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+--- Thread: Simpler way to remove link [uri->segment()] (/showthread.php?tid=38275)
Simpler way to remove link [uri->segment()] - El Forum - 02-03-2011
[eluser]carvingCode[/eluser]
I want to remove the link on the current page. The code blow allows me to do that, but seems a bit excessive, especially as I build out the rest of the options (approx 20).
Is there a simpler way to do this?
Code:
if ($this->uri->segment(2) == 'add_section') {
echo "<li>Add section</li>";
} else {
echo "<li><a >Add section</a></li>";
}
if ($this->uri->segment(2) == 'add_indicator') {
echo "<li>Add indicator</li>";
} else {
echo "<li><a >Add indicator</a></li>";
}
.
.
.TIA
Randy
Simpler way to remove link [uri->segment()] - El Forum - 02-03-2011
[eluser]cideveloper[/eluser]
Using the below code you just need to add to the $data array your uri segments and text and it will generate you list items
Code:
$data = array (
'add_section' => 'Add section',
'add_indicator' => 'Add indicator',
'add_garbage' => 'Add garbage'
);
$this->show_menu($data, $this->uri->segment(2));
function show_menu($data_array, $uri_segment) {
foreach ($data_array as $key => $val){
if ($uri_segment == $key) {
echo "<li>$val</li>";
} else {
echo "<li>" . anchor($key,$val) . "</li>";
}
}
}Simpler way to remove link [uri->segment()] - El Forum - 02-04-2011
[eluser]carvingCode[/eluser]
Thanks. Indeed a simpler way.
I made a slight modification to handle a 'controller/function' type of menu call.
Code:
function show_menu($data_array, $first_uri_segment, $second_uri_segment) {
foreach ($data_array as $key => $val) {
if ($second_uri_segment == $key) {
echo "<li>" . $val . "</li>";
} else {
echo "<li>" . anchor($first_uri_segment . "/" . $key, $val) . "</li>";
}
}
}
$this->cc_functions->show_menu($data, $this->uri->segment(1), $this->uri->segment(2));I placed the function in a custom library in my project, hence the 'cc_functions' in the function call statement.
Thanks!
Randy