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calling function name? i need help
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hi;
please help to check this code. i have a fucntion name to use in my select dropdown list, and I problem in calling the the function name;
here is the whole code of my function.
function getAllR($Region_id) {
//returns an array with city info
$query="SELECT * FROM cityinfo where Region_index='$Region_id'";
$result=mysql_query($query);
if (!$result)
return false;
else {
$i=0;
$returnArray=array();
while ($row=mysql_fetch_array($result)) {
$returnArray[$i]=new city();
$returnArray[$i]->City_id=$row['City_id'];
$returnArray[$i]->Region_index=$row['Region_index'];
$returnArray[$i]->displayName=$row['City_name'];
$i++;
}
return $returnArray;
}
}here the whole code of my dropdown list
$cities=region::getAllR();
if (!isError($cities)){
foreach ($cities as $element => $value) {
echo '<option value="'.$value->City_id.'"';
if ($value->City_id=='$city')
echo ' selected="selected"';
echo '>'.$value->displayName.'</option>';
}
}
}•
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what is wrong in this code? why it does not display the item for the dropsown list?
please i need your help, thanks advance
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Join Date: Aug 2007
Posts: 805
Reputation: 
Solved Threads: 110
Practically a Posting Shark
Just taking a quick look, you have not passed in a parameter to the region::getAllR() call. Note that the signature of the function is function getAllR($Region_id) which contains a single parameter with no default value set. Therefore you must call the function like so: region::getAllR($aRegionId);
There are no stupid questions, only those too stupid to ask for help.
echo is a web developer's best friend. •
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Originally Posted by darkagn 
Just taking a quick look, you have not passed in a parameter to the region::getAllR() call. Note that the signature of the function is function getAllR($Region_id) which contains a single parameter with no default value set. Therefore you must call the function like so: region::getAllR($aRegionId);
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