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I am trying to make a nested query, in which I am using the result of the first query in second query the function like is returning an undefined value.I know there is problem with asynchronous nature of nodejs but i can't figure out the solution to convert my code into synchronous.

My code:

function like(article_id,user_id){
  var status;
  pool.query('SELECT * FROM likes WHERE article_id=$1 AND user_id=$2',[article_id,user_id], function (err, result) {
     if (err) {
        res.status(500).send(err.toString());
     } else {
            if (result.rows.length === 0) {
              status=`like`;
            } else {
              status=`unlike`;
            }  

      }
   })`
    return status;
}

app.get('/article', function (req, res) {
        pool.query("SELECT * FROM article",function (err, result) {
        if (err) {
            res.status(500).send(err.toString());
        } else {
            if (result.rows.length === 0) {
                res.status(404).send('Article not found');
            } else {
                var middle='';
                numRows=result.rows.length;
                for(var i=numRows-1;i>=0;i-=1){
                    var articleData = result.rows[i];
                    middle=middle + createTemplate(articleData);
                     var r=like(result.rows[i].article_id,req.session.auth.userId.toString());
                     middle+=r;
                     middle+= `</button>`;
                      middle+=comment_section(result.rows[i].article_id.toString())+`<hr><hr>`;
                }
                res.send(middle);
            }
        }
      });
});

Note: createTemplate(articleData) and comment_section(result.rows[i].article_id.toString()) are synchronous.

pirs
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ASHISH CHAUHAN
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  • You can't convert your code into synchronous. You have to program correctly for asynchronous responses. That is all described in the one you're marked a dup of. – jfriend00 Oct 28 '17 at 05:37
  • Use a callback in `like(article_id, user_id, callback)` and `like(result,function(r){ middle+=r; /*...*/ res.send(middle); });` – pirs Oct 28 '17 at 05:40

0 Answers0