Selecting row from DB using mysql/php

Question

I'm trying to get a row from the DB using php, i've made an html form that's supposed to take a book title from users and gets the review from the DB about this book, and then post it in an input text, the form's action leads to the following function :

function GetReview($BookTitle)
{
require'DB.php';
if(empty($_POST['BookTitle']))
    {
        echo " You must enter a book name!";
        return false;
    }

$BookTitle = mysql_real_escape_string($BookTitle);

$q="Select Reviews from Users_Booklist where (Book_Title like '%" .$BookTitle."%');";
if(!mysql_query($q,$con))
{
die("Error".mysql_error());
}
else
{
$row = mysql_fetch_row($q);
?>
<html>
<head><title>Delete Review </title>
</head>
<body> 
<br>
<form name="DeleteReview " action="DeleteReviewsFunction.php" method="post">
Review: <input type="text" name="Review" size="200" value="<?php echo $row[0]; ?>"/>            
<input type="submit" value="Delete Review" />
</form>
</body>
</html>
<?php
}
}
GetReview($_POST['BookTitle'])

However, it leads me to the next form with nothing in the input text and this warning:

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in     C:\AppServ\www\GetReview.php on line 20

I've searched and tried different code but still same result. Could anyone please tell me where the error is???... Thanks

The first example of br2.php.net/manual/en/function.mysql-fetch-row.php shows the reason — Tiago Peczenyj
– Tiago Peczenyj, Commented May 13, 2012 at 16:35
You're trying to fetch a result form your query STRING. You need to fetch from the value returned by the mysql_query() call. — Marc B
– Marc B, Commented May 13, 2012 at 16:42
Please don't use mysql_* functions in new code. They were removed from PHP 7.0.0 in 2015. Instead, use prepared statements via PDO or MySQLi. See Why shouldn't I use mysql_* functions in PHP? for more information. — vascowhite
– vascowhite, Commented May 13, 2012 at 18:10

bfavaretto · Accepted Answer · 2012-05-13 16:33:37Z

3

$qq = mysql_query($q,$con);
if(!$qq) {
// (...)

$row = mysql_fetch_row($qq);

answered May 13, 2012 at 16:33

bfavaretto

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ACarter · Accepted Answer · 2012-05-13 16:35:18Z

0

I'm not going to be a lot of help, but your question seems to be where the error is occuring, and I can tell you that.

It's in the $row = mysql_fetch_row($q); line.

You can tell this because the error record starts with mysql_fetch_row(), and the above line is the only mention of mysql_fetch_row() in the code.

answered May 13, 2012 at 16:35

ACarter

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bitfox · Accepted Answer · 2012-05-13 17:16:01Z

0

Check the SQL query by printing the output of $q variable with:

echo $q;

Now, try to execute it from your MySQL client. Collect the results (if there are) and check for errors.

A suggestion: If you want, you can use a tool like ezSQL that can be very useful (especially for code organization)

answered May 13, 2012 at 17:16

bitfox

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