Node-mysql wordpress在所有查詢和循環完成後獲取帖子和標籤並呈現頁面

Question

有人可以幫助我在所有查詢完成後才呈現頁面嗎？

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 
     }}) 
     }; 
    }}) 
    }}); 
}); 

即使我檢查i的值，我也無法獲得變量中的新值。有人可以建議什麼是在nodejs和node-sql中進行這種調用的理想方式嗎？

Answer 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}); 
     }}); 
     }; 
    }}); 
    }}); 
}); 

做這項工作，但可以做一個更好的解決方案。