Wordpress, foreach on taxonomy shows duplicates












0















I have the below code, it works initially with regards to the fact that it displays the products for all taxonomies. However if a product is set to 2 taxonomies then it will display twice on the page as opposed to showing the first instance of the product.



<?php if ( $terms && !is_wp_error( $terms ) ) {
foreach ( $terms as $term ) {

$args = array(
'post_type' => 'products',
'posts_per_page' => -1,
'orderby' => 'menu_order',
'tax_query' => array(
array(
'taxonomy' => 'product_cat',
'field' => 'slug',
'terms' => $term->slug,
),
),
'order' => 'asc',
);

runQuery($args);
}
} ?>


Here is the runQuery function:



<?php $x = 0;
function runQuery($args) {
global $x;
$query = new WP_Query( $args );

if ($query->have_posts()) : while ($query->have_posts()) : $query->the_post();
$cat_terms = get_the_terms($post->id, 'product_cat');
$datagroups = '';

foreach ($cat_terms as $key => $cat) {
if (count($cat_terms) == ($key + 1)) {
$datagroups .= '"' . $cat->name . '"';
} else {
$datagroups .= '"' . $cat->name . '", ';
}
}
?>

HTML Here that is displayed;

<?php $x ++;
endwhile;
endif;
wp_reset_postdata();
}?>









share|improve this question



























    0















    I have the below code, it works initially with regards to the fact that it displays the products for all taxonomies. However if a product is set to 2 taxonomies then it will display twice on the page as opposed to showing the first instance of the product.



    <?php if ( $terms && !is_wp_error( $terms ) ) {
    foreach ( $terms as $term ) {

    $args = array(
    'post_type' => 'products',
    'posts_per_page' => -1,
    'orderby' => 'menu_order',
    'tax_query' => array(
    array(
    'taxonomy' => 'product_cat',
    'field' => 'slug',
    'terms' => $term->slug,
    ),
    ),
    'order' => 'asc',
    );

    runQuery($args);
    }
    } ?>


    Here is the runQuery function:



    <?php $x = 0;
    function runQuery($args) {
    global $x;
    $query = new WP_Query( $args );

    if ($query->have_posts()) : while ($query->have_posts()) : $query->the_post();
    $cat_terms = get_the_terms($post->id, 'product_cat');
    $datagroups = '';

    foreach ($cat_terms as $key => $cat) {
    if (count($cat_terms) == ($key + 1)) {
    $datagroups .= '"' . $cat->name . '"';
    } else {
    $datagroups .= '"' . $cat->name . '", ';
    }
    }
    ?>

    HTML Here that is displayed;

    <?php $x ++;
    endwhile;
    endif;
    wp_reset_postdata();
    }?>









    share|improve this question

























      0












      0








      0








      I have the below code, it works initially with regards to the fact that it displays the products for all taxonomies. However if a product is set to 2 taxonomies then it will display twice on the page as opposed to showing the first instance of the product.



      <?php if ( $terms && !is_wp_error( $terms ) ) {
      foreach ( $terms as $term ) {

      $args = array(
      'post_type' => 'products',
      'posts_per_page' => -1,
      'orderby' => 'menu_order',
      'tax_query' => array(
      array(
      'taxonomy' => 'product_cat',
      'field' => 'slug',
      'terms' => $term->slug,
      ),
      ),
      'order' => 'asc',
      );

      runQuery($args);
      }
      } ?>


      Here is the runQuery function:



      <?php $x = 0;
      function runQuery($args) {
      global $x;
      $query = new WP_Query( $args );

      if ($query->have_posts()) : while ($query->have_posts()) : $query->the_post();
      $cat_terms = get_the_terms($post->id, 'product_cat');
      $datagroups = '';

      foreach ($cat_terms as $key => $cat) {
      if (count($cat_terms) == ($key + 1)) {
      $datagroups .= '"' . $cat->name . '"';
      } else {
      $datagroups .= '"' . $cat->name . '", ';
      }
      }
      ?>

      HTML Here that is displayed;

      <?php $x ++;
      endwhile;
      endif;
      wp_reset_postdata();
      }?>









      share|improve this question














      I have the below code, it works initially with regards to the fact that it displays the products for all taxonomies. However if a product is set to 2 taxonomies then it will display twice on the page as opposed to showing the first instance of the product.



      <?php if ( $terms && !is_wp_error( $terms ) ) {
      foreach ( $terms as $term ) {

      $args = array(
      'post_type' => 'products',
      'posts_per_page' => -1,
      'orderby' => 'menu_order',
      'tax_query' => array(
      array(
      'taxonomy' => 'product_cat',
      'field' => 'slug',
      'terms' => $term->slug,
      ),
      ),
      'order' => 'asc',
      );

      runQuery($args);
      }
      } ?>


      Here is the runQuery function:



      <?php $x = 0;
      function runQuery($args) {
      global $x;
      $query = new WP_Query( $args );

      if ($query->have_posts()) : while ($query->have_posts()) : $query->the_post();
      $cat_terms = get_the_terms($post->id, 'product_cat');
      $datagroups = '';

      foreach ($cat_terms as $key => $cat) {
      if (count($cat_terms) == ($key + 1)) {
      $datagroups .= '"' . $cat->name . '"';
      } else {
      $datagroups .= '"' . $cat->name . '", ';
      }
      }
      ?>

