PHP - Search Function for MySQL Database Doesn't Work

Question

I need to retrieve something from database but the search does not work. When I input in search bar the folloowing: alex/bobo/stano, nothing happens, it does not return any results. Could someone point to me where the problem lies?

nume     |  prenume   |    id
stano        bobo           1
alex         bobo           2

Here is the code:

    // conectare la baza de date
    session_start();
    $db = mysqli_connect("localhost", "root", "", "inregistrare");
    $output = '';
    //conectare
    if (isset($_POST['cauta'])) {
        $cauta1 = $_POST['cauta'];
        $cauta1 = preg_replace("#[^0-9a-z]#i","",$cauta1);

        $query = mysqli_query($db, "SELECT * FROM users WHERE nume LIKE '%$cauta1%' OR prenume LIKE '%$cauta1%'");
        $count = mysqli_num_rows($db, $query);
        if ($count == 0) {
            $output = 'Nu a fost gasit elevul';
        } else {
            while ($row = mysqli_fetch_array($query)) {
                $nume = $row['nume'];
                $prenume = $row['prenume'];
                $id = $row['id'];

                $output .= '<div>'.$nume.' '.$prenume.'</div>';
            }
        }
    }

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Cauta Elev</title>
</head>

<body>
<form method="post">
	<input type="text" name="cauta" placeholder="Cauta Elevul"/>
	<input type="submit" value="Cauta"/>
</form>
<?php print("$output"); ?>
</body>
</html>

if(isset($_POST['cauta'])) Maybe is not set? Hint: check your form code again ;) — Damien Pirsy
– Damien Pirsy, Commented Oct 19, 2016 at 15:46
ops, i changed in the form, but now : Notice: Use of undefined constant cauta - assumed 'cauta' in .. on line 8 — Rares Coco
– Rares Coco, Commented Oct 19, 2016 at 15:52
You're just ASSUMING that nothing ever goes wrong. you don't check if you connected to the db properly, you don't check if the query succeeded (or even ran). That is exactly the WRONG attitude to have. Never EVER assume success when dealing with external resources. Always assume failure, check for failure, and treat success as a pleasant surprise. — Marc B
– Marc B, Commented Oct 19, 2016 at 17:38

krasipenkov · Accepted Answer · 2016-10-19 15:51:07Z

2

You missed $ for nume and prenume vars in the $output. The correct one is:

$output .= '<div>'.$nume.' '.$prenume.'</div>';

answered Oct 19, 2016 at 15:51

krasipenkov

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5 Comments

Rares Coco Over a year ago

Notice: Use of undefined constant cauta - assumed 'cauta' in .. on line 8 Warning: mysqli_query() expects at least 2 parameters, 1 given in ... on line 11 Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in .. on line 12

krasipenkov Over a year ago

You have to pass $db as first param of mysqli_query

krasipenkov Over a year ago

And use $_POST['cauta']

Rares Coco Over a year ago

Warning: mysqli_num_rows() expects exactly 1 parameter, 2 given in .. on line 12

krasipenkov Over a year ago

Remove $db from mysqli_num_rows. Pass only $query

Mehedi · Accepted Answer · 2016-10-19 17:25:54Z

1

I think it should be your final code.

<?php
    // conectare la baza de date
    session_start();
    $db = mysqli_connect("localhost", "root", "", "inregistrare");
    $output = '';
    //conectare
    if(isset($_POST['cauta'])){
        $cauta1 = $_POST['cauta'];
        $cauta1 = preg_replace("#[^0-9a-z]#i","",$cauta1);

        $query = mysqli_query($db, "SELECT * FROM users WHERE nume LIKE '%$cauta1%' OR prenume LIKE '%$cauta1%'");
        $count = mysqli_num_rows($query);
        if($count == 0){
            $output = 'Nu a fost gasit elevul';
        }else{
            while($row = mysqli_fetch_array($query)){
                $nume = $row['nume'];
                $prenume = $row['prenume'];
                $id = $row['id'];

                $output .= '<div>'.$nume.' '.$prenume.'</div>';             
            }       
        }       
    }
?>

<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>Cauta Elev</title>
    </head>

    <body>
        <form method="post">
            <input type="text" name="cauta" placeholder="Cauta Elevul">
            <input type="submit" value="Cauta">

        </form>
        <?php print("$output");
        ?>
    </body>
</html>

answered Oct 19, 2016 at 17:25

Mehedi

212 bronze badges

2 Comments

Rares Coco Over a year ago

yes, works, it.s a way to add a function when get info from database to capitalize first letter?

Mehedi Over a year ago

To show in capitalize first letter you can use $output .= '<div>'.ucfirst($nume).' '.ucfirst($prenume).'</div>';

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