CODEWITHSUNDEEP
phpjquery
Ray Hebert
Ray Hebert

Reputation: 61

How to Define Variable in jquery from PHP

I am able to pass the server name to PHP which runs the query successfully. In the html file I want the rating option to change depending on what value was returned from the PHP file. In my file I set it to D but I need to change that to relflect what is being returned from PHP.

server.html

<html>
    <head>
        <script type="text/javascript" src="http://code.jquery.com/jquery-1.8.3.min.js"></script>
        <script type="text/javascript" >
        $(document).ready(function(){
            var id = $('#existingserver').val();            
            $('#assetCenter').click(function(){
                var id = $('#textfield').val();
                $.get('servertest.php',{q:id}, function(htmlData){
                    $('#txtHint').html(htmlData);
                    var rating = $(htmlData).find("td[data-col=rating]").text();
                    alert(rating);
                    });
            });
        });

        </script>
    </head>
    <body>
        <form>
            <label>Existing Server</label><input type="text" name="existingserver" id="textfield" maxlength="15"/>

            <input type="checkbox" id="assetCenter" >Select to Pull Asset Center Data<br>
            <br />
            <br />
            Rating
            <select name="rating" id="rating" >
                <option value="A">A</option>
                <option value="B">B</option>
                <option value="C">C</option>
                <option value="D">D</option>
            </select>
        </form>
        <br />

        <div id="txtHint"><b>Server info will be listed here.</b></div>
    </body>
</html>

servertest.php

<?php
$q=$_GET["q"];

$con = mysql_connect('localhost', 'assignip', 'assignip');
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("ipreservation", $con);

$sql="SELECT * FROM acdata WHERE servername = '".$q."'";

$result = mysql_query($sql);

echo "<table border='1'>
<tr>
<th>Servername</th>
<th>Contact</th>
<th>Classification</th>
<th>Rating</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['servername'] . "</td>";
  echo "<td>" . $row['contact'] . "</td>";
  echo "<td>" . $row['classification'] . "</td>";
  echo "<td>" . $row['rating'] . "</td>";
  echo "</tr>";
  echo "<td data-col='rating'>" . $row['rating'] . "</td>";
  }

  echo "</table>";

mysql_close($con);
?>

database fields: servername, contact, classification, rating Data: Server1, Ray, Production, A

Upvotes: 0

Views: 1073

Answers (2)

McGarnagle
McGarnagle

Reputation: 102743

The short answer is to use a JQuery selector to get the rating:

// get the text of the 4th td
var rating = $(htmlData).find("td").eq(3).text();

$("#rating").val(rating);

However, you might notice this approach is kind of brittle (aka tightly coupled) -- if something changes in the UI, say you re-order the columns, then the above logic will break.

I'd suggest returning the data from the server as JSON, and then applying a client-side template to get the HTML table. At the very least, give the columns a name like this:

echo "<td data-col='rating'>" . $row['rating'] . "</td>";

And then you can select on the client side by referencing the name:

var rating = $(htmlData).find("td[data-col=rating]").text();

Upvotes: 1

RonaldBarzell
RonaldBarzell

Reputation: 3830

In your PHP code, change the output so it displays the value you want to show, without any html tags.

Next in your html code, change your $(document).ready callback so that 'D' is replaced with the response text of the call (that is what PHP will return).

Upvotes: 1

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