<tr id="<?php echo $id?>">
<input type="hidden" value="<?php echo $id?>"/>
<td> <?php echo $id;?> </td>
<td id="fname"> <?php echo $firstname[$key];?> </td>
<td id="lname"> <?php echo $lastname[$key];?> </td>
<td id="tage"> <?php echo $age[$key];?> </td>
</tr>
jQuery:
$('.edit').click (function(){
var id = $(this).parent().data('id');
var fname = $('#id').find('#fname');
alert(fname);
});
Want to get the text inside <td>
In alert it returns an object.
What am I doing wrong?
You are using a string instead of an jquery object for your selector.
$(id).find('#fname')
You might consider this, if the value you're retrieving is that of an id, so it will need the hash to be a valid selector:
$('#' + id).find('#fname')
Also, valid markup requires id's to be unique on the page.
Related
I am trying to post the value of fields in a table through Ajax to process.php that will process the data.
the table:
<form name="items">
<table width="90%" border="1">
<tbody>
<tr>
<td width="26%">Item Name</td>
<td width="22%">If other then give name</td>
<td width="22%">Quantity</td>
<td width="16%">$/Unit</td>
<td width="14%">Total</td>
</tr>
<?php for ($i = 1; $i <= 10; $i++) {
?>
<tr>
<td>
<select name='itemname<?php echo $i;?>' id='itemname<?php echo $i;?>'>
<option value="other">other</option>
<?php
$qry_item_name = "SELECT DISTINCT item_name FROM bus_name_details";
$result_item_name = mysql_query($qry_item_name);
while($row_item_name = mysql_fetch_array($result_item_name)) {
$option .="<option>" . $row_item_name['item_name'] . "</option>";
echo $option;
}
?>
</select>
</td>
<td>
<input type="text" name="other<?php echo $i;?>" id="other<?php echo $i;?>"></td>
<td>
<input type="text" name="quan<?php echo $i;?>" id="quan<?php echo $i;?>" value="0"></td>
<td><input type="text" name="unit<?php echo $i;?>" id="unit<?php echo $i;?>" onkeyup="calculateTotal('<?php echo $i;?>')"></td>
<td><span name="total<?php echo $i;?>" id="total<?php echo $i;?>"></span></td>
</tr>
<?php }?>
</tbody>
</table>
</form>
<span id="subm3" class="subm3" >Submit </span>
The Ajax:
<script>
$(document).on('click', '.subm3', function() {
var datas = {};
for ($i = 1; $i < 2; $i++) {
// drop down list of items
datas["item_name"+$i]= $('#itemname'+$i+' option:selected').attr('value')
// other name
datas["other"+$i]= $('input[name=other'+$i+']').val()
// quantity
datas["quan"+$i]= $('input[name=quan'+$i+']').val()
// price per unit
datas["unit"+$i]= $('input[name=unit'+$i+']').val()
}
$i = $i_limit-1;
$.ajax({
url: 'parts/process.php?order=3&items='+$i,
type: 'POST',
contentType: 'application/json',
data: JSON.stringify(datas)
});
});
</script>
then the process :
<?php
if ($order==3){
$loop = $_GET[items];
$other1= $_POST["other1"];
echo "other1 = ".$other1;
}
?>
On chrome inspector i get that the data was sent like this as source :
{"item_name1":"other","other1":"Water Valve","quan1":"2","unit1":"2"}
and like this as parsed :
item_name1: "other"
other1: "Water Valve"
quan1: "2"
unit1: "2"
but the response from process.php (the one that got the values) seems to only have read the get values but not the post values :
other1 =
i am pretty much new to ajax and hope someone could help point me to where i am going wrong here. This might not even be the easiest or best way of doing what i am trying to do, but it kind of makes sense to my brain at the moment so i am taking this route. Any help is appreciated
can you try data:datas insted of data: JSON.stringify(datas) in ajax request
ref: how to send multiple data with $.ajax() jquery
The problem occurs when I insert without data. So I want your help in solving this problem.
