{"id":16620,"date":"2022-10-21T17:51:48","date_gmt":"2022-10-21T12:21:48","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-reducing-arrays-of-floats-from-datetime-closed\/"},"modified":"2022-10-21T17:51:48","modified_gmt":"2022-10-21T12:21:48","slug":"solved-reducing-arrays-of-floats-from-datetime-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-reducing-arrays-of-floats-from-datetime-closed\/","title":{"rendered":"[Solved] Reducing arrays of floats from datetime [closed]"},"content":{"rendered":"

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So your problem was of using dateinterval<\/a> object to get the time difference. But them you just add the hours and minutes up but this is “base 60” (as 60 minutes in 1 hour) so you cannot add them up with +<\/code>.<\/p>\n

Example<\/strong>: 1 hour and 15 minute (1.15<\/code>) plus 1 hour and 45 minute (1.45<\/code>) are 2.6<\/code> numerically speaking but 3 hours (as from your comment). <\/p>\n

In order to solve that you need to convert to numerical. You can use the following function:<\/p>\n

function convertToDecimal($hours) {\n    $parts = explode('.', (string)$hours);\n    $min = (count($parts) == 2) ? $parts[1] \/ 60 : 0;\n    return $parts[0] + $min;\n}\n<\/code><\/pre>\n
when calling convertToDecimal(1.15)<\/code> will return 1.25<\/code>. add that to you reduce function as: <\/p>\n
carry[$item[\"recruiter\"]] += convertToDecimal($item[\"hours\"]);\n<\/code><\/pre>\n
Hint – asking question properly is not easy task – now that you have your solution try edit your question so it will include all (and only<\/strong>) the data needed to unnderstand your problem and reproduce it.<\/p>\n
Better solution include change the way you extract the time difference. Consider the following method:<\/p>\n
$shiftStartTime = \"2019-01-11T07:00:00\";\n$shiftEndTime = \"2019-01-11T15:15:00\";\n\n$startLong = (new DateTime($shiftStartTime))->getTimestamp();\n$endLong = (new DateTime($shiftEndTime))->getTimestamp();;\n\necho ($endLong - $startLong) \/ (60 * 60);\n<\/code><\/pre>\n
This will give you 8.25<\/code> as you counting the second and not setting the hours using DateInterval.<\/p>\n
Hope that helps!<\/p>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            4<\/span> <\/p><\/div>\n<\/div>\n
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solved Reducing arrays of floats from datetime [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
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Example: 1 hour and 15 minute (1.15) plus 1 hour and 45 … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[339],"class_list":["post-16620","post","type-post","status-publish","format-standard","hentry","category-solved","tag-php"],"yoast_head":"\n[Solved] Reducing arrays of floats from datetime [closed] - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-reducing-arrays-of-floats-from-datetime-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Reducing arrays of floats from datetime [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] So your problem was of using dateinterval object to get the time difference. 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