PHP/Smarty giving 'Array' instead of correct value?

Question

Here is my code, from the search results page:

{if $listing.SalaryType!=''&& $listing.Salary!=''}<dt>Salary:</dt>
<dd>{$listing.Salary} {$listing.SalaryType}</dd>{/if}

Which gives this output:

Screenshot

It says 'Array' instead of the correct value. Any idea where I can start looking to solve this problem? Where is the 'Array' coming from?

Here is some more information. On the job details page, this code:

<div class="smallListingInfo"><strong>[[FormFieldCaptions!Salary]]:</strong> {display property=Salary} [[$listing.SalaryType]]</div>

Gives this output:

second example

You can see the salary value works in this instance, how is that different?

Using the smarty debug console I got the following:

Title => Services Senior Manager
EmploymentType => Contractor
JobCategory => Accounting,Admin-Clerical
Occupations => empty
Country => United States
Salary => Array (5)
  add_parameter => 2
  value => 30000
  currency_sign => £
  course => 1
  currency_code => USD
State => California
SalaryType => per month
City => Los Angeles

How do I get the value to show instead of 'Array'?

Busydude · Accepted Answer · 2010-11-09 10:46:49Z

1

$listing.Salary seems to be an Array.

Insert {debug} in your HTML code for dumping the debug console wich shows you the assigned variables.

Search for "debug" in the Smarty documentation (sorry, i can't post links here right now)

How to access elements of arrays in Smarty can also be found in the docs, just search for "variables"

answered Nov 9, 2010 at 10:46

Busydude

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2 Comments

Busydude Over a year ago

{if $listing.SalaryType!=''&& $listing.Salary!=''}<dt>Salary:</dt> <dd>{$listing.Salary.value} {$listing.SalaryType}</dd>{/if}

Busydude Over a year ago

optional you may use {$listing.Salary.currency_sign}

deceze · Accepted Answer · 2010-11-09 09:53:29Z

1

It means the {$listing.Salary} value is an array instead of the value you expect. When PHP is asked to treat an array as a string, it'll use the value "Array". Fix your data structures and/or your template to use the right value.

$foo = array();
echo $foo; // echoes "Array"

answered Nov 9, 2010 at 9:53

deceze♦

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Comments

Eric · Accepted Answer · 2010-11-09 10:00:08Z

0

{$listing.Salary} is an array.

In the PHP code, the object/array you pass to the template : 'Salary' index or attribute should return an integer and not an array.

answered Nov 9, 2010 at 10:00

Eric

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