Need assistance with an HTML form, PHP and insertting data into MySQL

[imthewuman]

I am using Yahoo as my webhosting company for my website using MySQL 3.23.49 and PHP Version 4.3.6.

I am trying to create a simple HTML form that a user fills out then calls upon a php file to display the message that I received their information and then I would like it to add the information to mySQL database.

I am able to INSERT a first name and last name into the database using the GUI which I did successfully, however, I am not sure what I am doing wrong with the php code. If someone could assist I would greatly appreciate it! I have searched on google, w3schools, this forum and could not find my answer.

I am a newbie, so please be gentle. THANK YOU IN ADVANCE!

HTML FORM:

<form method=post action="create_db.php">
<table width="70%" align="center">
<tr>
<td width="7%" height="26" bgcolor="#FFFFFF"> <div align="center"><font color="#FFFFFF" face="Arial"></font></div></td>
<td width="86%" bgcolor="#660033"> <div align="center"><font color="#FFFFFF" face="Arial"><strong>Personal
Information </strong></font></div></td>
<td width="7%" bgcolor="#FFFFFF">&nbsp;</td>
</tr>
</table>
<table width=516 border=0 align="center" cellpadding=0>
<tr valign="bottom">
<td width="102" height="33" align=right> <div align="left"><font face=arial size=-1>
</font><font face=arial size=-1><b>&nbsp;First Name:</b></font>
</div></td>
<td width="147" align=right><div align="left">
<input name="firstname" type=textsize=20>
</div></td>
<td width="106" align=left> <div align="left"><font face=arial size=-1>
</font><font face=arial size=-1><b>Last Name:</b></font>
</div></td>
<td width="151" align=left><input name="lastname" type=text size=20></td>
</tr>
<tr valign="bottom">
<td height="33" align=right> <div align="left"><font face=arial size=-1>&nbsp;<b>Address:</b></font>
</div></td>
<td height="33" align=right><div align="left">
<input name="address" type=textsize=20>
</div></td>
<td align=left> <font face=arial size=-1><b>City:</b></font>
</td>
<td align=left><input name="city" type=text size=20></td>
</tr>
<tr valign="bottom">
<td height="33" align=right> <div align="left"><font face=arial size=-1>&nbsp;<b>State:</b></font>
</div></td>
<td height="33" align=right><div align="left">
<input name="state" type=text size=20>
</div></td>
<td align=left> <font face=arial size=-1><b>Zip:</b></font>
</td>
<td align=left><input name="zip" type=text size=20></td>
</tr>
<tr valign="bottom">
<td height="33" align=right> <div align="left"><font face=arial size=-1>&nbsp;<b>Phone:</b></font>
</div></td>
<td height="33" align=right><div align="left">
<input name="phone" type=text size=20>
</div></td>
<td align=left> <font face=arial size=-1><b>Fax:</b></font>
</td>
<td align=left><input name="fax" type=text size=20></td>
</tr>
<tr valign="bottom">
<td height="33" align=right> <div align="left"><font face=arial size=-1>&nbsp;<b>Email:</b></font>
</div></td>
<td height="33" align=right><div align="left">
<input name="email" type=text size=20>
</div></td>
<td align=left> <font face=arial size=-1><b>Business Name:</b></font>
</td>
<td align=left><input name="busname" type=text size=20></td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td height="33" align=right valign="bottom"> <div align="left"><font face=arial size=-1><b>URL:</b></font>
</div></td>
<td height="33" align=right valign="bottom"><div align="left">
<input name="url" type=text id="url" size=20>
</div></td>
</tr>
<tr>
<td colspan=4 valign=top align=center> <font face=arial size=-1>&nbsp;
</font> <p>
<input type="submit" value="Send">
<input type="reset" value="Clear the form">
</td>
</tr>
</table>
</div>
</form>

PHP CODE:

<?php
$db=mysql_connect("localhost", "un", "pw") or die("Could not connect to localhost."); mysql_select_db("clients", $db) or die("Could not find clients.");

//the above host, un and pw was replaced with my actual information

$querymySQL = "INSERT INTO users (firstname, lastname) VALUES ('$firstname', '$lastname')";
if(!$querymySQL) error_message(mysql_error());

echo "Hello, $firstname";
echo "<BR>";
echo "Your information has been received!";
?>

[planetsim]

Your going to need to use mysql_query() at the moment MySQL cannot see your variable as the Query.

[imthewuman]

are you saying to change the following and it should work?

1. $mysql_query() = "INSERT INTO users (firstname, lastname) 2. VALUES ('$firstname', '$lastname')";
2. if(!$mysql_query() ) error_message(mysql_error());

-LINES 1 AND 3 HAVE CHANGED. Is this what you are referring to? If not can you give an example?

[SumitGupta]

OOOPs..You need to know "how to use functions in general"...anyway...

Your Solution is :

The function mysql_query() takes your query as parameter and if set your mysql_connect and mysql_select_db properly as you have done than it run the query in selected database...

So your query will execute with
$query=""INSERT INTO users (firstname, lastname) 2. VALUES ('$firstname', '$lastname')";
mysql_query($query) or die("error in Query execution");

simple.....

And for my first line function never comes on left side of the equal that means function can only take value through parameters and can only return values..

so anything like

abc()=xyz is always wrong

though you can compare the values like
if (abc()==xyz()) this is right

hope i am clear to you

have a nice day