Need to show data in javascript from db

Question

I'm working with a php code where I get variables from an input and using ajax

code:

 $.ajax({
  type: "GET",
  url: "controller/appointment/src_agenda.php",
  data: {
    function: "professional",
    variable: professional,
  },
  dataType: "html",
  fail: function( jqXHR, textStatus, errorThrown ) {
    alert(jqXHR);
  },
  error: function (jqxhr, ajaxOptions, thrownError) {
    alert(JSON.stringify(jqxhr));
    alert(jqxhr.status);
    alert(thrownError);
  },
  success: function(data){
    console.log(data);
  }
});

I get an array in the success function:

 object(mysqli_result)#3 (5) {
   ["current_field"]=>
   int(0)
   ["field_count"]=>
   int(6)
   ["lengths"]=>
   NULL
   ["num_rows"]=>
   int(76)
   ["type"]=>
   int(0)
 }

If I use php foreach it shows all the rows, but I tried javascript data.forEach and it says it doesn't exist. I also tried for(var element of data) but it shows each character as an individual.

How can I show each row from this query result in javascript?

This is the code of the model:

 $sql = "SELECT medical_agenda_id, day, `date`, time_start, time_end, hour_status_id FROM medical_agenda WHERE hour_status_id>='0' AND professional_id='".$id."' ;";
 var_dump($sql);
 $res = $this->db->query($sql);
 return $res;

I need to put the result in the success function

You need to fetch the result in PHP before returning it.

M. Eriksson
– M. Eriksson

2018-05-09 14:00:55 +00:00
Commented May 9, 2018 at 14:00 — M. Eriksson
– M. Eriksson, Commented May 9, 2018 at 14:00

Romeo Sierra · Accepted Answer · 2018-05-14 02:50:22Z

1

return $res; would return plenty of unwanted data. Rather, get the resultset's data into an associative array and echo json_encode() it.

data_type better be JSON instead of html. Because, that will let you get the minimal amount of data in the payload. And you can decide how you present them in the front end.

It should be something like follows.

while($row = mysqli_fetch_assoc($res)){//Using procedural style here
    $data[] = $row;
}

echo json_encode($data);

Plus, var_dump($sql) also should be taken out. Otherwise, that will also be included in the HTTP response, which will make the JSON string invalid.

With all the garbage out,

$sql = "SELECT medical_agenda_id, day, `date`, time_start, time_end, hour_status_id FROM medical_agenda WHERE hour_status_id>='0' AND professional_id='".$id."' ;";

//var_dump($sql); Take this out!

$res = $this->db->query($sql);

//return $res; Take this out! You will return the data by echoing it to the Web Server

while($row = mysqli_fetch_assoc($res)){//Using procedural style here
    $data[] = $row;
}

echo json_encode($data); //This is the output tho the client's request.

should do the trick.

This is the MINIMAL that you could do to get your code working. There are many many optimizations that you need to do, before you put it in production.

edited May 14, 2018 at 2:50

answered May 9, 2018 at 13:59

Romeo Sierra

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2 Comments

Ricardo Javier Chavarra Gonzal Over a year ago

It gave me on the console the data perfect, I need it now to generate links to interact with the user.

Ricardo Javier Chavarra Gonzal Over a year ago

So he can see a table with the data in order with the information in the specific place. I can't understand if I must use data[date] or data.date in the javascript

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