show data through ajax in php

Question

I want to create a search function. This page already has some details which I retrieved from DB in div. I want to when I click search button, hide that all retrieved details and show searched details. I used ajax for it. Now problem is when I click search button previous retrieved data is hidden, but not show as search results.

    <form action="../PHP/searchrmvvac.php" method="post">
                    <div class="search hidden-xs hidden-sm">
                        <input type="text" placeholder="Search" id="search" name="search">
                    </div>
                    <div>
                        <input type="button" value="Search" id="searchrmvcom" name="searchrmvcom">
                        <script>
                            $("#searchrmvcom").click(function () {
                                var comname=$('#search').val();
                                $.ajax({
                                    type:"post",
                                    url:"../PHP/searchrmvvac.php",
                                    data:{comname:comname},
                                    success:function (data3) {
                                        $('#rmvcomdiv').hide();
                                        $('#ela').html(data3)
                                    }
                                });
                            });
                        </script>
                    </div>
                    </form>

searchrmvvac.php

    <?php
    session_start();
    require('../PHP/dbconnection.php');
    $output=$_POST['comname'];
    $sql="select * from company where company_name='$output' and activation_code=1";
    $res=mysqli_query($conn,$sql);
    if(mysqli_num_rows($res)>0) {
   echo '
while ($row = mysqli_fetch_assoc($res)) {
    ?>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
    <table class="table table-striped table-bordered table-list">
        <thead>

        <tr>
            <th>Action</th>
            <th>ID</th>
            <th>Registration number</th>
            <th>Company Name</th>
            <th>Email</th>
        </tr>
        </thead>
        <tbody>

            <tr>
                <td align="center"><button type="submit" class="btn btn-default myButton" value="$row['/companyid/']" id="accept" name="accept">Remove</button></td>
                <td>$row['/companyid/']</td>
                <td>$row['/government_reg_no/']</td>
                <td>$row['/company_name/']</td>
                <td>$row['/email/']</td>
            </tr>

        </tbody>
    </table>

    <?php
}
';
}
?>[![enter image description here][1]][1]

line 27 is

    <td align="center"><button type="submit" class="btn btn-default myButton" value="$row['/companyid/']" id="accept" name="accept">Remove</button></td>

What's your question? Please clearly explain your problem,comprising of your *expected* Vs. *current* output etc. — Rajdeep Paul, May 06 '17 at 19:49
is there code missing or typo ? `echo ' while ($row = mysqli_fetch_assoc($res)) {` -> do you try to echo a loop ? — OldPadawan, May 06 '17 at 19:55
please use `error_reporting(E_ALL); ini_set('display_errors', 1);` on top of your page and let us know what PHP tells you, if error... — OldPadawan, May 06 '17 at 19:56
@RajdeepPaul there is no error but show nothing when click search button — Danith Kumarasinghe, May 06 '17 at 20:07
there **must** be an error, check my 1st comment (any decent IDE would have showed you this at first sight though ^^) — OldPadawan, May 06 '17 at 20:14
`$row['/companyid/']` this also is reeeaally weird ! and you want to include as many times `jQuery library` as you have results for your search ? you have many many issues here... — OldPadawan, May 06 '17 at 20:15
@OldPadawan i got error like this ' Notice: Use of undefined constant companyid - assumed 'companyid' ' and Division by zero in C:\wamp\www\myIMP\PHP\searchrmvvac.php on line 27 in this manner i got so long error.line 27 is — Danith Kumarasinghe, May 06 '17 at 20:19
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/143579/discussion-between-danith-kumarasinghe-and-oldpadawan). — Danith Kumarasinghe, May 06 '17 at 20:30
@DanithKumarasinghe I've given an answer below. Hopefully this will resolve your issue. — Rajdeep Paul, May 06 '17 at 20:38

score 0 · Accepted Answer · edited May 23 '17 at 11:54

There are few issues with your code, such as:

You're not including/referring to any jQuery library in your code, so $("#searchrmvcom").click( ... won't work in the first place.
There's no point embedding your JavaScript/jQuery code in your HTML form, keep them separate.
There's no JavaScript/jQuery code in searchrmvvac.php page, so there's no point including the jQuery library here.
You're embedding PHP variables $row['...'] in your HTML table in the wrong way. Plus you're echoing things in the wrong way.

Your form page would be like this:

<form action="../PHP/searchrmvvac.php" method="post">
    <div class="search hidden-xs hidden-sm">
        <input type="text" placeholder="Search" id="search" name="search">
    </div>
    <div>
        <input type="button" value="Search" id="searchrmvcom" name="searchrmvcom">
    </div>
</form>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
<script>
    $("#searchrmvcom").click(function () {
        var comname=$('#search').val();
        $.ajax({
            type:"post",
            url:"../PHP/searchrmvvac.php",
            data:{comname:comname},
            success:function (data3) {
                $('#rmvcomdiv').hide();
                $('#ela').html(data3)
            }
        });
    });
</script>

Subsequently, your searchrmvvac.php page would be like this:

<?php
    session_start();
    require('../PHP/dbconnection.php');
    $output=$_POST['comname'];
    $sql="select * from company where company_name='$output' and activation_code=1";
    $res=mysqli_query($conn,$sql);
    if(mysqli_num_rows($res)>0) {
        while ($row = mysqli_fetch_assoc($res)) {
            ?>
            <table class="table table-striped table-bordered table-list">
                <thead>

                <tr>
                    <th>Action</th>
                    <th>ID</th>
                    <th>Registration number</th>
                    <th>Company Name</th>
                    <th>Email</th>
                </tr>
                </thead>
                <tbody>

                    <tr>
                        <td align="center"><button type="submit" class="btn btn-default myButton" value="<?php echo $row['companyid']; ?>" id="accept" name="accept">Remove</button></td>
                        <td><?php echo $row['companyid']; ?></td>
                        <td><?php echo $row['government_reg_no']; ?></td>
                        <td><?php echo $row['company_name']; ?></td>
                        <td><?php echo $row['email']; ?></td>
                    </tr>

                </tbody>
            </table>
            <?php
        }
    }
?>

Sidenote: Learn about prepared statement because right now your query is susceptible to SQL injection attack. Also see how you can prevent SQL injection in PHP.

show data through ajax in php

1 Answers1