Image Question

[James0816]

I have my page set up to pull image locations from the table and of course display accordingly. Here's my problem.....When I first load the page there is no image to display (of course since no record has been selected at this time). What does get displayed is:

<img src='

Of course that looks bad. Here is my code pulling from the selected value:

<img src='<?php echo mapPath($Test3->Value("picloc"), "./")?>' alt='No Picture Available' height="247" width="415" border="0">

How can I get it to either display a default image or nothing at all until a selection is made?

Thx,
Jim

[rachel2004]

If you want to display a "default" image, you'd probably want to nest your <img> tag into an if/else conditional:

if (!empty($request_for_pic))
{
    echo "<img src=\"some_pic_here.jpg\">";
}
else
{ // no request yet for pic, so display default
    echo "<img src = \"my_default_graphic.jpg\">";
}

[James0816]

Tried this but it errored on me:

<?php
if (!empty($Test3->Value("picloc")))
{
echo mapPath($Test3->Value("picloc"), "./")' alt='No Picture Available' height="247" width="415" border="0">;
}
else
{ // no request yet for pic, so display default
echo "<img src = \"Images\default.jpg">";
}
?>

The error message is a parse error on the !empty line.

[rachel2004]

I'm not sure what you're trying to echo on your first echo statement, but the problem lies in the use of single and double quotes, not with your !empty... boolean expression.

Try using concatenation with the dot (.) between your string (in quotes) and your function mapPath().

Also, your "alt" needs double quotes.

You need to escape the double quotes in your second echo statement right after the .jpg:

default.img\">";

[James0816]

You'll have to forgive me as I am not familiar with the world of PHP but trying to learn it.

The exact error message is:

Parse error: parse error, unexpected '(', expecting ')' in E:\TestSite folder\TestSite\Test2.php on line 103

with line 103 being the line with the !empty statement on it.

the first echo statement is the value in the datasource ($Test3 is my datasource and picloc is the field name).

[rachel2004]

See if this works:

<?php
if (!empty($Test3->Value("picloc")))
{
    $path = mapPath($Test3->Value("picloc");
    echo '<img src = '.$path.' alt="No Picture Available" height="247" width="415" border="0">';
}
else
{ // no request yet for pic, so display default
    echo "<img src = \"Images\default.jpg\">";
}
?>

[James0816]

unfortunately it didn't go. same error message.

recon it won't let you use that datasource value if you don't know it as yet?

[rachel2004]

What do you get if you do:
echo $path;

[James0816]

same thing...will not get past the !empty statement.

[rachel2004]

We need to find out if $path is storing a value:

echo $path;

Put that line directly below your:

$path = mapPath( ... stuff

Let me know what you get when you do echo $path.

My other suggestion is to check if there is a problem before line 103 with some other part of your code where you forgot a curly brace, a closing parenthesis, or a semicolon, closing double quote, etc.

You can check that by completely commenting out the block of code that you just inserted:

/*
if (!empty($Test3->Value("picloc")))
{
    $path = mapPath($Test3->Value("picloc");
    echo '<img src = '.$path.' alt="No Picture Available" height="247" width="415" border="0">';
}
else
{ // no request yet for pic, so display default
    echo "<img src = \"Imagesdefault.jpg\">";
}
*/

If you do that, do you still get the error message?

[James0816]

I'm getting somewhere now....I think....here's what I just did:

if (!empty($Test3->picloc))

since i'm not actually displaying a value but checking for one.

loaded the page and it looked like it kinda worked i didn't change the value for the default pic yet and got the little red "x" symbol. I then clicked on a link and it displayed the record information but again the little red "x".

So now I need to get down the format for displaying the image properly and it might work out? Since I can't put in a PHP within a PHP that doesn't work.

[James0816]

Here's the recently modified code. It doesn't error out but it doesn't display the pics:

<?php
if (!empty($Test3->picloc))
{
$path = mapPath($Test3->Value("picloc"));
echo '<img src = '.$path.' alt="No Picture Available" height="247" width="415" border="0">';
}
else
{ // no request yet for pic, so display default
echo "<img src = \"Images\default.jpg\">";
}
?>

[James0816]

ok...so my possibly working theory was incorrect.

I put in echo "not working" for when a value is not found but that's all that displays even after selecting an item. So eveidentally I need to have the $Test3->value("picloc") item in the if (!empty) statement.

:bemused:

[James0816]

WooHoo! I got it to work....here's my code:

<?php
$tester=$Test3->Value("picloc");

if(empty($tester)) {
$path = "Images//default.jpg";
}
else {
$path = $tester;
}
echo '<img src = '.$path.' alt="No Picture Available" height="247" width="415" border="0">';

/*
?>

Does this look like acceptable coding practices?

[rachel2004]

I cleaned it up a little bit. Check out those double forward slashes I remarked about in the code:

<?php

if(empty($tester)) {
    $path = "Images//default.jpg"; // recheck the double forward slash in your path...
    $alt = "Some default caption here.";
}else {
    $path = $Test3->Value("picloc");
    $alt = "Some caption for this picture";
}
echo '<img src = '.$path.' alt="'.$alt.'" height="247" width="415" border="0">';

?>