Problem Inserting Data In A Mysql Database Using PHP

Dapernia · 09-16-2007, 05:01 PM

Hello My Friends:
I am learning PHP and MySql using some tutorials and manuals that i've found in internet. I am using Ubuntu 7.04 and I've already installed Apache2, PHP5, MySQL5.0 and everything is working fine. I 've created a small database with PHPMyAdmin called `ficha_laboral` and one table called `datos_personales` with 6 fields.

Ok, everything looks fine until here, but when i try to insert some data in the database through the web browser (firefox), MySql doesn't insert anything in the database. i am using two scripts to do this, the first script is the form where i write all the data, and the second script inserts that data in the database.

This is the form code i'm using:

[HTML]
<html>
<body>
<form method="post" action="http://127.0.0.1/cursophp/add_reg.php">
Nombre :<input type="Text" name="nombre"><br>
Apellido :<input type="Text" name="apellido"><br>
C.I. :<input type="Text" name="ci"><br>
Profesion :<input type="Text" name="profesion"><br>
Fecha de Nacimiento :<input type="Text" name="nacimiento"><br>
Estado Civil :<input type="Text" name="estado"><br>
<input type="Submit" name="enviar" value="Aceptar información">
</form>
</body>
</html>
[/HTML]

and this is the PHP Script i am using (it's name is add_reg.php)

PHP Code:



<?php 
   // process form
   $link = mysql_connect("127.0.0.1", "root","1234");
   mysql_select_db("ficha_laboral",$link);
   $sql = "INSERT INTO `datos_personales` (`Nombre`,`Apellido`,`CI`,`Profesion`,`Fecha_Nacimiento`,`Edo_Civil`) VALUES ('$nombre', '$apellido', '$ci', '$profesion','$nacimiento','$estado');";
   $result = mysql_query($sql);
   echo "¡Gracias! Hemos recibido sus datos.\n"; 
?>

when i execute the script I don't get any error, and everything looks fine, but the database keeps blank.
I've been the whole day trying to find out what is wrong in this code. Can you please help me out???
Thank you very much...

chrism01 · 09-16-2007, 07:43 PM

You need to check for errors, see here: http://au2.php.net/manual/en/function.mysql-error.php

paulsm4 · 09-16-2007, 07:46 PM

I'm suspicious of the back quotes (``).

Please:
1. Look at this example:
http://www.w3schools.com/php/php_mysql_insert.asp

2. Add a "die" clause with "mysql_err ()" (in case the connect fails).

3. Remove the back quotes from your "insert" line.

4. Add another "die" with a corresponding "mysql_err()" to your mysql_query().

5. "echo" the value of $result and the values of $nombre, $apellido, etc

6. Don't forget to call "mysql_close ()" to free the connection.

'Hope that helps .. PSM

owenjh · 09-16-2007, 10:09 PM

The back ticks are fine. If the code you posted for add_reg.php is all the code you have then I would try something like:

Code:

<?php
$nombre = $_POST['nombre'];
$apellido = $_POST['apellido'];
$ci = $_POST['ci'];
$profesion = $_POST['profesion'];
$nacimiento = $_POST['nacimiento'];
$estado = $_POST['estado'];

   // process form
   $link = mysql_connect("127.0.0.1", "root","1234");
   mysql_select_db("ficha_laboral", $link);
   $sql = "INSERT INTO `datos_personales` (`Nombre`,`Apellido`,`CI`,`Profesion`,`Fecha_Nacimiento`,`Edo_Civil`) VALUES ('$nombre', '$apellido', '$ci', '$profesion','$nacimiento','$estado');";
   $result = mysql_query($sql);
   echo "¡Gracias! Hemos recibido sus datos.\n"; 
?>

Since you aren't assigning any variables it won't add the row.

Hope this helps

- Owen

p.s Its not the most secure code in the world, but it is good for a simple tutorial

Dapernia · 09-17-2007, 08:59 AM

Thanks Owenjh!!. Now It's working fine!!!
It's kind of weird that none of the tutorials i found had that assignment.

owenjh · 09-17-2007, 10:05 AM

Check out php.net/mysql_escape_string

You should use this to help with the data of the input fields since if some one puts in a quote it will break the query and leave your script open for exploitation.

I think it will be something like this for all your variables:

Code:

$ci = mysql_escape_string($_POST['ci']);

But, its been a while so without testing it I wouldn't know.