CODEWITHSUNDEEP
phphtmlmysql
aaron head
aaron head

Reputation: 7

how to insert data from a select box using php

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

Upvotes: 0

Views: 10629

Answers (3)

Farshad
Farshad

Reputation: 1485

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>

Upvotes: 2

c4pone
c4pone

Reputation: 797

  • 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.

Upvotes: 1

Jack
Jack

Reputation: 302

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

print_r($_POST,1);

to check all variables existence in it

Upvotes: 0

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