don't show image placeholder

desertdirk

I have a repeat region that contains an image which is dynamically linked to my data base. The database hold the path to the actual image, not the image itself.

I want to display the page so that if there is no data in the database table, nothing will display. As it is now, I get the image placeholder icon (box with red X) So how do I get rid of the box?

I have this:

  <?php  $row_rshousemaster['pict1'] = str_replace("../", "", $row_rshousemaster['pict1']); ?>
            <?php if (($row_rshousemaster['pict1'] != "[NULL]") || ($row_rshousemaster['pict1'] != "")) { ?>
            <img src="<?php echo $row_rshousemaster['pict1']; ?>"  alt="" name="pict1"> 
            <?php  } ?>

I used both "null" and "" in my if statement because I wasn't sure which would be returned.
But I still get the image icon
Any suggestions?
thanks
dd

stolzyboy

<?
if ($row_rshousemaster['pict1'] == "")
{
//don't spit out image tag
}
else
{
//spit out the image tag
}
?>

if your not sure what it brings back, do this

echo $row_rshousemaster['pict1'];

that will show you each time what it equals

desertdirk

yarg

<?<? 
if ($row_rshousemaster['pict1'] == "") 
{ 
//don't spit out image tag 
} 
else 
{ 
<img src="<?php echo $row_rshousemaster['pict1']; ?>"> //spit out the image tag 
} 
?>

gives me a parse error

Parse error: parse error, unexpected '<' in

am I putting it in the wrong place?
Sorry I am sure this is so simple
DD

stolzyboy

change this

to this

echo "<img src=\"" . $row_rshousemaster['pict1']; . "\">";

desertdirk

echo "<img src=\"" . $row_rshousemaster['pict1']; . "\">";

gave me a parse error, but I took out the last "." which did work.

Can you tell me what that does? Is it refrencing the variable outside the if-else statement?

So much to learn

Many thanks stolzyboy

desertdirk

Well I spoke too soon.
This last bit of code will not show the image place holder when there is no image, but it won't show the image now when there is one.

<?php

if ($row_rshousemaster['pict1'] == "[NULL]") 
{ 
//don't spit out image tag 
} 
else 
{ 
echo "<img src=\"" . $row_rshousemaster['pict1'];  "\">";
//spit out the image tag 
} 
?>

stolzyboy

what do you get when you simply echo out

echo $row_rshousemaster['pict1'];

what do you get when there is an image there and what do you get when their isn't

do you get graphics/image23423.gif when there is and nothing when there isn't, i think your [NULL] may be wrong, unless you are specifically storing that in the db, you should just be able to check for "" (nothing) unless as i said, you stored it like that or you specified a default, in that case, check the default value

desertdirk

When I echo $row_rshousemaster['pict1']; I get "[NULL]" when there is no data (actually the data is "NULL") and I get "/house/100_001.jpg" when there is data.

As far a the null issue. I allowed null in my table but did not set it as a default and did not specify it in any way. It may be a quirk of the "MYSQL Control Center" that i am using to manage my data base.

Other than that, I am still not getting an image when there is one listed in the data base. And when I try to add a line following the ELSE statement as in:

<?php 

if ($row_rshousemaster['pict1'] == "[NULL]") 
{ 
echo "<img src=\"" . $row_rshousemaster['pict1'];  "\">";//don't spit out image tag 
} 
else 
{ 

<img src="<?php echo $row_rshousemaster['pict1']; >"> //spit out the image tag 
} 
?>

I get a parse error: unexpected "<"

I think I mentioned that this bit of code exsist in a repeat region (do-while loop) if that makes any difference.
DD

desertdirk

OK
finally got it I think. Changed my database so that it there is "" instead of [NULL]

now I can go back to :

<?php
if ($row_rshousemaster['pict1'] !="") 
{ 
  ?> 
  <img src="<?php echo $row_rshousemaster['pict1']; ?>">
  <?php   

} 
?>

Thanks for your help Stolzyboy