A specific question...and some general help needed.

TheWart

Hey all, I just signed up, after reading some of the helpful tutorials, and was wondering if you could help me with a little issue I am having.

I am helping this guy fix his website , as he has created it and updated it over the years using MS frontpage, and now it is a beast to update. Specifically, I am talking about the part on the right hand column, where you can list plants that are in the Southeast US. It is currently in a table in Frontpage, and to update it and sort it by Common Name, Scientific Name, etc, requires a lot more work than I thought it should.

So I decided to try and help, and read up a little, and created a MySQL database that has the same info as that table, except for the pictures.

Now, from what I can gather, there seems to be some argument over whether or not storing images directly into the database is a good thing. From my little knowledge of MySQL databses, I would think it is not the best idea, as I bet a lot of pictures would make it pretty bloated.

Anyway, I tried to enter it directly into the database anyway, to see if it would even work, and all I get is a huge string of characters outputted on the webpage. If you could tell me why, I would be greatly appreciative.

More importantly though, how would I go about simply make the DB link to the images on the filesystem? I think it would be something along the lines of making a standard text field in the MySQL table, and putting the path to the image as the entry, and then in the php code do something like <img src=$path1></img> I dunno, I may be way off, but for the life of me, I cannot find a guide on the net.

Phew, that was long.

I would greatly appreciate any and all help.

Thanks in advance.

MrPixel

Originally posted by TheWart
Anyway, I tried to enter it directly into the database anyway, to see if it would even work, and all I get is a huge string of characters outputted on the webpage. If you could tell me why, I would be greatly appreciative.

Mostly likely cause is that you are ouputing the raw graphic binary. You'll need to process that (probably with the GD library functions (imagecreate() and family)), and then output it with the appropriate headers. This can be more trouble than it's worth, and will create an extra load for the server (it'll have to process a bunch of photos as well as the database access script) ... for me, this is why it's not such a great idea - especially when you can simply just server the graphic straight up.

More importantly though, how would I go about simply make the DB link to the images on the filesystem? I think it would be something along the lines of making a standard text field in the MySQL table, and putting the path to the image as the entry, and then in the php code do something like <img src=$path1></img> I dunno, I may be way off, but for the life of me, I cannot find a guide on the net.

I do something like this. However, since all the photos associated with a given table are generally in the same directory, I don't store the path in the database. If you want to allow several different paths, you could include a path column, or perhaps easier, base the path on the databse columns instead of the other way around - this avoids having to create and process an extra column, while still giving you a reasonable amount of flexibility in organizing the graphics on the filesystem.

TheWart

Sorry to bring this back from the dead, but I have not had a chance to really work on it since I posted.

What I did, was in order to test was create a new table within my database called 'pictures'

In this table, I have 3 fields: id, name, pathtoimg.

The id is auto-increment, the name is self explanatory, and the pathtoimg is the path to the pictures. I have entered like: 'C:\site\test.jpg'

I have this as 'pics.php'

<?
//connect to database using the mysql_connect function()
//This is your host and then your username and then finally your password
$db = mysql_connect("localhost","user","password");

//The function mysql_select_db() sets the current
//database to the one specified in this case "test"
mysql_select_db("bioimages" ,$db);
//mysql_query() execute the specified MySQL query in this case
//we are are storing this in the variable $sql
$sql = mysql_query("SELECT * FROM herbplants order by 'Scientific Name'" ,$db);

//now we wil start producing a table
echo ("<table border ='1'>");
//this part displays our headings
echo ("<tr><td>ID Number</td><td>Scientific Name</td><td>Common Name</td><td>Family Name</td><td>Page Number in Guide</td><td>Start Bloom</td><td>Stop Bloom</td><td>Origin</td><td>Synonyms</td><td>Picture</td></tr>");

//the mysql_fetch_row() function fetches the next row in the resultset
//and stores this in an array , in this case $tablerows
while ($tablerows = mysql_fetch_row($sql))

{
//we now finish our table off by inserting values into each
//column . $tablerows[1] is equivalent to our url field and
//$tablerows[2] is the description field
echo("<tr><td>$tablerows[0]</td><td>$tablerows[1]</td><td>$tablerows[2]</td><td>$tablerows[3]</td><td>$tablerows[4]</td><td>$tablerows[5]</td><td>$tablerows[6]</td><td>$tablerows[7]</td><td>$tablerows[8]</td></tr> ");
}
echo "</table>";
//close database (good practice)
mysql_close($db);
?>

Now it shows the path output as a field, but how do i make it 'translate' the path into a picture?

I would be very appreciative if someone could help me out, as I kind of need to figure this out tonight.

Thanks a ton.

-aaron