You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
{{ message }}
This repository has been archived by the owner on Jul 17, 2022. It is now read-only.
Hi everybody,
I'm currently trying to have an individual style (e.g. tr style = 'color: red;') for a row in the result list depending on the value of one column (column name: "risk") in my mysql table.
I tried to find a solution within the template in a way like it is done with not displaying one column ($style = (strtolower($columnName) === 'id') ? 'display: none;' : '';)
I thought about having something like: if ($risk === 'high') {$style = 'color: red;'}
Of course, there is no variable called $risk yet, so the question is where to find the result of the mysql query in which the column "risk" is read and its value can be used for my if condition.
Any ideas anyone?
I think it must be done in this part of the code somewhere:
Hi everybody,
I'm currently trying to have an individual style (e.g. tr style = 'color: red;') for a row in the result list depending on the value of one column (column name: "risk") in my mysql table.
I tried to find a solution within the template in a way like it is done with not displaying one column (
$style = (strtolower($columnName) === 'id') ? 'display: none;' : '';
)I thought about having something like:
if ($risk === 'high') {$style = 'color: red;'}
Of course, there is no variable called $risk yet, so the question is where to find the result of the mysql query in which the column "risk" is read and its value can be used for my if condition.
Any ideas anyone?
I think it must be done in this part of the code somewhere:
Thanks a lot for your help!
The text was updated successfully, but these errors were encountered: