Smarty substr string skip first characters

Question

The variable `$siteroot` contains a string that looks like: http://www.example.nl/folder/ To get the part http://www.example.nl/ you use: `{$siteroot|substr:0:22}` But i want to skip this part and return /folder/ The length of the folder is unknown.

Accepted Answer

```
{$siteroot|substr:22}
```

Should work

Smarty substr string skip first characters

Solution

Problem