find in nested array and print the corrosponding names - php

Question

I want to print the names of people when supplied the did.
I have used this to get the array :
$getdepts_another=mysqli_query($connection,
"select did,name,photo,dabout,fname,lname
from dept
where (acid='1' and name !='Admin')");
$deparray=array();
while($res = mysqli_fetch_row($getdepts_another))
{
array_push($deparray, $res);
}
$_SESSION['deps_array']=$deparray;
This is my array:
array(9) {
[0]=> array(6) {
[0]=> string(1) "2"
[1]=> string(8) "Accounts"
[2]=> string(29) "1851530771msstore_400x400.jpg"
[3]=> string(0) ""
[4]=> string(0) ""
[5]=> string(0) ""
}
[1]=> array(6) {
[0]=> string(1) "3"
[1]=> string(14) "Human Resource"
[2]=> string(9) "photo.png"
[3]=> string(0) ""
[4]=> string(0) ""
[5]=> string(0) ""
}
[2]=> array(6) {
[0]=> string(1) "4"
[1]=> string(9) "Documents"
[2]=> string(9) "photo.png"
[3]=> string(0) ""
[4]=> string(0) ""
[5]=> string(0) ""
}
[3]=> array(6) {
[0]=> string(1) "5"
[1]=> string(5) "Sales"
[2]=> string(9) "photo.png"
[3]=> string(0) ""
[4]=> string(0) ""
[5]=> string(0) ""
}
[4]=> array(6) {
[0]=> string(1) "6"
[1]=> string(7) "General"
[2]=> string(9) "photo.png"
[3]=> string(0) ""
[4]=> string(0) ""
[5]=> string(0) ""
}
[5]=> array(6) {
[0]=> string(1) "7"
[1]=> string(10) "Marketting"
[2]=> string(9) "photo.png"
[3]=> string(0) ""
[4]=> string(0) ""
[5]=> string(0) ""
}
And this is my function to search and print the names of did
function search($ids, $array) {
$idArray = explode(",", $ids);
$result = Array();
foreach($idArray as $id)
{
foreach ($array as $key => $val) {
if ($val['did'] === $id) {
echo "name:".$val['fame'];
array_push($result, $val['fame']);
}
}
}
return $result;
}//end function
$res = search('3,6', $array);
If I use search(1,2); it should print Accounts,Human Resource.

Answer 1

First of all, try using mysql_fetch_assoc instead of mysqli_fetch_row. That may be why there are no assoc keys in the array to search for.
$getdepts_another=mysqli_query($connection,
"
SELECT did,name,photo,dabout,fname,lname
FROM dept
WHERE (acid='1' and name !='Admin')
"
);
$deparray = array();
while($res = mysqli_fetch_assoc($getdepts_another))
{
array_push($deparray, $res);
}
$_SESSION['deps_array']=$deparray;
Secondly, in your search function, you are returing the key fame instead of fname.
function search($ids, $array) {
$idArray = explode(",", $ids);
$result = Array();
foreach($idArray as $id)
{
foreach ($array as $key => $val) {
if ($val['did'] === $id) {
echo "name:".$val['fname'];
array_push($result, $val['fname']);
}
}
}
return $result;
}//end function
$res = search('3,6', $array);
Hope this helps.