Check Box to store into MySQL database

[phpapprentice]

I am a beginner in PHP. Need Help...

The form is working fine but the check box is not, when I choose steak, it will add steak into the FOOD field.

when I choose steak and pizza, it will add only pizza.

I want to add both steak and pizza into one field separated by comma.

 

 

 

this is my form

 

*************

<form action="insert.php" method="post">

<table cellspacing="0" cellpadding="2" width="98%" border="0">

<tbody><tr><td>Firstname:</td><td><input name="firstname" /></td></tr>

<tr><td>Lastname:</td><td><input name="lastname" /></td></tr>

<tr><td>Age:</td><td><input name="age" /> </td></tr>

<tr><td> </td><td><input type="submit" /> </td></tr></tbody></table>

<p /><p /><p /><pre> </pre><pre>Gender: Male<input type="radio" name="gender"value="male" />

Female<input type="radio" name="gender" value="female" /></pre><br />

Please choose type of residence::<br />Steak:<input type="checkbox" name="food value="steak"/>:<br />

Pizza:<input type="checkbox" name="food" value="pizza"/>:<br />

Chicken:<input type="checkbox" name="food" />:<br /></form>

 

*******************************

 

this is my insert.php

***********************************

<?php

$con = mysql_connect("localhost","mydatabase","mypassword");

if (!$con)

  {

  die('Could not connect: ' . mysql_error());

  }mysql_select_db("noyce", $con);

 

  $sql="INSERT INTO Persons (FirstName, LastName, Age, gender, food)

VALUES

('$_POST[firstname]','$_POST[lastname]','$_POST[age]','$_POST[gender]','$_POST[food]')";if (!mysql_query($sql,$con))

  {

  die('Error: ' . mysql_error());

  }

echo "1 record added";mysql_close($con)

?>

 

 

*****************************

my database - SELECT

 

**********************

 

mysql> select * from Persons;

+----------+-----------+----------+------+--------+---------+

| personID | FirstName | LastName | Age  | gender | food    |

+----------+-----------+----------+------+--------+---------+

|        1 | Rommelito | Santos  |  29 | NULL  | NULL    |

|        2 |              |            |    0 | NULL  | NULL    |

|        3 | Mike        |            |  29 | NULL  | NULL    |

|        4 | Alvin        | Garcia    |  20 | NULL  | NULL    |

|        5 | John        | Berry    |  19 | NULL  | NULL    |

|        6 | Jessica    | Alba      |  30 |        | NULL    |

|        7 | Sharon    | Stone    |  40 | on    | NULL    |

|        8 | Robin      | Padilla    |  45 | on    | NULL    |

|        9 | William    | Martinez  |  45 | on    | on      |

|      10 | Jessa      | Zaragosa |  40 | female | on      |

|      11 | Hanna      | Montana  |  15 | male  | pizza  |

|      12 | Bon        | Jovi        |  40 | male  | steak  |

|      13 | Vilma      | Santos    |  50 | female | chicken |

|

+----------+-----------+----------+------+--------+---------+

13 rows in set (0.01 sec)

 

 

[dennismonsewicz]

change every one of your checkbox inputs to look something like this

 

<input type="checkbox" name="food[]" value="pizza"/>

 

Notice the brackets ([]) after the word food

 

Then just post them like this

 

$food = $_POST['food'];

[Barand]

your process code would be

 

$food = join (', ', $_POST['food']);     // eg "steak, pizza"

// insert food into db

[phpapprentice]

where will i put this code?

$food = $_POST['food'];

I mark it red in the code.

 

 

Do i have to add it in my insert.php

 

my INSERT.PHP

***********************

<?php

$con = mysql_connect("localhost","mydatabase","mypassword");

if (!$con)

  {

  die('Could not connect: ' . mysql_error());

  }mysql_select_db("noyce", $con);

$food = $_POST['food'];

  $sql="INSERT INTO Persons (FirstName, LastName, Age, gender, food)

VALUES

('$_POST[firstname]','$_POST[lastname]','$_POST[age]','$_POST[gender]','$_POST[food]')";if (!mysql_query($sql,$con))

  {

  die('Error: ' . mysql_error());

  }

echo "1 record added";mysql_close($con)

?>

************************************

[dennismonsewicz]

using barand's code use it like this:

 

<?php
$con = mysql_connect("localhost","mydatabase","mypassword");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }mysql_select_db("noyce", $con);
[color=red]$food = join (', ', $_POST['food']); [/color]
  $sql="INSERT INTO Persons (FirstName, LastName, Age, gender, food)
VALUES
('$_POST[firstname]','$_POST[lastname]','$_POST[age]','$_POST[gender]','[color=red]$food[/color]')";if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";mysql_close($con)
?>

[dennismonsewicz]

sorry use it like this (forget the tags)

 

<?php
$con = mysql_connect("localhost","mydatabase","mypassword");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }mysql_select_db("noyce", $con);
$food = join (', ', $_POST['food']);
  $sql="INSERT INTO Persons (FirstName, LastName, Age, gender, food)
VALUES
('$_POST[firstname]','$_POST[lastname]','$_POST[age]','$_POST[gender]','$food')";if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";mysql_close($con)
?>

[phpapprentice]

PROBLEM SOLVED

 

Thanks dennismonsewicz and Barand.

I appreciate it very much.

 

MILLION THANKS!!!!!!

 

phpapprentice