Node-mysql wordpress 在所有查询和for循环完成后获取帖子和标签并渲染页面答案

【问题标题】：Node-mysql wordpress Get post and tags and render page after all queries and for loops are completeNode-mysql wordpress 在所有查询和for循环完成后获取帖子和标签并渲染页面
【发布时间】：2015-04-12 15:34:28
【问题描述】：

只有在所有查询都完成后，有人才能帮我呈现页面吗？

app.get("/", function(req, res) {
  connection = mysql.createConnection(dbconfig);
  connection.connect(function(err) { if(!err) {
    //Get all posts
    connection.query('SELECT * from wp_posts WHERE post_status = "publish" AND post_type = "html5-blank" ORDER BY post_date DESC', function(err, rows, fields) { if(!err) {
      //Get tags
      for (var i in rows) {
        connection.query('SELECT slug FROM wp_terms INNER JOIN wp_term_taxonomy ON wp_term_taxonomy.term_id = wp_terms.term_id INNER JOIN wp_term_relationships ON wp_term_relationships.term_taxonomy_id = wp_term_taxonomy.term_taxonomy_id WHERE taxonomy = "post_tag" AND object_id =' + rows[i].ID, function(err, _rows, fields) { if(!err) {
          var slugs = '';
          for(var y in _rows) {
            slugs += '#' + _rows[y].slug + ' ';
          };
          rows[i]['hashtags'] = slugs;
          console.log(i) //Returns last index
          if(i == rows.length) res.render('content', {_page: '', _posts: rows}); //Executes multiple times
        }})
      };
    }})
  }});
});

即使我检查 i 的值，我也无法获取变量中的新值。有人可以建议在 nodejs 和 node-sql 中进行此类调用的理想方法是什么？

【问题讨论】：

标签： mysql node.js wordpress

【解决方案1】：

app.get("/", function(req, res) {
  connection = mysql.createConnection(dbconfig);
  connection.connect(function(err) { if(!err) {
    //Get all posts
    connection.query('SELECT * from wp_posts WHERE post_status = "publish" AND post_type = "html5-blank" ORDER BY post_date DESC', function(err, rows, fields) { if(!err) {
      //Get tags
      var z = 0;
      for (var i in rows) {
        connection.query('SELECT slug FROM wp_terms INNER JOIN wp_term_taxonomy ON wp_term_taxonomy.term_id = wp_terms.term_id INNER JOIN wp_term_relationships ON wp_term_relationships.term_taxonomy_id = wp_term_taxonomy.term_taxonomy_id WHERE taxonomy = "post_tag" AND object_id =' + rows[i].ID, function(err, _rows, fields) { if(!err) {
          var slugs = '';
          for(var y in _rows) {
            slugs += '#' + _rows[y].slug + ' ';
          };
          rows[z]['hashtags'] = slugs;
          z++;
          console.log(z);
          if(z == rows.length) res.render('content', {_page: '', _posts: rows});
        }});
      };
    }});
  }});
});

可以完成这项工作，但可以使用更好的解决方案。

【讨论】：