update values in textbox using ajax

Question

actualy im just a newbie programmer and i made my own update javascript function for my program but sadly the codes doesnt update..can anyone help me assist in order my codes work? please.

i want to do is if i change the values in the textboxes and click update it will update.

enter image description here

script code:

<script type="text/javascript">
$(document).ready(function () {
       $('#updates').click(function (e) {

           e.preventDefault();
           var $id1 = $('#id1').val();
           var $id2 = $('#id2').val();
           var $name1 = $('#name1').val();
           var $name2 = $('#name2').val();
           var $optA1 = $('#optA1').val();
           var $optA2 = $('#optA2').val();
           var $optB1 = $('#optB1').val();
           var $optB2 = $('#optB2').val();
           var $other_qual1 = $('#other_qual1').val();
           var $other_qual2 = $('#other_qual2').val();
           var $interview1 = $('#interview1').val();
           var $interview2 = $('#interview2').val();
           var $total1 = $('#total1').val();
           var $total2 = $('#total2').val();

           $.ajax({
               type: "POST",
               url: "update.php",
               data: {
                   id1_text: $id1,
                   id2_text: $id2,
                   name1_text: $name1,
                   name2_text: $name2,
                   optA1_text: $optA1,
                   optA2_text: $optA2,
                   optB1_text: $optB1,
                   optB2_text: $optB2,
                   other_qual1_text: $other_qual1,
                   other_qual2_text: $other_qual2,
                   interview1_text: $interview1,
                   interview2_text: $interview2,
                   total1_text: $total1,
                   total2_text: $total2
               },
               cache: false,
               success: function (data) {
                   alert('data has been updated!');
               }
           });
       });
   });
</script>

update.php page:

<?php

    mysql_connect("localhost", "root", "") or die("cant connect to database!");
    mysql_select_db("test") or die("cant find database!");
    $id1         = @$_POST['id1_text'];
    $id2         = @$_POST['id2_text'];
    $name1       = @$_POST['name1_text'];
    $name2       = @$_POST['name2_text'];
    $optA1       = @$_POST['optA1_text'];
    $optA2       = @$_POST['optA2_text'];
    $optB1       = @$_POST['optB1_text'];
    $optB2       = @$_POST['optB2_text'];
    $other_qual1 = @$_POST['other_qual1_text'];
    $other_qual2 = @$_POST['other_qual2_text'];
    $interview1  = @$_POST['interview1_text'];
    $interview2  = @$_POST['interview2_text'];
    $total1      = @$_POST['total1_text'];
    $total2      = @$_POST['total2_text'];

    $query1 = mysql_query("UPDATE score SET name=$name1, score1=$optA1, score2=$optB1, other_qual=$other_qual1, interview=$interview1, total=$total1 WHERE id=$id1");
    $resource1 = mysql_query($query1) or die(mysql_error());

    $query2 = mysql_query("UPDATE score SET name=$name2, score1=$optA2, score2=optB2, other_qual=$other_qual2, interview=$interview2, total=$total2 WHERE id=$id2");
    $resource2 = mysql_query($query2) or die(mysql_error());


?>

html:

<form>
    Search batchcode: <input id="query" name="search" type="text"><input id="send_search_form" type="button" value="Go" /><br>
Batch #: <label id="batchcode" class="empty_batchcode"></label>
    <table>
        <tr>
            <td>ID:<br>
            <input readonly id="id1" class="search_form_input" name="id1" type="text"><br>
            <input readonly id="id2" class="search_form_input" name="id2" type="text"><br></td>

            <td>Name:<br>
            <input id="name1" class="search_form_input" name="name1" type="text"><br>
            <input id="name2" class="search_form_input" name="name2" type="text"><br></td>

            <td>Score 1:<br>
            <input id="optA1" class="search_form_input" name="optA1" type="text"><br>
            <input id="optA2" class="search_form_input" name="optA2" type="text"><br></td>

            <td>Score 2:<br>
            <input id="optB1" class="search_form_input" name="optB1" type="text"><br>
            <input id="optB2" class="search_form_input" name="optB2" type="text"><br></td>

