PHP in HTML

I have a line of HTML code which inserts an image placeholder into a table and I have the img src stored in a PHP variable. For some reason I can not get the variable to echo anything. So when I view the page the sorce code shows <img src ="">. What am I doing wrong? (I have echoed the variable out elsewhere on the page and it echo's fine.)
<td> <img src="<? echo $FlagImage ?>" alt="Flag Image" name="FlagPlaceHolder" width="150" height="150" id="FlagPlaceHolder" /> </td>

Open in new window

dannyg280Asked:
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Roger BaklundCommented:
Try using full PHP tags:
<img src="<?php echo $FlagImage; ?>" alt=...

Open in new window

0
dannyg280Author Commented:
That did not affect it... I have php variable echoed throughout the page.. all others echo correctly. I have tried echoing the same variables that echo correctly on other parts of the page in the <img src...> and I still get nothing... here's a bit more of my code below: All the other vaiables are echoing fine...
<table width="487">
  <tr>
    <td colspan="4"><div align="center" class="style12">
      <p class="style4">Audit Report For Store #<? echo $StoreNum ?></p>
      <p class="style4"><? echo $City ?>, <? echo $State ?>&nbsp;</p>
    </div></td>
    <td> <img src='<?php echo $FlagImage; ?>' alt="Flag Image" name="FlagPlaceHolder" width="150" height="150" id="FlagPlaceHolder" /> </td>

Open in new window

0
Loganathan NatarajanLAMP DeveloperCommented:
Verify you have value in this variable, $FlagImage
0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
CompTIA Security+

Learn the essential functions of CompTIA Security+, which establishes the core knowledge required of any cybersecurity role and leads professionals into intermediate-level cybersecurity jobs.

Loganathan NatarajanLAMP DeveloperCommented:
May be try to echo $FlagImage , out side the img ... and check it
0
dannyg280Author Commented:
I have... when I echo it out I get:  ../HAN/images/RedFlag.png

UPDATE: when I echoed the $City vairable in the <img src > it did appear in the souce code on the webpage..  and as I said, when I echo $ImageFlag elsewhere on the page it echos out fine... I'm at a loss... I've tried changing the variable to just RedFlag.png and typing in the path and just echoing the image name but I get the same result... everything except the echoed value appears :(
0
dannyg280Author Commented:
This seems odd.... I have an external php file that I call VarConversions which I include() in the page. This file pulls my vairables from a database and changes the values for disply on the webpage. I have the code below in the VarConversions.php script...
If I manually define $FlagImage without using "if" staements, then it works fine, but when I use the if statements to define the variable it will not echo in the img src.
(Remember, the $FlagImage varibable DOES echo correctly outside of the img src, so I know the conversion is happening properly)
//Flag Image Coversion
if ($FlagColor =="G") {
$FlagImage = "../HAN/images/GreenFlag.png";
}
else if ($FlagColor =="Y") {
$FlagImage = "../HAN/images/YellowFlag.png";
}
else if ($FlagColor =="R") {
$FlagImage = "../HAN/images/RedFlag.png";
}

Open in new window

0
lharrispvCommented:
Try:

<img src=<?php echo "'" + $FlagImage; + "'"?> alt="Flag Image" name="FlagPlaceHolder" width="150" height="150" id="FlagPlaceHolder" />
0
dannyg280Author Commented:
I figured out my problem... and I apoligize for wasting your time (I will be assigning points to all who tried to help). The issue was my fault, the query I run to pull the $FlagColor (and other variables) from the database was actually running after I inserted the $FlagImage varible into the script.. so when I tried to have the image appear at the top of the screen, the variable did not have a value. When I tested it by echoing the varible on another part of the page, I was echoing it out further down, after the query had run.
   Thanks to all who offered help.
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
PHP

From novice to tech pro — start learning today.