Trying to create a sql query using checkboxes

Question

I'm a bit of a beginner at sql but I believe this should work. I'm doing it step by step so this isn't the complete code yet but I've hit a problem. Basically what I'm trying to do is use the array created from the form:

<form action="search.php" method="post">
<input type="checkbox" name="features[]" value="Texture" >Textures<br />
<input type="checkbox" name="features[]" value="Items" >Items<br />
<input type="checkbox" name="features[]" value="GUI" >Gui<br />
</form

And create what will be end up being a query for my database. Here's the PHP code:

<?php

    $button = $_POST['submit'];
    $features = $_POST['features'];

    if ($button){

    //connect
    mysql_connect("server","username","password");
    mysql_select_db("rdm");

    foreach ($features){

    $x++;

    if ($x=1)
    $construct .= "feature LIKE '$features[$x]'";
    else
    $construct .= " OR feature LIKE '$features[$x]'";
    }

    $construct = "SELECT * FROM Textures WHERE '$construct';

    echo $construct;

    ?>

My understanding is this should echo out the variable $construct, which will contain the text "SELECT * FROM Textures WHERE feature LIKE ... and then the first value in the features array, followed by the other values.

Any help as to why this isn't working would be fantastic! As I say, I'm pretty new to this so sorry if this is a silly mistake.

Thanks in advance!

You forgot your closing double-quote on the line before your echo. — jrn.ak
– jrn.ak, Commented Feb 14, 2011 at 19:41
As a side note, I guess you should escape your input values: xkcd.com/327 — Albin Sunnanbo
– Albin Sunnanbo, Commented Feb 14, 2011 at 19:50
You're assigning a value with if ($x=1) instead of comparing values. — jrn.ak
– jrn.ak, Commented Feb 14, 2011 at 19:59

Marc B · Accepted Answer · 2011-02-14 19:57:22Z

3

Your query string will be generated wrong:

if ($x=1)

This isn't a comparison, it's setting x to 1, so your query will have no 'OR' conditions in it. A better method to do your query generation is:

$query_conditions = array()
foreach($_POST['features'] as $feature) {
     $query_conditions[] = "'" . mysql_real_escape_string($feature) . "'"
}
$where_in_clause = implode(',', $query_conditions);
$query = "SELECT * FROM Textures WHERE feature IN ($where_in_clause)";

answered Feb 14, 2011 at 19:57

Marc B

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Cristian Radu · Accepted Answer · 2011-02-14 19:46:28Z

0

Try this:

// implode the features array into a string, adding quotes
$featuresString = "'".implode("','", $features)."'";
// build the query
$construct = "SELECT * FROM Textures WHERE feature IN (".$featuresString.")";

answered Feb 14, 2011 at 19:46

Cristian Radu

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Chris Baker · Accepted Answer · 2011-02-14 19:55:22Z

0

You have a couple of syntax errors in there. Notably, the line with SQL does not close the double quote or use a semi-colon at the end of the line. There is also a missing bracket at the end of your if block.

This code will accomplish the task as stated:

$button = isset($_POST['submit']) ? true : false;
$features = isset($_POST['features']) ? $_POST['features'] : false;

if ($button && is_array($features)) {
        $construct = 'SELECT * FROM Textures WHERE feature IN  ("'.implode('","', $features ).'")';
    echo $construct;
}

answered Feb 14, 2011 at 19:55

Chris Baker

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