      HTML Here that is displayed;

      <?php $x ++;
      endwhile;
      endif;
      wp_reset_postdata();
      }?>






      php wordpress foreach






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      asked Nov 22 '18 at 10:22









      Daniel VickersDaniel Vickers

      354215




      354215
























          1 Answer
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          0














          I have figured this out, I essentially just had to check the current post in the loop to see if it had already been displayed:



          <?php $x = 0;
          $displayed = ;

          function runQuery($args) {
          global $displayed;
          global $x;
          $query = new WP_Query( $args );

          if ($query->have_posts()) : while ($query->have_posts()) : $query->the_post();
          $cat_terms = get_the_terms($post->id, 'product_cat');
          $datagroups = '';

          if ( in_array( get_the_ID(), $displayed ) ){
          continue;
          }
          // update array with currently displayed post ID
          $displayed = get_the_ID();

          foreach ($cat_terms as $key => $cat) {
          if (count($cat_terms) == ($key + 1)) {
          $datagroups .= '"' . $cat->name . '"';
          } else {
          $datagroups .= '"' . $cat->name . '", ';
          }
          }
          ?>


          Source that may help others: https://wordpress.stackexchange.com/questions/285091/avoid-duplicate-post-from-same-taxonomy






          share|improve this answer























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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

            votes









            0














            I have figured this out, I essentially just had to check the current post in the loop to see if it had already been displayed:



            <?php $x = 0;
            $displayed = ;

            function runQuery($args) {
            global $displayed;
            global $x;
            $query = new WP_Query( $args );

            if ($query->have_posts()) : while ($query->have_posts()) : $query->the_post();
            $cat_terms = get_the_terms($post->id, 'product_cat');
            $datagroups = '';

            if ( in_array( get_the_ID(), $displayed ) ){
            continue;
            }
            // update array with currently displayed post ID
            $displayed = get_the_ID();

            foreach ($cat_terms as $key => $cat) {
            if (count($cat_terms) == ($key + 1)) {
            $datagroups .= '"' . $cat->name . '"';
            } else {
            $datagroups .= '"' . $cat->name . '", ';
            }
            }
            ?>


            Source that may help others: https://wordpress.stackexchange.com/questions/285091/avoid-duplicate-post-from-same-taxonomy






            share|improve this answer




























              0














              I have figured this out, I essentially just had to check the current post in the loop to see if it had already been displayed:



              <?php $x = 0;
              $displayed = ;

              function runQuery($args) {
              global $displayed;
              global $x;
              $query = new WP_Query( $args );

              if ($query->have_posts()) : while ($query->have_posts()) : $query->the_post();
              $cat_terms = get_the_terms($post->id, 'product_cat');
              $datagroups = '';

              if ( in_array( get_the_ID(), $displayed ) ){
              continue;
              }
              // update array with currently displayed post ID
              $displayed = get_the_ID();

              foreach ($cat_terms as $key => $cat) {
              if (count($cat_terms) == ($key + 1)) {
              $datagroups .= '"' . $cat->name . '"';
              } else {
              $datagroups .= '"' . $cat->name . '", ';
              }
              }
              ?>


              Source that may help others: https://wordpress.stackexchange.com/questions/285091/avoid-duplicate-post-from-same-taxonomy






              share|improve this answer


























                0












                0








                0







                I have figured this out, I essentially just had to check the current post in the loop to see if it had already been displayed:



                <?php $x = 0;
                $displayed = ;

                function runQuery($args) {
                global $displayed;
                global $x;
                $query = new WP_Query( $args );

                if ($query->have_posts()) : while ($query->have_posts()) : $query->the_post();
                $cat_terms = get_the_terms($post->id, 'product_cat');
                $datagroups = '';

                if ( in_array( get_the_ID(), $displayed ) ){
                continue;
                }
                // update array with currently displayed post ID
                $displayed = get_the_ID();

                foreach ($cat_terms as $key => $cat) {
                if (count($cat_terms) == ($key + 1)) {
                $datagroups .= '"' . $cat->name . '"';
                } else {
                $datagroups .= '"' . $cat->name . '", ';
                }
                }
                ?>


                Source that may help others: https://wordpress.stackexchange.com/questions/285091/avoid-duplicate-post-from-same-taxonomy






                share|improve this answer













                I have figured this out, I essentially just had to check the current post in the loop to see if it had already been displayed:



                <?php $x = 0;
                $displayed = ;

                function runQuery($args) {
                global $displayed;
                global $x;
                $query = new WP_Query( $args );

                if ($query->have_posts()) : while ($query->have_posts()) : $query->the_post();
                $cat_terms = get_the_terms($post->id, 'product_cat');
                $datagroups = '';

                if ( in_array( get_the_ID(), $displayed ) ){
                continue;
                }
                // update array with currently displayed post ID
                $displayed = get_the_ID();

                foreach ($cat_terms as $key => $cat) {
                if (count($cat_terms) == ($key + 1)) {
                $datagroups .= '"' . $cat->name . '"';
                } else {
                $datagroups .= '"' . $cat->name . '", ';
                }
                }
                ?>


                Source that may help others: https://wordpress.stackexchange.com/questions/285091/avoid-duplicate-post-from-same-taxonomy







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 22 '18 at 10:45









                Daniel VickersDaniel Vickers

                354215




                354215






























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