Question

I am getting this error here is my code

if(isset($_POST['submit']))
    {
     $projTit=mysql_escape_string($_POST['projecttitle']);
     $projCat=mysql_escape_string($_POST['projectcategory']);
     $budget=intval(mysql_escape_string($_POST['budget']));
     $description=mysql_escape_string($_POST['editor1']);
     $query=sprintf("insert into projects value('%s','%s','%s',%d)",
            $projTit,$description,$projCat,$budget);
     if (!mysql_query($query)){
         die('Error: ' . mysql_error());
         }
        echo '<p class="record">Your Record has been Added<p>';
    }

    ?>

I have tried to write %d in '' but still not working.

Answer 1

Your table has five columns. However, you are not providing a value for the Project_Id column. This gives you the error you mentioned.

I understand that you’re not providing a value since it’s likely a PRIMARY KEY that autoincrements. To tell MySQL that you are intentionally not passing a value for that column, you should add NULL as the first value.

value(NULL, '%s','%s','%s',%d)

However, you really should specifically name which columns you are inserting into in case you add a new column at a future date.

INSERT INTO projects (col1, col2, col3, col4) value ('%s','%s','%s',%d)