Notice: Trying to get property of non-object in C:\xampp\htdocs\php\search.php on line 39

Question

I am trying to create search menu in my website, i am using php and mysqli ,i have created all the credential stuffs which are all required to it but still it showing me some errors,can anyone help me from this. analyze.php

<!DOCTYPE html>
<html>
  <head>
    <title>Gtec NetWork</title>
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <!-- Bootstrap -->
    <link href="asset/bootstrap/css/bootstrap.min.css" rel="stylesheet" media="screen">
  </head>
  <body>
    <div class="navbar">
        <div class="navbar-inner">
            <a class="brand" href="#">GTEC Network</a>
            <ul class="nav">
                <li><a href="#">Home</a></li>
                <li><a href="#">Boys</a></li>
                <li><a href="#">Girls</a></li>
            </ul>
        </div>
    </div>
    <div class="container-fluid">
        <div class="row">
            <div class="span12">
                    <form method="POST" action="search.php">
                    <label>Search
                            <input type="text" name="keywords" >
                            <input type="submit" class="btn btn-primary lg" value="Search">
                            </label>

                    </form>
            </div>
        </div>
    </div>
  </body>
 </html>

search.php

<?php  
    require_once 'lib/db.php';
?><?php
    if (isset($_GET['keywords'])) {
        $keywords = $db->escape_string($_GET['keywords']);
        $query = $db->query("
            SELECT *
            FROM user 
            WHERE  registernumber LIKE '%{$keywords}%'
            OR username LIKE '%{$keywords}%' 
            "
            );
        $run = mysqli_query($query);

    }

?>
    <!DOCTYPE html>
<html>
  <head>
    <title>Gtec NetWork</title>
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <!-- Bootstrap -->
    <link href="asset/bootstrap/css/bootstrap.min.css" rel="stylesheet" media="screen">
  </head>
  <body>
    <div class="navbar">
        <div class="navbar-inner">
            <a class="brand" href="#">GTEC Network</a>
            <ul class="nav">
                <li><a href="#">Home</a></li>
                <li><a href="#">Boys</a></li>
                <li><a href="#">Girls</a></li>
            </ul>
        </div>
    </div>
    <div class="container">
        <div class="row">
            <div class="col-md-4">
                Found<?php echo $query->num_rows;?> results.
                <?php  
                    if ($query->num_rows) {
                        while($r = $query->mysqli_fetch_array())
                        {
                        ?>

                        <div class="gylphicon glyphicon-heart">
                            <a href="#"><?php echo $r->username;?></a>
                        </div>
                        <?php   }
                    }?>

            </div>
        </div>
    </div>
    </body>
</html>

The problem is

Found Notice: Undefined variable: query in C:\xampp\htdocs\php\search.php on line 39

Notice: Trying to get property of non-object in C:\xampp\htdocs\php\search.php on line 39 results. Notice: Undefined variable: query in C:\xampp\htdocs\php\search.php on line 41

Notice: Trying to get property of non-object in C:\xampp\htdocs\php\search.php on line 41

Please help me

In your html code `method` is "POST" and in your php code you search for `GET` var ? Change the method in your html form to `GET` — e-Learner, Jul 04 '15 at 01:00
Why does this line appear in the code? **`$run = mysqli_query($query);`** ... What is that supposed to do? What the *plastic*? — spencer7593, Jul 04 '15 at 01:06

dlegall · Accepted Answer · 2015-07-04T01:16:39.887

0

$query is declared inside the if-block

if (isset($_GET['keywords'])) { ... }

So, if $_GET['keywords'] isn't set, $query won't be defined later in the script.

Maybe you can do something like this :

if (isset($_POST['keywords'])) {
    $keywords = $db->escape_string($_POST['keywords']);
    $query = $db->query("
        SELECT *
        FROM user 
        WHERE  registernumber LIKE '%{$keywords}%'
        OR username LIKE '%{$keywords}%' 
        "
        );

} else {
    $query = $db->query("
        SELECT *
        FROM user
        "
        );
}

edited Jul 04 '15 at 01:16

answered Jul 04 '15 at 00:49

dlegall

412
2
5

So what to do for that? – Raja Raman TechWorld Jul 04 '15 at 00:50
Do not use $query in your body if $_GET['keywords'] isn't defined. – dlegall Jul 04 '15 at 00:51
Then where i query from database – Raja Raman TechWorld Jul 04 '15 at 00:52
You're not querying anything if $_GET['keywords'] isn't defined, see your own code. So, if you're not querying anything, you can't display any result, right? – dlegall Jul 04 '15 at 00:53
I am asking about where i can put select * user ,mysql query? – Raja Raman TechWorld Jul 04 '15 at 00:55
Warning: mysqli_query() expects parameter 2 to be string, object given in C:\xampp\htdocs\php\search.php on line 33 Notice: Trying to get property of non-object in C:\xampp\htdocs\php\search.php on line 34 – Raja Raman TechWorld Jul 04 '15 at 01:08
Please just remove the mysqli_query(). It's not necessary in this case, didn't see it. – dlegall Jul 04 '15 at 01:11
And, as pointed out by @e-Learner, you're using the method POST to send the form. So you must replace $_GET['keywords'] by $_POST['keywords'] – dlegall Jul 04 '15 at 01:16

score 0 · Answer 2 · answered Jul 04 '15 at 01:06

As already pointed out by @dlegall, what is immediatlely sure is: the errors you're getting come from the fact that keywords is currently not set.

From there, two points:

Anyway your code must prevent this case, so you cannot simply use any $query->... in the bottom (HTML) part without surveying it.
So you might use something like:

<div class="col-md-4">
  <?php if (@$query) { ?>
    Found<?php echo $query->num_rows;?> results.
    <?php  
        if ($query->num_rows) {
            while($r = $query->mysqli_fetch_array())
            {
            ?>
            <div class="gylphicon glyphicon-heart">
                <a href="#"><?php echo $r->username;?></a>
            </div>
            <?php   }
        }?>
  <?php } else { ?>
    No keywords available.
  <?php } ?>
</div>

This way your application is now capable to respond to any situation.
Now you may test for what is currently causing keywords to be empty: this may be a totally different issue.

Notice: Trying to get property of non-object in C:\xampp\htdocs\php\search.php on line 39

2 Answers2