How do you specify single print blocks when looping through a 'for' loop?

How do you specify single print blocks when looping through a 'for' loop?

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How do you specify single print blocks when looping through a 'for' loop?
Tag : php , By : Fernando
Date : November 24 2020, 09:00 AM

I hope this helps . I have a carousel created from images in a page template. For each page, the template will try to loop through a for and print out the images as needed. , This code is cleaner, and more adaptable.
    // Set up an empty array to hold all of the slides
    $slideCheck = array();
    // We can only have 10 slides.  Change this number to allow for more or less.
    $max_slides = 10;
    // Loop through and get all the slides.
    for( $i = 0; $i <= $max_slides; $i++ ) {
        // Set up the key
        $key = 'slide' . $i;
        // Get the slide url
        $url = get_post_meta($post->ID, $key, $single = true);
        // Get the slide text
        $text = get_post_meta($post->ID, $key . '_label', $single = true);
        // Check to make sure we at least have one of the values
        if( ! empty( $url ) || ! empty( $text ) ) {
            // If we do, add them to our array
            $slideCheck[] = array(
                'url'  => $url,
                'text' => $text
    // If we have one or more slides
    if( ! empty( $slideCheck ) ) { ?>
        <section id="myCarousel" class="carousel slide" data-ride="carousel">

            <?php if( sizeof( $slideCheck )  > 1 ) { // Here we are checking if we have more than one slide. ?>
            <ol class="carousel-indicators">
                <?php for( $i = 0; $i < sizeof( $slideCheck ); $i++ ) { // Make this section dynamic, as we don't know how many slides we'll actually have ?>
                <li data-target="#myCarousel" data-slide-to="<?php echo $i; ?>" class="<?php echo $i == 1 ? 'active' : ''; ?>"></li>
                <?php } ?>
            <?php } ?>

            <div class="carousel-inner">
                    foreach( $slideCheck as $key => $slide ) {
                        // Some check to see if we are on the second slide, per your old code??? Not sure what it does.
                        $class = ($key == 1 ) ? 'active' : '';
                        <div class="item <?php echo $class; ?>">
                            <?php if( ! empty( $slide['url'] ) ) { ?>
                            <img src="<?php echo $slide['url']; ?>" class="slider-images" />
                            <?php } if( ! empty( $slide['text'] ) ) { ?>
                            <div class="carousel-caption">
                                <h2><?php echo $slide['text']; ?></h2>
                            <?php } ?>
                    <?php } // end foreach ?>
            <?php if( sizeof( $slideCheck ) > 1 ) { // Again, if we have more than one slide... ?>
            <a class="left carousel-control" href="#myCarousel" role="button" data-slide="prev">
                <span class="glyphicon glyphicon-chevron-left" aria-hidden="true"></span>
                <span class="sr-only">Previous</span>
            <a class="right carousel-control" href="#myCarousel" role="button" data-slide="next">
                <span class="glyphicon glyphicon-chevron-right" aria-hidden="true"></span>
                <span class="sr-only">Next</span>
            <?php } ?>
    <?php } // end if ?>

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looping, want to print out information about an array of dictionary objects at the end of the loop

Tag : iphone , By : user121405
Date : March 29 2020, 07:55 AM
With these it helps Try using [NSString stringWithFormat:@"Something about value: %@", dictionaryItem]

with single how loop to print last 3rd element from given single linked list?

Tag : cpp , By : user181706
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further With single loop how can we print last 3rd element from given single linked list lets take there are 10 node in give linked list and i am supposed to find last 3rd node? below code is for inserting a node at beginning now how can i print last 3rd element with sing loop? please help me , Keep three pointers:
node * prevPrev = null;
node * prev = null;
node * current = head;
prevPrev = prev;
prev = current;
current = current->next;

While loop with try-catch blocks doesn't wait for input in the try block after looping once

Tag : java , By : Singularity
Date : March 29 2020, 07:55 AM
Any of those help The behaviour you're observing stems from the fact that scanner (assuming you're working with java.util.Scanner) only advances if nextInt() is successful. That means that you have to skip over the faulty input (i.e. by using next()) before proceding.

Tag : javascript , By : xie renhui
Date : March 29 2020, 07:55 AM
To fix this issue Although Nina's answer is better, for completeness, I post a corrected version of the code that you started with:
let n = 8;
for (let i=0; i<=n; i+=2) {
  for (let j=i; j<i+2 && j <= n; j++) {

How to define blocks of elements in a vector to zero and looping over the next blocks?

Tag : r , By : orneka
Date : March 29 2020, 07:55 AM
To fix the issue you can do I have a large vector which I would like to multiply with a matrix. This is easy to do in r. The issue here is that I would like to define certain blocks of elements within the vector to zero, multiply this with the matrix which then creates a vector, and then repeat for the next block of elements leaving the first block with the original values. , First, I define my vector and matrix.
# Data definitions
a <- c(1:6)
b <- matrix(1:36, nrow = 6, ncol = 6)
# Define blocks of zeroes in vector 
define_block <- function(v, n = 2){
  # Check that v is a multiple of n
  if(length(v)%%n)warning("Vector not a multiple of block length")

  # Define blocks of zeroes
  lapply(1:(length(v)/n), function(x)replace(v, ((x-1)*n + 1):(x*n), 0))
# Create lists of blocks
block_list <- define_block(a)
# [[1]]
# [1] 0 0 3 4 5 6
# [[2]]
# [1] 1 2 0 0 5 6
# [[3]]
# [1] 1 2 3 4 0 0
# Run through block list and bind result into a matrix
do.call(rbind, lapply(block_list, function(x)x%*%b))

#     [,1] [,2] [,3] [,4] [,5] [,6]
#[1,]   86  194  302  410  518  626
#[2,]   66  150  234  318  402  486
#[3,]   30   90  150  210  270  330
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