how to insert data from a select box using php

Question

I have come across a problem when inserting values from a html select in to a mysql database. I can't seem to get the values to insert for some reason; I have looked for help on this but they keep giving me errors. Also, can some please tell me what the difference between mysql and mysqli?

php code

<?php
$con = mysql_connect("localhost","barsne","bit me");
if (!$con) {
    die('Could not connect: ' . mysql_error());
}

mysql_select_db("testing", $con);

$sql="INSERT INTO client_details (id, f_name, l_name, phone, email, job_est) VALUES
   ('', '$_POST[f_name]', '$_POST[l_name]', '$_POST[phone]', '$_POST[email]', '$_POST[job_est]')";

if (!mysql_query($sql,$con)) {
    die('Error: ' . mysql_error());
}

echo "Thank you for booking a with us we will contact you in the next 24 hours to confirm your booking with a time and date";

mysql_close($con)
?>

html code

<form method="post" action="processing_booking.php">
    <h4><u>Basic Contact Details</u></h4>
    <label>First Name</label>
    <input type="text" name="f_name">
    <label>Last Name:</label>
    <input type="text" name="l_name" id="l_name">
    <label>Phone Number:</label>
    <input type="text" name="phone" id="phone">
    <label>Email:</label>
    <input type="text" name="email" id="email">

    <h4><u>Job Details</u></h4>
    <label>You Would Like To Book A:</label>
    <select name="job_est">
        <option value="select">--SELECT--</options>
        <option value="job">Job</option>
        <option value="est">Estimation</option>
    </select>

    <label>Service Your Booking:</label>
    <select name="job_type">
        <option value="select">--SELECT--</option>
        <option value="gardening">Gardening</option>
        <option value="landscaping">Landscaping</option>
        <option value="painting">Painting & Decorating</option>
        <option value="decking">Decking & Fencing</option>
    </select>

    <label>Any Additional Information </label>
    <textarea name="extra_info"></textarea>
    <input type="submit" value="lets get your dreams started">
</form>

sorry wirting is not my strong point

The MySQL family of PHP is deprecated and support thereof will disappear. Please look into PDO or Mysqli. The code is also wide open to SQL injection. — chwcz7F
– chwcz7F, Commented Aug 26, 2014 at 9:24
Can you please add the starting form tag also to the question — Veerendra
– Veerendra, Commented Aug 26, 2014 at 9:25
you should never write $_POST values directly in your sql query. Even when it is a selectbox the value can be easily changed with tools like chrome developer toolbar tampadata etc. — c4pone
– c4pone, Commented Aug 26, 2014 at 9:31
have you noticed that you're using $_POST[f_name] istead of $_POST['f_name'] ... — may saghira
– may saghira, Commented Aug 26, 2014 at 9:34

Farshad · Accepted Answer · 2014-08-26 10:50:14Z

2

Note that $_POST['name'] you missed single quotations

you must prevent from injection with

// String Type Fields 
$name = strip_tags($_POST['name']);
$f_name= strip_tags($_POST['f_name']);
$l_name= strip_tags($_POST['l_name']);
$phone= strip_tags($_POST['phone']);
$email= strip_tags($_POST['email']);

// Int Type Fields
if (isset($_POST['job_est']) && is_numeric($_POST['job_est']))
   $job_est= $_POST['job_est'];
else 
   $job_est= 0;

then use in your query

other point is if your id field is primary and auto increment , you can define your query as below :

$sql="INSERT INTO client_details (f_name, l_name, phone, email, job_est) VALUES
   ( '".$f_name."', '".$l_name."', '".$phone."', '".$email."', ".$job_est.")";

for strings you must use '".$variable."' and for integer or numeric you must use ".$variable."

other point is you must change your select element to below because you have noticed in your comments your job_est field type is int(11)

<select name="job_est">
    <option value="0">--SELECT--</option>
    <option value="1">Gardening</option>
    <option value="2">Landscaping</option>
    <option value="3">Painting & Decorating</option>
    <option value="4">Decking & Fencing</option>
</select>

edited Aug 26, 2014 at 10:50

answered Aug 26, 2014 at 9:39

Farshad

1,4951 gold badge9 silver badges12 bronze badges

Sign up to request clarification or add additional context in comments.

14 Comments

Farshad Over a year ago

put strip_tags before $sql="

Farshad Over a year ago

what is the field type of your job_est filed in your db ?

aaron head Over a year ago

the job_est is int(11) let me guess as a newibe i made one of the most simplest mistakes

Farshad Over a year ago

so you must set the value of select box to integer and so you dont need strip tags the job_est and simply check if is_nummeric($_POST['job_est']). I have updated my answer

aaron head Over a year ago

how would i put if is_nummeric($_POST['job_est']) in to my script

|

c4pone · Accepted Answer · 2014-08-26 09:53:06Z

1

The mysql functions are deprecated. Use the mysqli or pdo class instead.

mysqli: https://php.net/manual/en/class.mysqli.php

pdo: https://php.net/manual/en/book.pdo.php
Make a var_dump($_POST) and you will see what the problem is. By the way you missed the single quotes $_POST['value']
Never ever write $_POST or $_GET data directly in your sql queries. Always validate them before, because even the value of a selectbox can get easily changed with tampadata or chrome developer tools.

edited Aug 26, 2014 at 9:53

answered Aug 26, 2014 at 9:36

c4pone

7977 silver badges20 bronze badges

Comments

Jack · Accepted Answer · 2014-08-26 09:36:41Z

0

Well you can obviously start printing the results of your $_POST array with :

print_r($_POST,1);

to check all variables existence in it

answered Aug 26, 2014 at 9:36

Jack

3022 silver badges11 bronze badges

Collectives™ on Stack Overflow

how to insert data from a select box using php

3 Answers 3

14 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

14 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related