Month: December 2022

[Solved] How do I change my JavaScript to run on page load instead of event change

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You have non-unique ID’s for elements – jQuery will not work. Change ID to class. Also remove any reference to event (since there is no such thing on DOM ready). I have converted your JS code to actual jQuery: $(document).ready(function() { $(‘.numberofdaysperchosenfrequency’).each(hideDays); $(‘.numberofdaysperchosenfrequency’).on(‘change’, hideDays); }); function hideDays() { let parent = $(this).parent(); parent.find(‘.daycol’).show(); parent .find(`.daycol:not([data-id=”${$(this).val()}”])`) … Read more

[Solved] Grouping by a column in R [closed]

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We can try data.table. Convert the ‘data.frame’ to ‘data.table’ (setDT(df2), grouped by ‘RouteId’, if any of the run-length-type id of the logical vector (StopType==’Load’) is greater than 2, we get the Subset of Data.table (.SD). This will give the rows with ‘RouteId’ 103. library(data.table) setDT(df2)[,if(any(rleid(StopType==’Load’) >2)) .SD ,.(RouteId)] # RouteId StopOrder StopType #1: 103 1 … Read more

[Solved] Fatal error: Call to a member function url() on a non-object on line 8

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Here’s your code corrected: <?php $subpages = $site->children()->visible(); foreach($subpages as $subpage) { $image_url = $subpage->image()->url(); $title = $subpage->title()->html(); echo ‘<div class=”col-md-4″>’; echo ‘<h2>’ . $title . ‘</h2>’; echo ‘<a href=”‘ . $subpage->url() . ‘” title=”‘ . $title . ‘”>’; echo ‘<img src=”‘ . $image_url . ‘” alt=”‘ . $title . ‘” class=”img-responsive img-thumbnail”>’; echo ‘</a>’; … Read more

[Solved] copy from unsigned char to unsigned char

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The reason for the warnings and error is that strncpy() accepts char * arguments, which is different from unsigned char *. Assuming the two arrays in the structs are the same size simply do memcpy(d1.d.something, c->do_something, sizeof(d1.d.something)); If you can’t assume the two arrays are the same size, you’ll need to write code to check … Read more

[Solved] I have a Java homework problem that I cannot seem to figure out. Using regular expression and/or API classes like Pattern, Matcher is not allowed [closed]

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I have a Java homework problem that I cannot seem to figure out. Using regular expression and/or API classes like Pattern, Matcher is not allowed [closed] solved I have a Java homework problem that I cannot seem to figure out. Using regular expression and/or API classes like Pattern, Matcher is not allowed [closed]

[Solved] LNK1120 & LNK2019 in C

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You don’t say which compiler you are using. At a guess it is Microsoft 2019. The delay function is called Sleep (uppercase s) and it is in milliseconds. It lives in the Windows.h file. #define WIN32_LEAN_AND_MEAN #include <Windows.h> #include <stdio.h> … /* Sleep 100ms */ Sleep(100); Not that it makes a lot of difference on … Read more

[Solved] cant access json array using php [closed]

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Look at documentation: json_decode( string $json, ?bool $associative = null, int $depth = 512, int $flags = 0 ): mixed and you are passing second argument as true, so it returns array and not object. And you missing intermediate keys. So final solution would be: $update = file_get_contents(‘php://input’); $update = json_decode($update, true); $callbak = $update[‘result’][0][‘callback_query’][‘data’]; … Read more

[Solved] C++ Storing constant varibale names within a vector [closed]

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Simply use a new local variable inside your reading loop like this: while(stream >> token) { if(token == “#define”) { constantVariable addconstant; stream >> token; addconstant.constantName = token; stream >> token; addconstant.constantValue = token; constant.push_back(addconstant); } } But take care with checking the input stream. It should not be done as easy as you did … Read more

[Solved] Removing all occurrences of any characters in the word ‘dust’ in the string [closed]

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Use a regular expression. import re result = re.sub(r'[dust]’, ”, string) The regexp [dust] matches any of those characters, and all the matches are replaced with an empty string. If you want to remove only the whole word dust, with possible repetitions of the letters, the regexp would be r’d+u+s+t+’. If only u can be … Read more