CODEWITHSUNDEEP
phpjqueryjsonajax
user8878793
user8878793

Reputation: 60

PHP passing JSON data using Ajax for inArray() comparison

I'm trying to check an user input in a html form to a list of allowable values in a database using Ajax and then I want to display the corresponding message. This is the code I'm using:

js

$(document).ready(function(e){

  $.ajax({
    type: 'POST',
    url: 'datafile.php',
    dataType : "json",
    success: function(result) {
          $("#postal_code").change(function(){
            var postal_code = $("#postal_code").val().slice(0,-2);
            if ($.inArray(postal_code,result) !== -1){
             console.log('success');
            } else {
              console.log('failure');
            }
        });
      }
    });
  });

datafile.php

<?php
require_once("./db_con.php");

$query = mysqli_query($con, "SELECT postal_code FROM table");
$rows = mysqli_fetch_all($query);
echo json_encode($rows);
mysqli_free_result($query);
mysqli_close($con);
exit();
?>

The problem I'm having is the input value never matches any of the values in the result variable even through I know that they are in the database. I think the issue is related to the fact that the result variable contains arrays in side another array, e.g. [["value1"],["value2"],["value3"]].

Upvotes: 1

Views: 68

Answers (1)

AbraCadaver
AbraCadaver

Reputation: 78994

I can't really speak to the JavaScript portion, but since you are only selecting one column you can extract it into a single dimension. mysqli_fetch_all fetches a numerically indexed array by default:

$rows = array_column(mysqli_fetch_all($query), 0);

Or fetch and extract by column name:

$rows = array_column(mysqli_fetch_all($query, MYSQLI_ASSOC), 'postal_code');

Upvotes: 1

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