            <td>Other Qualification:<br>
            <input id="other_qual1" class="search_form_input" name="other_qual1" type="text"><br>
            <input id="other_qual2" class="search_form_input" name="other_qual2" type=
            "text"><br></td>

            <td>Interview:<br>
            <input id="interview1" class="search_form_input" name="interview1" type="text"><br>
            <input id="interview2" class="search_form_input" name="interview2" type="text"><br></td>

            <td>Total:<br>
            <input id="total1" class="search_form_input" name="total1" type="text"><br>
            <input id="total2" class="search_form_input" name="total2" type="text"><br></td>
        </tr>
    </table>
    <input type="button" value="update" id="updates" />
</form>

What exactly doesn't update? Your database? The form elements (you didn't write any code that updates the elements, so of course they cannot update)? Please explain your problem properly. We cannot look into your head. — Felix Kling
– Felix Kling, Commented Feb 3, 2014 at 5:18

Barmar · Accepted Answer · 2014-02-03 05:23:53Z

2

You're missing quotes around the name in your SQL query.

$query1 = mysql_query("UPDATE score SET name='$name1', score1=$optA1, score2=$optB1, other_qual=$other_qual1, interview=$interview1, total=$total1 WHERE id=$id1");

It would be better if you used PDO or mysqli, so you could use parametrized queries instead of string substitution. Your code is in serious danger of SQL injection.

answered Feb 3, 2014 at 5:23

Barmar

789k57 gold badges554 silver badges669 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Deryck · Accepted Answer · 2014-02-03 05:40:10Z

1

Neither of these will work.

$resource1 = mysql_query($query1) or die(mysql_error());
$resource2 = mysql_query($query2) or die(mysql_error());

Because the referenced $query1 and $query2 are the actual query. You need to change them to simple strings.

Corrected PHP:

$query1 = "UPDATE score SET name=$name1, score1=$optA1, score2=$optB1, other_qual=$other_qual1, interview=$interview1, total=$total1 WHERE id=$id1";
$resource1 = mysql_query($query1) or die(mysql_error());

$query2 = "UPDATE score SET name=$name2, score1=$optA2, score2=optB2, other_qual=$other_qual2, interview=$interview2, total=$total2 WHERE id=$id2";
$resource2 = mysql_query($query2) or die(mysql_error());

Or even better, update to something more current, like so:

$mysqli = new mysqli("localhost", "root", "", "test");
if ($mysqli->connect_errno) {
    echo "Failed to connect to MySQL: (Error code: " . $mysqli->connect_errno . ")... " . $mysqli->connect_error;
}
$query1 = "UPDATE score SET name='$name1', score1='$optA1', score2='$optB1', other_qual='$other_qual1', interview='$interview1', total='$total1' WHERE id='$id1'";
$resource1 = $mysqli->query($query1) or die(mysqli_error($mysqli));
$query2 = "UPDATE score SET name='$name2', score1='$optA2', score2='$optB2', other_qual='$other_qual2', interview='$interview2', total='$total2' WHERE id='$id2'";
$resource2 = $mysqli->query($query2) or die(mysqli_error($mysqli));
mysqli_close($mysqli);

@Barmar also has a point - you need quotes around those variables in case the have spaces or other characters that might result in syntax error.

edited Feb 3, 2014 at 5:40

answered Feb 3, 2014 at 5:25

Deryck

7,6662 gold badges27 silver badges44 bronze badges

5 Comments

xplody Over a year ago

k il pOSt the html...i change the query as your advise still it wont update.

Deryck Over a year ago

did you use the first set of PHP I posted or the set I just updated with mysqli?

xplody Over a year ago

nvm the query just need '' on query values its working now thanks :D

Deryck Over a year ago

well I'm glad it works now but that was from @Barmar you should select his answer instead.

xplody Over a year ago

i used ur corrected php code...do you still have time? can you help me that if the its not updated it will say "failed to update!"

Collectives™ on Stack Overflow

update values in textbox using ajax

2 Answers 2

Comments

5 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

5 Comments

Your Answer

Sign up or log in

Post as a guest

Related