php code:
1. if (isset($data['city_id']))
2. {
3. $city_id = "city_id='". $data['city_id']. "', ";
4. }
And I get:
Notice: Undefined index: city_id on line 3
How can this be?
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u_mulder
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Make sure there's no typo in `city_id` – u_mulder Sep 28 '17 at 07:30
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`print_r($data)` and show all result – Bilal Ahmed Sep 28 '17 at 07:30
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In $data array in this situation isn't index 'city_id' ant this is ok, so I use "isset": if (isset($data['city_id'])) – Evaldas Jablonskas Sep 28 '17 at 07:33
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1Erase all 4 lines of code and type from scratch again and the problem will be gone. $5 – Hanky Panky Sep 28 '17 at 07:37
1 Answers
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Just ran your code sample and it works perfectly, I do not get 'undefined index' error - taking us to the big apple
<?php
$data['city_id']='New York';
if (isset($data['city_id']))
{
$city_id = "city_id='". $data['city_id']. "', ";
echo $city_id;
}
?>
output: city_id='New York',
Surely, without the $data['city_index']='New York'; I just get a blank screen, as the if condition is not met - no errors.
lovelace
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Surely, without the `$data['city_index']='New York';` I just get a blank screen, as the if condition is not met - no errors. – lovelace Sep 28 '17 at 07:38
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when You write: $data['city_id']='New York'; not necessary use: if (isset($data['city_id'])) , because You know exactly that this index is. I always use "isset" and this helped me. But now I have this notice. – Evaldas Jablonskas Sep 28 '17 at 08:24