Images

[Anon]

Can anyone help me figure out the best way to load images into a mysql db? I use Access on my PC then usually export as a .txt file but that won't work with images... will it? Should I export as something else or use a different syntax for the "load data" command?

[[deleted]]

giggle
This is exactly why you should not store images in a database.

I don't know if access will make the binary data ASCII safe, but I somehow doubt that very much.

I think your safest bet is to write some PHP to read the data directly from the access database (don't you just lovce M$ products right now? :-)

[Anon]

I wish I didn't have to store images but for the info stored in my db, icons are a must (according to my boss). If it's so bad to have images stored in the db.. is there anyway I can have a field in mysql that calls the directory for the image saved elsewhere on the server?

[cheald]

That's what I do. I just have a specified image directory, and store filenames in the database.

Makes them more accessable, more easily updated, and the DB easier to maintain.

[Anon]

Can you show me a quick example of how you store/retrieve the file reference?

[Anon]

The way I do it is I store a pointer in the database. there are two ways

1. <img src="images/image.jpg"> store this in the database or:

2. image.gif. just put this in the database and put the rest of the tag in the page like this:

<img src="images/<? echo "$image"; ?>">

I do it the second way so I don't have to type as much in the db, but the problem with the second way is if there is not graphic, it will still look for one, so you would have to have a blank.gif and put that in the db. I then keep all the pics in a separate folder.

[Anon]

Ok... How would I fit the "echo" statement into my existing output... I have been trying to tweek it but keep falling short.

if ($myrow = mysql_fetch_array($result))
{

echo "<TABLE BGCOLOR=#808000 BORDER=5 CELLPADDING=2 CELLSPACING=2>\n";

echo "<tr><th>Trip Name</th><th>Region</th><th>Country</th><th>Description</th><th>Cost</th><th>More Info</th><th>Icon</th><th>Icon</th>";

do{
$trip_id=urlencode($myrow["trip_id"]);
printf("<tr><td bgcolor=#C0C0C0>%s</td><td bgcolor=#C0C0C0>%s</td><td bgcolor=#C0C0C0>%s</td><td bgcolor=#C0C0C0>%s</td><td bgcolor=#C0C0C0>%s</td><td bgcolor=#C0C0C0>%s</td><td bgcolor=#C0C0C0>%s</td></tr>\n",
$myrow["trp_name"], $myrow["region"], $myrow["country"], $myrow["description"], $myrow["cost"], "<a href='more_info.php?trip_id=$trip_id'>More Info</a>", "<img src="$myrow['icon']">");

}
while ($myrow=mysql_fetch_array($result));
echo "</table>\n";
}
else
{
echo "Sorry, no records were found.";
}

[btaylor]

in your file it looks like your images are stored at the same level as your page. Depending on what you put in your database,
a normal image tag would look like this:
<img src="images/image_name.jpg"> so are you missing the images/

<img src="images/$myrow['icon']">

brett

[Anon]

I am getting this.

Parse error: parse error, expecting T_STRING' orT_VARIABLE' or `T_NUM_STRING' in /usr/local/etc/httpd/htdocs/idema/New_site/take_out/search_options2.php on line 37

echo "<TABLE BGCOLOR=#808000 BORDER=5 CELLPADDING=2 CELLSPACING=2>\n";

echo "<tr><th>Trip Name</th><th>Region</th><th>Country</th><th>Description</th><th>Cost</th><th>More Info</th><th>icon</th>";

do{
(THIS IS 37)printf("<tr><td bgcolor=#C0C0C0>%s</td><td bgcolor=#C0C0C0>%s</td><td bgcolor=#C0C0C0>%s</td><td bgcolor=#C0C0C0>%s</td><td bgcolor=#C0C0C0>%s</td><td bgcolor=#C0C0C0>%s</td></tr>\n", $myrow["trp_name"], $myrow["region"], $myrow["country"], $myrow["description"], $myrow["cost"], "<a href='more_info.php'>More Info</a>", "<img src='www/htdocs/images/$myrow["icon"]'>");

}
while ($myrow=mysql_fetch_array($result));
echo "</table>\n";
}
else
{
echo "Sorry, no records were found.";
}

[Anon]

try assigning your array values to a varible like this.

$region = $myrow["region"];

then print the $region varible in line 37