SQL queries not working properly through PHP

Question

please do not hate because i have searched far and wide for a cure to my problem here. I am using xampp to run this php code in question. The problem is that the search function works perfectly but for some reason my if statements varying what order it is sorted by do not. All help is appreciated. All love no hate. Thanks.

    <html>
<head>  

<title>SeedBase</title>
<link rel="stylesheet" type="text/css" href="CSSstyle.css">

</head>

<body>
<h1 class="logo">SeedBase</h1>

<ul div class="navbar">
    <li><a href="HTMLhome.HTML">Home     |</a></li>
    <li><a href="PHPseeds.php">Seeds     |</a></li>
    <li><a href="HTMLcontact.html">Contact</a></li>
</ul>


<div class="search">
<form method="POST" action="PHPseeds.php">
    <p>Search SeedBase:<input type="text" name="searchbox" value="seed"></p>
        <p><select name="selector">
        <optgroup label = "Sort By">
            <option value="def" selected>Defult</option>
            <option value="asc" >Quantity Asc</option>
            <option value="desc">Quantity Desc</option>
        </optgroup>
    </select></p>

        <p><input type="submit" name="button" value="GO"></p>
</form>
</div>





<?php
echo "<phpcss>";
include 'db.inc.php';

$row = 0;
$find = $_POST['searchbox'];
$choice = $_POST['selector'];
$output = "";

    if ($find != '') {
        if ($choice="def"){
        $sql =  "SELECT * FROM seeds WHERE  name LIKE '%$find%' OR description LIKE  '%$find%' "; 
        }
        else if ($choice="asc"){
        $sql =  "SELECT * FROM seeds WHERE  name LIKE '%$find%' OR description LIKE  '%$find%' ORDER BY quantity ASC "; 
        }
        else if ($choice="desc"){
        $sql =  "SELECT * FROM seeds WHERE  name LIKE '%$find%' OR description LIKE  '%$find%' ORDER BY quantity DESC "; 
        }
    }

    if ($find == ''){
        if ($choice="Defult"){
        $sql =  "SELECT * FROM seeds "; 
        }
        else if ($choice="asc"){
        $sql =  "SELECT * FROM seeds WHERE  name LIKE '%$find%' OR description LIKE  '%$find%' ORDER BY quantity ASC "; 
        }
        else if ($choice="desc"){
        $sql =  "SELECT * FROM seeds WHERE  name LIKE '%$find%' OR description LIKE  '%$find%' ORDER BY quantity DESC "; 
        }

    }
$result = $pdo->query($sql);
?>



<div class="DBpic">
<?php
    echo "<table>";
foreach ($result as $row)
{
    $seed = $row['name'];
    $description = $row['description'];
    $quantity = $row['quantity'];
    $image = $row['image'];
    $output ="<tr><tr> $seed <tr> $quantity cases" ;

    echo "<table>";
    echo  '<br><img src="data:image/jpeg;base64,' . base64_encode( $row['image'] ) . '" /><br>'. $output;
    echo "</table>";

    }
    echo "</table>";



?>
</div>




<?php
if ($row == 0)
{
echo "nothing could be found for that search term";
}
echo "</phpcss>";
?>


</body>
</html>

`$choice="def"`, and all the rest of the `if`s. You're assigning, not comparing. — aynber, Mar 28 '18 at 13:16
Also [How can I prevent SQL injection in PHP?](https://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) — Alex K., Mar 28 '18 at 13:17
Still use select but not select "*" star, meaning do not select ALL columns, just the ones you need. — Brad, Mar 28 '18 at 13:22

score 0 · Answer 1 · answered Mar 28 '18 at 13:17

0

In second If Condition you have to use !=

 if ($find != '')

answered Mar 28 '18 at 13:17

Jay Shankar Gupta

5,707
1
7
26

Why? How do you know? – schlonzo Mar 28 '18 at 13:20
because in second if also he is using $find which must not be empty – Jay Shankar Gupta Mar 28 '18 at 13:21
The second `if` could be done as an `else` as it's building the queries if no conditions have been entered. – Nigel Ren Mar 28 '18 at 13:26

SQL queries not working properly through PHP

1 Answers1