PHP mySQL select via html select

Question

I'm trying to do a select from a table based on the post value of an HTML select box. I'm getting no results at all, I'm echoing out the post value no problem. The statement works on it's own but won't when I use the select form to populate it. This is just my test I will be adding other options to the dropdown box.

<?php
if(isset($_POST['value'])) {
if($_POST['value'] == 'Militaria') {
    $query = "SELECT * FROM listings WHERE category1=Militaria";  
}
else {  
    // query to get all records  
    $query = "SELECT * FROM listings";  
}  
}
$sql = mysql_query($query);  
while ($row = mysql_fetch_array($query)){ 
    echo 'Description:' . $row['description'];
}
mysql_close($con);    
?>

Here is the html form I'm using, can anyone tell me where I'm going wrong, should I do it a different way etc, I'm new to php? Thanks!!

<form action='<?php echo $_SERVER['PHP_SELF']; ?>' method='post' name='form_filter' > 
<select name="value"> 
    <option value="all">All</option> 
    <option value="Militaria">Militaria</option>
</select> 
<br /> 
<input type='submit' value = 'Filter'> 
</form>

try var_dump($_POST) and see if the there are values inside $_POST — Lal
– Lal, Commented May 16, 2015 at 11:51

yozh · Accepted Answer · 2015-05-16 11:55:38Z

2

mysql_fetch_array() should receive resorce as a parameter. Try mysql_fetch_array($sql).

answered May 16, 2015 at 11:55

yozh

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Comments

kamlesh.bar · Accepted Answer · 2015-05-16 11:58:50Z

Quote around 'Militaria' and mysql_fetch_array($sql)

<?php

        if(isset($_POST['value'])) {
            if($_POST['value'] == 'Militaria') {
                $query = "SELECT * FROM listings WHERE category1='Militaria'";  
            }
            else {  
                // query to get all records  
                $query = "SELECT * FROM listings";  
            }  

            $sql = mysql_query($sql);  
            while ($row = mysql_fetch_array($sql)){ 
                echo 'Description:' . $row['description'];
            }
            mysql_close($con); 
        }
    ?>

    <form action='<?php echo $_SERVER['PHP_SELF']; ?>' method='post' name='form_filter' > 
    <select name="value"> 
        <option value="all">All</option> 
        <option value="Militaria">Militaria</option>
    </select> 
    <br /> 
    <input type='submit' value = 'Filter'> 
    </form>

Bhavesh G · Accepted Answer · 2015-05-16 12:34:31Z

0

You have two mistakes in your php code.

1st : quote around Militaria. The query should be, $query = "SELECT * FROM listings WHERE category1='Militaria'";

2nd : mysql_fetch_array accepts executed query's result as parameter. It should be, $row = mysql_fetch_array($sql)

Final code:

<?php
if(isset($_POST['value'])) {
if($_POST['value'] == 'Militaria') {
    $query = "SELECT * FROM listings WHERE category1 = 'Militaria'";  
}
else {  
    // query to get all records  
    $query = "SELECT * FROM listings";  
}  
}
$sql = mysql_query($query);  
while ($row = mysql_fetch_array($sql)){ 
    echo 'Description:' . $row['description'];
}
mysql_close($con);    
?>

edited May 16, 2015 at 12:34

answered May 16, 2015 at 12:14

Bhavesh G

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