[SOLVED] Help with defining an instance for if statement

[grozanc]

Hello,

 

I'm trying to use an if statement to either display a jpg or flv image that has the link stored in a database. The file names will vary, but the .jpg or .flv will always be the same. How do I tell the if statement to ignore everything before the .jpg or .flv file? Below is an example of what I'm using.

 

if ($row['image1'] == jpg) { echo "<img name=mainimage src='".$row['image1']."'><br>"; }

 

Basically, what should I put in place of "jpg" so it will distinguish between file types?

 

Thanks,

Gary

[Dragen]

if(strrchr($row['image1'], '.') == '.jpg'){
    echo "<img name=mainimage src='".$row['image1']."'><br>";
}else{
    //it's a .flv
}

[craygo]

if (strrchr($row['image1'], ".") == ".jpg") { echo "<img name=mainimage src='".$row['image1']."'><br>"; }

 

strrchr will return everything from the last instance of the given character. in this case the "."

 

Ray

[Dragen]

if (strrchr($row['image1'], ".") == ".jpg") { echo "<img name=mainimage src='".$row['image1']."'><br>"; }

 

strrchr will return everything from the last instance of the given character. in this case the "."

 

Ray

yes, which should always be '.jpg' or '.flv' in this case.