been advised on so many occasions not to use joins. Joins are relatively slower compared to subqueries. This is what I have read.

Subqueries are even better.. :)

select `user_id`, (select count(`comment`) from `comments_table` where `comments_table`.`user_id`=`users_table`.`user_id`) AS `xcnt` FROM `users_table`
GROUP BY...

I modified your query a little bit and I don't know why it doesn't return users who have post = 0.

SELECT COUNT(
COMMENT ) AS `post_count` , (

SELECT `user_id`
FROM `users_table` u
WHERE...

Nice queries.. But still, that wouldn't solve what OP is looking for.. The still doesn't return the count as 0, if a user doesn't have any posts. I created dummy tables to create the same scenario...

After

$result = mysql_query("SELECT * FROM clients WHERE clientID=$_POST[bookingID]");

Use
if(mysql_num_rows($result) > 0 ) {
// record exists.. Do whatever you want to do with that record...

Use mysql_num_rows($result)
For example,

$query = "select *from table where id='".$_POST['id']."'";
$result = mysql_query($query);
if(mysql_num_rows($result) > 0 ) {
//record exists
} else...

Emm.. You can mark this thread as solved if all your questions have been answered. If you are extremely happy with Maxmumford's help, you can add to his reputation. Click on 'Add to MaxMumford's...

It wont. You have to save it in varchar datatype, which again would cause you problems. For example, you can't sort the fields on date. You can't calculate the difference (or add) x number of days to...

My suggestion is, store date in YYYY-mm-dd format itself. You can change the format using php's date function. As maxmumford has already mentioned, you can convert that date to unixtimestamp and use...

Hmm.. Okay! From what I have read so far, this is what STR_TO_DATE do.

STR_TO_DATE(date_string, format);

STR_TO_DATE function will format string and insert it to the date column. The second...

echo your query. Execute it in phpmyadmin. See what's the exact error.
P.S. I tried to work with str_to_date and I simply can't get it to work ! :-/

<?php

// open database connection code and then my code as follows

$sql="UPDATE bookings SET startdate...

How does your code look like ?


P.S. I will be out of town for 3 days, so, If not me, someone else will help you with your problem :)

First, query the table and get all the details of that particular bookingID and then display it. You can also give an option for the user saying something like, "Are you sure want to delete ?" :)...

<?php
// open database connection code and then my code as follows

//$id = mysql_insert_id(); << This line wouldn't work ie., $id will be null because there isn't any insert statement before...

I can't explain you any better :( Post all the relevant code here. And please use [code] tags.

Well, You can do it in quite easily. Before redirecting them to the page where they book a room, use $last_inserted_mysql_id = mysql_insert_id(); to get their clientID.
After getting their...

Where are you inserting the value to client table ?

last_insert_id() (http://dev.mysql.com/doc/refman/5.0/en/getting-unique-id.html) in mysql or mysql_insert_id() (http://in2.php.net/mysql_insert_id) in php gets the value of the autoincrement field...

Great! Cheers :)

Yeah. Try it out. :)

You are doing it wrong.. $result = mysql_query($query); This will execute the query and return the result resource. You have to then use mysql_fetch_array...

When adding the data to the table, use nl2br (http://in.php.net/nl2br) function (if you are using php).

If you 'investigate' properly in phpmyadmin, there is an option exactly below the table's structure to add a new column. You can specify the columnname, its datatype and mention "auto_increment"...