WPIntell

Source evidence

Incompatible with PHP 7?

Easy External Links · support · 2017-02-16T01:09:00+00:00

complaintsentiment
highseverity
0.97relevance
1replies
Evidence linked to opportunitycommercial context

Proof Health

Open evidence

Commercial opportunities need traceable source links before they are treated as build-worthy.

5 / 34 rows with source links

14.7% of this page's analysis has direct source links.

0 build-decision rows missing links

0 rows here require auditable proof before promotion.

29 rows with no attached evidence

0 rows have source counts but still need direct links.

Conversation

support
Rafael Fischmann unresolved
As soon as I enable it, my error_log is being spammed: php[error] [16-Feb-2017 01:08:09 UTC] PHP Notice: Undefined offset: 1 in /data/wp-content/plugins/wp-links/wp-links.php on line 175 php[error] [16-Feb-2017 01:08:09 UTC] PHP Notice: Undefined variable: wplinks_image in /data/wp-content/plugins/wp-links/wp-links.php on line 149 I hope you can fix this. Thanks. You should get rid of these error messages if you add $wplinks_image = isset($value[1]) ? $value[1] : null; in line 148 of wp-links.php right before the line if ( WPLINKS_NOFOLLOW == 'on' && !in_array($url_top_level, $WPLINKS_WHITELIST_ARRAY) ) return '<a href="'.$url.'" '.$wplinks_open.' rel="external nofollow" '.$wplinks_title.' '.$style.'>' . $matches[5] . '</a>'.$wplinks_image; and if you replace line 175 if ($results[1]) return true; with if(isset($results[1]) ? $results[1] : null) return true; This reply was modified 8 years, 11 months ago by cinematic .

Comments

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cinematic 2017-07-01T10:00:00+00:00

You should get rid of these error messages if you add $wplinks_image = isset($value[1]) ? $value[1] : null; in line 148 of wp-links.php right before the line if ( WPLINKS_NOFOLLOW == 'on' && !in_array($url_top_level, $WPLINKS_WHITELIST_ARRAY) ) return '<a href="'.$url.'" '.$wplinks_open.' rel="external nofollow" '.$wplinks_title.' '.$style.'>' . $matches[5] . '</a>'.$wplinks_image; and if you replace line 175 if ($results[1]) return true; with if(isset($results[1]) ? $results[1] : null) return true; This reply was modified 8 years, 11 months ago by cinematic .