This is my file:
students.php
<form id="student_form" method="POST" action="">
<?php
if(mysql_num_rows($q)>0){
?>
<table border="0" dir="ltr" align="center"cellpadding='0' cellspacing='0' >
<tr> <th>Student ID</th><th>Name</th><th>AVG</th><th>Joining Year</th><th>Message</th><th>Sending Message</th> </tr>
<?php while($row = mysql_fetch_array($q)){ ?>
<tr>
<td id="stud_id[]"> <?php echo $row['studant_ID']; ?></td>
<td> <?php echo $row['studant_Name']; ?></td>
<td> <?php echo $row['Average']; ?></td>
<td> <?php echo $row['year']; ?></td>
<td> <input id="message1[]" name="message1[]" type="text" size="25px" /></td>
<td><input name="submit[]" id="submit[]" type="submit" value="Send" /> </td>
</tr>
<?php }}
and this is my insert file:
insert_message.php
if (isset($_POST['message1']) && $_POST['message1']!='') {
$addss = mysql_real_escape_string($_POST['message1']);
}
if (isset($_POST['stud_id']) && $_POST['stud_id']!='') {
$std_id = mysql_real_escape_string($_POST['stud_id']);
}
//#######################################################################
$query1 = "INSERT INTO `message` (`rcvrid`, `senderid`, `msg`) VALUES ('$std_id', '22011111', '$addss'); ";
mysql_query($query1);
I connect between two file by jquery and ajax.
<script>
$("#student_form").on("submit", function(event) {
event.preventDefault();
$.ajax({
type: "POST",
url: "insert_message.php",
data: $(this).serialize(),
success: function(data) {
$("#inner_contant").append(data+"<br/>");//instead this line here you can call some function to read database values and display
},
});
});
</script>
Remove the form from the page
<tr>
<td id="stud_id"> <?php echo $row['studant_ID']; ?>
<input type="hidden" name="stud_id" value="<?php echo $row['studant_ID']; ?>"/>
</td>
<td> <?php echo $row['studant_Name']; ?></td>
<td> <?php echo $row['Average']; ?></td>
<td> <?php echo $row['year']; ?></td>
<td> <input id="message1" name="message1" type="text" size="25px" /></td>
<td><button class="submit" type="submit" />Send </button> </td>
</tr>
second:
your js should look like this:
$(".submit").on("click", function(event) {
event.preventDefault();
$.ajax({
type: "POST",
url: "insert_message.php",
data: {stud_id:$(this).closest('tr').find('input[name="stud_id"]').val(),message1:$(this).closest('tr').find('input[name="message1"]').val()},
success: function(data) {
$("#inner_contant").append(data+"<br/>");//instead this line here you can call some function to read database values and display
},
});
});
In insert_message.php
you need to echo a message to see if you where succesful in updating the database
echo json_encode(array('message'=>'Data updated/Error'));
First of all - all data passed to server in a $_POST array is taken from input fields with name attribute (unless you have some custom js handlers).
So
<td id="stud_id[]"> <?php echo $row['studant_ID']; ?></td>
does nothing.
If you want to store studant_ID somehow - use hidden field for example:
<td>
<input type="hidden" name="stud_id[]" value="<?php echo $row['studant_ID']; ?>" />
</td>
Next - what do you want from this buttons:
<input name="submit[]" id="submit[]" type="submit" value="Send" />
As they are all belong to one form they will do the same - send all fields to server. If you want every submit button send to server only a pair student_id, message you have to create a form for each pair (or do some js-handlers). Otherwise, on your server you'll have:
$_POST['stud_id'] array of all students ids from form
$_POST['message1'] array of all messages from form
If you want to process them all - do a foreach:
foreach ($_POST['stud_id'] as $key => $id) {
// find the message
$msg = $_POST['message1'][$key];
$query1 = "INSERT INTO `message` (`rcvrid`, `senderid`, `msg`) VALUES ('$id', '22011111', '$msg'); ";
mysql_query($query1);
}
And of course you should remove mysql_ functions and use newer apis - PDO or mysqli
<input type="text" name="a" id="a" />
CLICK
Now this is my question that can we pass an input tag value of HTML to 'mk' so that value can be used in the next page for process.
Though i have done my project in a different way but still i want to know wether is it possible to do so .
I have searched a lot but none of the question is same as i got so plz help me out.
And i dnt want to use form so i just want to know can we pass this value using href tag r not.
This is the code i have
<form method="post">
<table align="center" bgcolor="#FFFFCC" border="1">
<tr><th>ID</th><th>PRODUCT</th><th>PRICE</th><th>DATE OF POST</th><th>PHONE NUMBER</th><th>AUTHENTICATE</th></tr>
<?
$sel = mysql_query("SELECT * FROM addunauth WHERE adminauthorize = 'uncheck'") or die("CANNOT FETCH DATA FOR ADMIN " . mysql_error());
if (mysql_num_rows($sel) > 0) {
while ($data = mysql_fetch_array($sel)) {
?> <tr>
<td><? echo $data['id']; ?></td>
<td><? echo $data['product_name']; ?></td>
<td><? echo $data['product_sp']; ?></td>
<td><? echo $data['product_time']; ?></td>
<td><? echo $data['phoneNumber']; ?></td>
<td><input type="text" name="a" id="a" /></td>
<td align="center">
<a href="processadminTask.php?id=<? echo $data['id']; ?>&mk=" >
AUTHENTICATE
</a>
</td>
</tr>
<?
