Fetch data from db using ajax jquery

Question

I've got some issues fetching a value from db by using ajax in my jquery script.

index.php:

<?php include 'db/db_connection.php';?>
<?php include 'db/db_getvalues.php';?>

<div class="arrow">
  <img src="img/arrow_left.png" />
</div>

<div class="text-box"></div>

script.js

$(document).ready(function () {
  $(".arrow").click(function () {
    $.ajax({
      type: "GET",
      url: "../db/db_getvalues.php", // This is the correct path, but do jquery recognize "../db/" etc?
      dataType: "text",
      success: function (response) { // Here is where i'm lost, what do i need to write to update my innerHTML with the returned value from the db_getvalues.php-file?
        $(".text-box").html(response);
      }
    });
  });
});

db_getvalues.php // This file works, i've selected data directly from html-file

<?php
  function getdbvalues() {
    $query = 'SELECT * FROM mydb WHERE Id = 1';
    $fetch = mysql_query($query) or die ('Could not find tablerow');
    $row = mysql_fetch_assoc($fetch);

    $textString = $row['Text'];

    return $textString;
  }
?>

what output you are currently getting ?

Atul Nar
– Atul Nar

2014-11-22 15:38:42 +00:00
Commented Nov 22, 2014 at 15:38 — Atul Nar
– Atul Nar, Commented Nov 22, 2014 at 15:38

MH2K9 · Accepted Answer · 2014-11-22 15:42:19Z

2

Yes, AJAX recognize such URL. Your ajax seems ok but a little suggestion to change in db_getvalues.php. You should echoing the value instead of return. I suggest you to call the getdbvalues() function & to echo it like below.

In db_getvalues.php

function getdbvalues() {
    $query = 'SELECT * FROM mydb WHERE Id = 1';
    $fetch = mysql_query($query) or die ('Could not find tablerow');
    $row = mysql_fetch_assoc($fetch);
    $textString = $row['Text'];
    return $textString;
}

echo getdbvalues(); //Added

answered Nov 22, 2014 at 15:42

MH2K9

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Comments

Sami · Accepted Answer · 2014-11-22 15:58:01Z

You have to make sure 1. your click function working 2. your ajax request is success ful 3. php is producing some html to be received by ajax success function 4. your php code is giving you (echoing) required html

Check my lines having comments(1,2,3,4) added in your code

$(document).ready(function () {
  $(".arrow").click(function () {
  alert("I am called"); //1
    $.ajax({
      type: "GET",
      url: "../db/db_getvalues.php", 
      dataType: "text",
      success: function (response) { 
      alert(response+" __ url is fine");//2
        $(".text-box").html(response);
      }
    });
  });
});



<?php
  echo "php is called"; //3
  function getdbvalues() {
    $query = 'SELECT * FROM mydb WHERE Id = 1';
    $fetch = mysql_query($query) or die ('Could not find tablerow');
    $row = mysql_fetch_assoc($fetch);

    $textString = $row['Text'];

    return $textString;
  }
echo getdbvalues(); //4
?>

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