looping through a result set. How?

[fjm]

Hi all.

I have pictures that I want to include in a php page based on a boolean value in the db. if the value is 1, show this pic. if the value is 2, show that pic, if the value is 3 show that pic. What I need is for ALL of the pics that have values to show, NOT just one of them. If the db has values for 6 pics, then I need to see 6 pics.

I cannot get this to work. When I put my html into the where loop, it loops like 26 pictures, It duplicates them over and over. I know that it is something that I am doing wrong but I can't figure out what I need to do.

I was thinking to put all of the values into an array and then call them out? I can get my data into the array, but I don't know how to extract them. I am only working off of one field in the database. Each html line requires a different value from the db field.

[PHP] if($data['mydbfield'] == "165"){
echo "<img src="blabla" />;
}else{
echo "<img src="SOME OTHER IMAGE" />;
if($data['mydbfield'] == "136"){
echo "<img src="blabla" />;
}else{
echo "<img src="SOME TOTALLY DIFFERENT OTHER IMAGE" />;
}
[/PHP]

If anyone would be good enough to give a hand here, I would really be grateful.

Thanks,

Frank

[bergy]

Something like this?

[PHP]
$query = "select * from table";
$result = mysql_query($query) or die("Error");
while($obj = mysql_fetch_object($result)){
switch($obj->some_column_name){
case "123": $img = "someimg.jpg"; break;
case "456": $img = "otherimg.jpg"; break;
default: $img = "defaultimg.jpg"; break;
}
echo '<img src="'.$img.'">';
}
[/PHP]

Normally when looping through the result set of a database you will use a while() loop and a function like mysql_fetch_object, mysql_fetch_row , mysql_fetch_array, or mysql_fetch_assoc .

Check out all of those functions, they have examples of use and you can decide whether you're more comforatable working with objects or arrays.

[bergy]

I just wanted to add (since you didn't specify the database type), that those functions are for mysql only, there are similar functions for ODBC and other databases your server/php version may support. Check out php.net for more info.

[fjm]

Something like this?

[PHP]
$query = "select * from table";
$result = mysql_query($query) or die("Error");
while($obj = mysql_fetch_object($result)){
switch($obj->some_column_name){
case "123": $img = "someimg.jpg"; break;
case "456": $img = "otherimg.jpg"; break;
default: $img = "defaultimg.jpg"; break;
}
echo '<img src="'.$img.'">';
}
[/PHP]

The way I am using the above code, it is only giving me the first image 2.jpg *if the variable is in the db*. For the case where no variable exists, I need it to show the off image.

Thansk

Normally when looping through the result set of a database you will use a while() loop and a function like mysql_fetch_object, mysql_fetch_row , mysql_fetch_array, or mysql_fetch_assoc .

Check out all of those functions, they have examples of use and you can decide whether you're more comforatable working with objects or arrays.

Hi Bergy,

Thanks for the help. Unfortunately, it doesn't work. I was up for about 16 hours and it just slipped my mind to mention that I am using mysql. I also should have mentioned that I am already using a while loop to iterate through the data.

Here is what I am trying to do. Using your code below:

[PHP]
case "123": $img = "someimg.jpg"; break;
[/PHP]

if 123 is TRUE, echo this image, if 123 is FALSE, echo that image. ... on to the next case.

Here is one case I have:

[PHP] case "2":
if($val == '2'){
echo '&nbsp;<a href="index.php?a=2"><img src="images/2.gif" class="img" alt="" /></a>';
}else{
echo '&nbsp;<img src="images/off_2.gif" class="img" alt="" />';
}
break;

case "3" ......[/PHP]