die - All output should be run through an escaping function

[kkmuffme]

Getting WordPress.Security.EscapeOutput.OutputNotEscaped for following code:

$var = 'hello';
echo $var == 'foo' ? 'bar' : die( 'world' );

echo 'example';

while using die( 'world' ); standalone, does not produce this error.

[meevly]

@kkmuffme it's not about die ( 'world' ), you get the error because of echo 'example'; not being escaped (and likely 'bar' as well). I don't think it's a bug.

[jrfnl]

@meevly Sorry, but that's not true. example and bar are hard-coded strings and don't need to be escaped.

I suspect the problem is with the ternary, but this will need further investigation. If I remember correctly, there are a number of other issues open about ternaries and output escaping.