Why isn't my PHP working in my search box

Question

I'm trying to get my search bar to link to a specific page when a user enters something I have this.. Keep in mind this isn't my whole code but it's giving me this error...

Notice: Undefined index: link in D:\xampp\htdocs\wd1_vtec_0100348514\pages\search.php on line 63

<?php 
//--- Authenticate code begins here ---
session_start();
//checks if the login session is true
if(!isset($_SESSION['username'])){
header("location:index.php");
}
$username = $_SESSION['username'];

// --- Authenticate code ends here ---


 include ('header.php'); ?> 
 <link rel="stylesheet" type="text/css" href="../css/style1.css">
 <?php
    mysql_connect("localhost", "root", "") or die("Error connecting to database: ".mysql_error());


    mysql_select_db("wd1_vtec_0100348514") or die(mysql_error());

?>


<html>
<head>
    <title>Search results</title>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <link rel="stylesheet" type="text/css" href="style.css"/>
</head>
<body>
<?php
    $query = $_GET['query']; 


    $min_length = 3;


    if(strlen($query) >= $min_length){ // if query length is more or equal minimum length then

        $query = htmlspecialchars($query); 
        // changes characters used in html to their equivalents, for example: < to &gt;

        $query = mysql_real_escape_string($query);
        // makes sure nobody uses SQL injection

        $raw_results = mysql_query("SELECT * FROM articles
            WHERE (`title` LIKE '%".$query."%') OR (`text` LIKE '%".$query."%')") or die(mysql_error());

        // * means that it selects all fields, you can also write: `id`, `title`, `text`
        // articles is the name of our table

        // '%$query%' is what I'm looking for, % means anything, for example if $query is Hello
        // it will match "hello", "Hello man", "gogohello", if you want exact match use `title`='$query'
        // or if you want to match just full word so "gogohello" is out use '% $query %' ...OR ... '$query %' ... OR ... '% $query'

       if(mysql_num_rows($raw_results) > 0){ // if one or more rows are returned do following

    while($results = mysql_fetch_array($raw_results)){
    // $results = mysql_fetch_array($raw_results) puts data from database into array, while it's valid it does the loop



echo "<p><a href='".$results['text']."'><h3>".$results['title']."</h3>".$results['text']."</‌p>";
    }

}
    else{ // if there is no matching rows do following
    echo "No results";
}

}

else{ // if query length is less than minimum
echo "Minimum length  is ".$min_length;
}

?>
</body>
<a class="btn btn-search" type="button" href="home.php" >Search Again</a>
</html>






       <div style="float:right">  <a class="btn btn-danger logout" href="logout.php" > Logout</a> </div>

        <div id="menu">
    <ul id="nav">
        <li><a href="home.php" target="_self" >Home</a></li>
        <li><a href="session1.php" target="_self" >Sessions</a>
            <ul>
                <li><a href="session1.php" target="_self" >Session 1</a></li>
                <li><a href="session2.php" target="_self" >Session 2</a></li>
                <li><a href="session3.php" target="_self" >Session 3</a></li>
                <li><a href="session4.php" target="_self" >Session 4</a></li>
                <li><a href="session5.php" target="_self" >Session 5</a></li>
                <li><a href="session6.php" target="_self" >Session 6</a></li>
                <li><a href="session7.php" target="_self" >Session 7</a></li>
                <li><a href="session8.php" target="_self" >Session 8</a></li>
                <li><a href="session9.php" target="_self" >Session 9</a></li>
                <li><a href="session10.php" target="_self" >Session 10</a></li>
                <li><a href="session11.php" target="_self" >Session 11</a></li>
                <li><a href="session12.php" target="_self" >Session 12</a></li>
                <li><a href="session13.php" target="_self" >Session 13</a></li>
                <li><a href="session14.php" target="_self" >Session 14</a></li>



            </ul>
            <li><a href="blog.php" target="_self" >Blog</a></li>


    </ul>

    </div>

    <?php include ('footer.php'); ?>

@BenjaminDiele I am wanting it to link to a page but for some reason it's not letting me do so — Sarah, Jul 23 '14 at 07:01
Do some debugging: use `var_dump($results); die()` immediately before your echo to see what it actually contains — Mark Baker, Jul 23 '14 at 07:02
@Sarah It's not letting you because there is no `link` in your `$results` array. — Benjamin Diele, Jul 23 '14 at 07:03
@BenjaminDiele I've looked in my DB and found a column that I want to link it to which has a name of '1' but now it's showing me a 404 — Sarah, Jul 23 '14 at 07:04
please check your code. its have syntax error. and there is no corresponding if condition of last else condition. please check — karan, Jul 23 '14 at 07:05
possible duplicate of [PHP: "Notice: Undefined variable" and "Notice: Undefined index"](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-and-notice-undefined-index) — Jocelyn, Jul 28 '14 at 00:41

score 0 · Answer 1 · answered Jul 23 '14 at 07:01

0

You should simply check if there is column link in your database. You show us one error what means that other columns (title and text) exist

You shouldn't also use mysql functions any more. They are deprecated. You should use mysqli or PDO

answered Jul 23 '14 at 07:01

Marcin Nabiałek

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Thanks for your help I've found in my column theres an ID called 1 so I used that but now it's giving me a 404 error. – Sarah Jul 23 '14 at 07:03
So it seems you create incorrect link or your PHP script doesn't work well. We don't know what's your site structure. And if ID for your link will be valid. Maybe you should add `/` at the beginning or use other column. – Marcin Nabiałek Jul 23 '14 at 07:04
you should append some page name before the ID – Khushboo Jul 23 '14 at 07:04
@Khushboo wait.. Why? – DarkBee Jul 23 '14 at 07:05
can you please echo your ID and paste the result here ? – Khushboo Jul 23 '14 at 07:07
@Khushboo I've just also tried fixing it but still no luck – Sarah Jul 23 '14 at 07:09
@Sarah Why you use `$results['text']` to create link to the article? It's text and it rather should be used to create anchor – Marcin Nabiałek Jul 23 '14 at 07:10
@MarcinNabiałek I thought that's how it should be done? – Sarah Jul 23 '14 at 07:11
no... in which column you have link in database? Please add that in href – Khushboo Jul 23 '14 at 07:12
@Khushboo I've linked it to the table session and have gone to sessionID name with the ID of '1'. That's what I've done basically – Sarah Jul 23 '14 at 07:14
but you are using Articles table to fetch the searched result. so, what exactly you want to display? Show that thing – Khushboo Jul 23 '14 at 07:20

Why isn't my PHP working in my search box

1 Answers1