Display only if added

69 Posts

GhostOfDixie Reply #1, 16 years, 10 months ago

I want to know if I can display an <img> tag only if I have actually added an image into that resource field.

I have a resource that has has an img field as an option, but if I don’t add one I want that img tag to not appear.

Any ideas how I go about it?

Thanks

116 Posts

j_p Reply #2, 16 years, 10 months ago

You can do this with PHX.

First install PHX, if you haven’t already:

http://modxcms.com/extras/package/?package=342

Then you can add this to your template where your <img> tag would usually be:

[+phx:if=`[*your-image-tv*]`:is=``:then=``:else=`<img src="[*your-image-tv*]" alt="" />`+]

There are probably other ways of doing it.

69 Posts

GhostOfDixie Reply #3, 16 years, 10 months ago

Thanks, but I managed to do it like this:

<?php
$image = isset($image) ? $image : "";
if ($image != "") {
    $output = "<img src=\"".$image."\" alt=\"manager Image\"  />";
} else {
    $output = "";
}
return $output;
?>

Now I am trying to do the same thing with tabular data. I basically have tabular data, but sometimes the format of the table is different so I want to be able to enter a custom table. I tried this:

<?php
$custom-table = isset($custom-table) ? $custom-table : "";
if ($custom-table != "") {
    $output = "";
} else {
    $output = "<table id="costs" summary="Monthly Costs">
<thead>
<tr>
<td class="label">Rent</td>
<td>£[*rent*]</td>
</tr>
<tr>
<td class="label">Service Charge</td>
<td>£[*service-charge*]</td>
</tr>
<tr>
<td class="label">Total Monthly Charge</td>
<td>£[*total-monthly-charge*]</td>
</tr>
</table>";
}
return $output;
?>

But it’s just not showing up.

This is how I am calling it:

[!getTable? &custom-table=``!]

195 Posts

jusicgn Reply #4, 16 years, 10 months ago

I think you have to enclose the HTML for the output in single quotes instead of double quotes.

Greetz
JUSI

69 Posts

GhostOfDixie Reply #5, 16 years, 10 months ago

Hmm, tried that and I get this error:

Parse error: syntax error, unexpected ’=’ in /home/alphahom/public_html/manager/includes/document.parser.class.inc.php(770) : eval()’d code on line 1

In fact I get that error with the double quotes.

116 Posts

j_p Reply #6, 16 years, 10 months ago

Shouldn’t

if ($custom-table != "")

be

if ($custom-table = "")

69 Posts

GhostOfDixie Reply #7, 16 years, 10 months ago

Well I want to show the custom table if the custom-table field is not empty, otherwise I want to show the main table.

69 Posts

GhostOfDixie Reply #8, 16 years, 10 months ago

Anybody know why I am getting this error:

Parse error: syntax error, unexpected '=' in

788 Posts

breezer Reply #9, 16 years, 10 months ago

I think you should try replacing the dash (-) in the variable name with an underscore and see if that helps:
change this:
$custom-table

to this:
$custom_table

Maybe give this snippet a try and see if it helps:

<?php
/*  
     [!getTable? &custom_table=``!]
*/
$output ='';

$custom_table = isset($custom_table) ? $custom_table : '';
if ($custom_table != '') {
    $output .= '';
} else {
    $output .= '<table id="costs" summary="Monthly Costs">
<thead>
<tr>
<td class="label">Rent</td>
<td>&#163;[*rent*]</td>
</tr>
<tr>
<td class="label">Service Charge</td>
<td>&#163;[*service-charge*]</td>
</tr>
<tr>
<td class="label">Total Monthly Charge</td>
<td>&#163;[*total-monthly-charge*]</td>
</tr>
</table>';
}
return $output;
?>

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69 Posts

GhostOfDixie Reply #10, 16 years, 10 months ago

Hi thanks,

I decided in the end to simplify the way I processed the data. But thanks anyway.