Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in

Question

struggling with my web design assignment. I've been following a tutorial to add in a search feature for my website, but I've been getting the following error:

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in /search.php on line 31

line 31 is (or was)

<pre>if(mysqli_num_rows($results) >= 1)</pre>

That was the original error. as per instructions in the comments, I've since revised the code:

<pre>



    <?php

//capture search term and remove spaces at its both ends if the is any
$searchTerm = trim($_GET['keyword']);

//check whether the name parsed is empty
if($searchTerm == "")
{
    echo "Enter the name/brand of what you're looking for.";
    exit();
}

//database connection info
$host = "localhost";
$db_name = "sookehhh_shopsy_db";
$username = "sookehhh_shopsy";
$password = "xxxx";



//connecting to server and creating link to database
$link = mysqli_connect($host, $username, $password, $db_name) or die('Could not connect: ' . mysqli_connect_error());

//MYSQL search statement
$query = "SELECT * FROM sookehhh_shopsy_db WHERE name LIKE '%" . mysqli_real_escape_string($link, $searchTerm)  . "%'";

// original query$query = "SELECT * FROM sookehhh_shopsy_db WHERE name LIKE '%$searchTerm%'";

$results = mysqli_query($link, $query);

//added suggestion below - not sure if correct place?
if (!$result) {
    die(mysqli_error($link)); 
}

/* check whethere there were matching records in the table
by counting the number of results returned */
if(mysqli_num_rows($results) >= 1)
{
    $output = "";
    while($row = mysqli_fetch_array($results))
    {
        $output .= "Product Name: " . $row['name'] . "<br />";
        $output .= "Price: " . $row['price'] . "<br />";
    }
    echo $output;
}
else
    echo "There was no matching record for that item " . $searchTerm;
?>
</pre>

made necessary changes and updated yet again -

now the only error message I'm getting here is "Table 'sookehhh_shopsy_db.sookehhh_shopsy_db' doesn't exist"

I'm assuming that I need to change the username, perhaps because it's too similar?

Anywho, thanks for your help so far, and I apologise for my complete ignorance.

I've been trying to teach myself, but unfortunately time is a luxury I just don't have at the moment.

Hrm. Which related question to close this as. Decisions, decisions... — Ignacio Vazquez-Abrams, May 26 '13 at 05:52
also you are using shopsy_db as both the db and table name.. is that in fact the case ... if it's making it to the fetch before puking it means the query is baked. Possible wrong table name? — Orangepill, May 26 '13 at 05:56
no, the database name I've just shortened to keep my username private. The real database name is username_shopsy_db — Chrissy, May 26 '13 at 05:58

chrislondon · Answer 1 · 2013-05-26T06:51:14.607

17

The problem is your query returned false meaning there was an error in your query. After your query you could do the following:

if (!$result) {
    die(mysqli_error($link));
}

Or you could combine it with your query:

$results = mysqli_query($link, $query) or die(mysqli_error($link));

That will print out your error.

Also... you need to sanitize your input. You can't just take user input and put that into a query. Try this:

$query = "SELECT * FROM shopsy_db WHERE name LIKE '%" . mysqli_real_escape_string($link, $searchTerm) . "%'";

In reply to: Table 'sookehhh_shopsy_db.sookehhh_shopsy_db' doesn't exist

Are you sure the table name is sookehhh_shopsy_db? maybe it's really like users or something.

edited May 26 '13 at 06:51

answered May 26 '13 at 05:56

chrislondon

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Thanks for the helpful information. Unfortunately, I'm still getting another error message. Please see revised post for details. – Chrissy May 26 '13 at 06:34
sorry that's my bad. I've updated my mysqli_real_escape_string function – chrislondon May 26 '13 at 06:39
It needs to be mysqli_real_escape_string($link, $searchTerm) – chrislondon May 26 '13 at 06:40
Thanks! Now I'm just getting " Table 'sookehhh_shopsy_db.sookehhh_shopsy_db' doesn't exist". – Chrissy May 26 '13 at 06:47
That means the table you are trying to select doesn't exist. You have the table name the same as the database name. Is that correct? Maybe you have a table like `users` or something? – chrislondon May 26 '13 at 06:49
oh god I'm such a noob. it finally clicked that the database name and table name are different things. Thanks for everything! I've added in the correct table name and now I'm just getting a blank page instead of error messages... which is good... but also bad.... – Chrissy May 26 '13 at 06:56

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in

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