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It works perfectly fine. But the $icon value in http://www..com/icons/$icon doesn't display it's self instade displays just the $icon not for example say the value of $icon is dude, then instade of displaying dude it just stays as $icon. If this makes sence to you then why is this?
It seems, and I'm going to have to repeat myself, that $icon isn't reterning the value of dude, but just returning the value of $icon. If I where to look at my souce code, it'll look like this:
Is it just me or does it seem like someone as some sorta grudge agiants me?
I not getting any error. My problem is is that my picture that is supposed to be displayed, is not. Cause the source is http://www..com/icons/$icon not http://www..com/icon/dude.
And maybe the fact that I didn't mention that the $array_block var is being displayed in an array in JS.
Ok finally I got it. But it I defenetly wouldn't of if it wheren't for you guys. Thanks you guys. This is something that I've been trying to figure out for a long time now. Thanks.
just remember that variables will be echoed out as what they are equal to when in double quotes but this will not happen in single quotes. it is generally the easiest to do it like:
because if you use a editor with color coding (which you should) you will be able to see that you have a variable there, otherwise it can be hard to spot.