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Select all options from a select box with MySQL


CristiM

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Hi guys my name is Chris i'm on this forum and in php programming so i have a few questions, i hope somebody can help me.

 

So:

 

i' m trying  to make a real estate website and i have some difficulties with the search form

 

the form has 5 fields: price_from; price_to; number_rooms, location and numb_square_meters

the data base has 5 columns: id, price, location, numb_square_meters, number_rooms

 

 

 

For the price fields i use an input type field

For numb_rooms i use a select type field with options like(vila, condo, 4 bedroom apartaments..something like that and ALL OPTIONS option)

For location and square_meters is the same way as for numb_rooms

 

The problem is this actually there are more than one:

1.When i choose the option "ALL OPTIONS" in the LOCATION field in my form i want it to retrieve every location from the SQL DB if not to retrieve one of the other options

2. If let say the user fills the price fields but the other ones remain with the ALL OPTIONS option selected i want it to retreive only the rows that is in that price range no matter the location number of rooms or the square meters

 

i want to point out the english is not my maternal language , so if you spot some  spelling or grammar errors i apologize.

 

This a part of my code...

 

<form method="post" action="index.php" enctype="multipart/form-data">

<input type="hidden" name="id">

Price <input type="text" name="price">

Price2 <input type="text" name="price2" />

 

 

 

Square_meters<input type="text" name="Square_meters">

 

Location<select name="Location">

<option>Bucuresti</option>

    <option>Iasi</option>

    <option>Brasov</option>

    <option>Focsani</option>   

</select>

Numb_rooms<select name="Numb_rooms">

<option>Garsoniera</option>

    <option>2 Camere</option>

    <option>3 Camere</option>

      <option>4 Camere</option>

    <option>Vila</option>   

</select>

<input type="submit" name="send" value="Search">

</form>

<?php

header("content-type: image.jpg");

 

 

 

if(isset($_POST['send']))

{

require_once("conexiune.php"); (this is my conect to sql database script)

$id = $_POST['id'];

$price = $_POST['price'];

$price2 = $_POST['price2'];

$square_m = $_POST['Square_meters'];

$location = $_POST['location'];

$numb_rooms = $_POST['Numb_rooms'];

$selectie = mysql_query("SELECT * FROM apartamente WHERE id = '$id' && Pret>='$price'&& Pret<='$price2' && Location = '$location' && Square_meters='$Square_meters' && Numb_rooms = '$Numb_rooms ' ");

 

 

while($row = mysql_fetch_array($selectie,MYSQL_ASSOC))

{

 

print"<tr><td>";

print $row['id'];

print"</td><td>";

print $row['Pret'];

print"</td><td>";

print $row['Nr_camere'];

print"</td><td>";

print $row['Suprafata'];

print"</td><td>";

print $row['Locatie'];

print"</td><td>";

print $row['Descriere'];

print"</td><td>";

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You have some serious security issues with your code, by the way. You really need to implement Input Validation and Output Escaping to your script, otherwise you're wide open to anyone who wants to attack the site.

 

Also, why do you have this in the middle of the code?

header("content-type: image.jpg");

Not only do I see no logical reason for it to be there, but it will not work in any case. See the "HEADER ERRORS" threat stickied at the top of this forum for more info.

 

PS: Please use the


tags around your code, as it helps make both your post and your code a lot easier to read. Thanks.

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