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I am a newbie in jquery, I have a stupid question for onclick function. Please help me.
I have a table, when viewer click "Add new" button, I will using jquery add a row to existing table. In new row, there are 2 textbox to input data and a button to transfer data from textbox to php file.
This is my code:
html code:
XML
    <INPUT type="button" value="Add Row" onclick="addRow('dataTable')" />

    <INPUT type="button" value="Delete Row" onclick="deleteRow('dataTable')" />

    <TABLE id="dataTable" width="350px" border="1">
        <thead>
            <td>No</td>
            <td>First name</td>
            <td>Last name</td>
        </thead>
    <?php

                 include("functions/connect_db.php");
                  $no=0;
        $sql="SELECT * FROM `fullnames`";
        $result=mysqli_query($conn,$sql) or die('Could not select Machine'.mysqli_error($conn));
        while ($set=mysqli_fetch_array($result,MYSQLI_ASSOC))  {
                        $no++;
            $first=$set['firstname'];
            $last=$set['lastname'];
     ?>
        <tbody>
         <tr>
         <td><?php echo $no; ?></td>
         <td><?php echo $first; ?></td>
         <td><?php echo $last; ?></td>
     </tr>
      <?php } ?>
        </tbody>
    </TABLE>
</pre>
script:

<pre><pre lang="text">
<script>
      function addRow(tableID) {

            var table = document.getElementById(tableID);

            var rowCount = table.rows.length;
            var row = table.insertRow(rowCount);

            var cell1 = row.insertCell(0);
            cell1.innerHTML = rowCount ;


            var cell2 = row.insertCell(1);
            var element1 = document.createElement("input");
            element1.type = "text";
            element1.id="first_input_"+rowCount;
            cell2.appendChild(element1);


            var cell3 = row.insertCell(2);
           var element2 = document.createElement("input");
            element2.type = "text";
            element2.id="last_input_"+rowCount;
            cell3.appendChild(element2);


            var cell4=row.insertCell(3);
            var element3 = document.createElement('input');
            element3.setAttribute('type','button');
            element3.id="#button_"+rowCount;
            element3.className="butt_class";
            element3.value='Save';
            cell4.appendChild(element3);

        }

$(document).ready(function() {
    $(".butt_class").click(function()
{
    var ID=$(this).attr('id');
    var first=$("#first_input_"+ID).val();
    var last=$("#last_input_"+ID).val();
if(first.length<0 && last.length>0)
{

    $.ajax({
    type: "POST",
    url: "add_new.php",
    data: dataString,
    cache: false

    });
}

})
})
<script>

When I click to Save button in each row, nothing is happen. I tried by replace all click Save function with alert but there is not change.
Pls help me.
P/S: Sorry because I didn't describe error for this code.
Thank you in advance!
Posted
Updated 17-Sep-14 18:18pm
v4
Comments
Kornfeld Eliyahu Peter 17-Sep-14 3:48am    
Your question is not clear. You hit us with the code dump, but do not describe your actual problem...Please improve!
ngocham2001 17-Sep-14 5:38am    
I am sorry because I forgot describe error in this code. Please help me for this case. Thank you very much.
Kornfeld Eliyahu Peter 17-Sep-14 6:10am    
You may not see errors but I think if you would start to debug your web page (in browser debug) you will encounter some...
[no name] 17-Sep-14 4:15am    
Is the error coming as undefined is not a function? Or you might be missing some references.
Please explain about where you are getting the error or something.
Thanks
ngocham2001 17-Sep-14 5:39am    
I described problem of the code in above question. Pls help me. Many thanks.

1 solution

You can simply see the problem why it is failing to handle the click event of the button in the console tab of the developer tool for any browser. Follow the steps below :-

1) Open the page where you have the button.
2) Press F12 and developers tool will get opened.
3) Open 'Console' tab and now click on the button.

If any error is there which restricting it to be handled properly that will be displayed in console tab and you can get help to fix the issue.

Hope this will definitely of help.
 
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