Question

EDIT: I had a typo in my original post....the issue is a bit more complicated...i had a variable passed in not a raw string.

I want to print out stories from a mysql database that are specific to a certain person:
so i have code that is similar to:

$stuff ="jamie"
    $query = "SELECT * FROM person_stories WHERE person =$stuff";
    $result = mysql_query($query) or die ("didnt work");
    while($row = mysql_fetch_array($result))
    {
    echo "<a href = 'PersonStoryPage.php?pid=$row[id]'>" .$row['title']. " </a>";
    }

I keep on getting “didnt work” …I know that my table person_stories is empty but is this the same thing as an error? The table will obviously not always be empty so I need to be able to use this block of code to go about business.

Help is appreciated!

EDIT 2: The actual error is:

Unknown column 'jamie' in 'where clause'

This is bizzare since it shouldn’t be interpreting jamie as the column!

Answer 1

You didn’t put single quotes around jamie. Try this:

$query = "SELECT * FROM person_stories WHERE person = 'jamie'"

Edit:

I see the post has been edited. It should now change from this:

$stuff ="jamie"
$query = "SELECT * FROM person_stories WHERE person =$stuff";

to something like this:

$stuff ="jamie"
$query = "SELECT * FROM person_stories WHERE person='" . mysql_real_escape_string($stuff) . "'";

This will not only solve your SQL syntax error, but also protect your app from a nasty SQL injection vulnerability.