}
} else {
?> <tr><td>-</td><td>-</td><td>-</td><td>-</td><td>-</td><td>-</td></tr>
<? }
?>
</table>
</form>
Now just tell me how to pass the input value
Your variable not parse inside single quotes. your <? echo '$data['id']' ?> should be <?php echo $data['id'] ?>
<input type="text" name="a" id="a" />
CLICK"
<a id="link" href="xxx.php?id=111&mk=">CLICK</a>
Edit:-
simply add one more hidden fields to store id value as well and get value in js and assign to href
<input type="text" name="idval" id="idval" value="100" />
<input type="hidden" name="mk" id="mkval" value="101" />
Js:-
var idval = $('#idval').val();
var mk = $('#mkval').val();
var link = 'xxx.php?id=' + idval + '&mk=' + mk;
$("#link").attr("href", link);
Working Demo
<? echo '$data['id']' ?>
Is wrong PHP Syntax
Use
<?php echo $data['id']; ?>
Like:
CLICK
Your
<? echo '$data['id']' ?>
is not correct. You can actually use something like this
<?= $data['id'] ?>
so you'll have
CLICK"
I think it is possible with javascript:
getElementById('id of html input').value
This will get you the input value and then you can assign it to 'mk'
So you can check with it
Try this
<?php
$data['id'] = 2;
?>
<html>
<input type="text" name="a" id="a" />
CLICK
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script >
$(document).ready(function(){
$('a[href]').click(function(event){
event.preventDefault();
event.stopPropagation();
var inputTagID = $('#a').val();
var hrefVal = this.toString().replace("&mk=", "&mk="+inputTagID);
window.location = hrefVal;
});
});
</script>
</html>
I have an html table built from a database query which loops through and creates a button for each camera name found and puts them in a table:
<?php
for($i=0;$i<$num_rows;$i++)
{
?>
<tr>
<td>
<input type="submit" class="play" data-hash="<?php echo $result_cameras[$i]["camera_hash"]; ?>" value="<?php echo $result_cameras[$i]["camera_name"]; ?>">
</td>
</tr>
<?php
}
?>
This resolves to something like this:
<tr>
<td>
<input type="submit" class="play" data-hash="0d3d0ac6e54a640c73f1149d4d0bbc38e99d10f5" value="Office Window">
</td>
</tr>
<tr>
<td>
<input type="submit" class="play" data-hash="b824cba374c3d5ab7806ad8260c939323c03147b" value="aaa">
</td>
</tr>
<tr>
<td>
<input type="submit" class="play" data-hash="ec9658f0c1855e2e2ac09ae284f5e6990dbf445d" value="laptop">
</td>
</tr>
Notice the data hash attribute is different for each button. I want to process this button with my jquery code:
$(".play").click(function(){
var camerahash = $('input').data('hash');
console.log($('input').data('hash'));
});
No matter which button I click I will always get the hash from the first button I click: 0d3d0ac6e54a640c73f1149d4d0bbc38e99d10f5. Any help is appreciated.
$('input').data('hash') gives you the data attribute of the first input in the selection use $(this).data('hash') to get the data attribute of the currently clicked input
You need to specify which input element to read.
Try something like:
$(".play").click(function(){
$this = $(this);
var camerahash = $this.data('hash');
console.log($this.data('hash'));
});
You are always calling the first object of .play. This would be a correct way:
$('.play').on('click', function(){
var camerahash = $(this).data('hash');
});
You could always grab them by using the .attr(data-hash) html5 attribute.
Try:
$('.play').on('click', function() {
var _hash = $(this).attr('data-hash');
console.log(_hash);
});
Your selector will return the first one that it comes to. Use this instead
<script>
$("input.play").click(function() {
alert($(this).data('hash'));
});
</script>
I have a parent page which has a drop down list in it. using the onchange event, data is posted to a second page using $.post(). This page uses the posted variables to output mysql data with a checkbox for each line. This page output is then inserted into the parent page using jquery $('#DIV name').html(output).show();
The user can then see the mysql table rows with corresponding checkboxes. They then select the checkboxes they want and say delete. Delete is a form submit button.
My question is, when they click delete how do I take the form data from the second page and post it with that of the parent page so that I can then use $_POST[] to get the checkbox info and delete the selected table rows?
example of the parent page code is:
javascript/jquery
<script type="text/javascript">
function get(row){ //row being processed, defined in onchange="get(x)"
('getpeopleinjobs.php',{ //data posted to external page
postvarposition: form["position"+row].value, //variable equal to input box, sent to external page
postvarjob: form["job"+row].value, //variable equal to input box, sent to external page
postvarperson: form["person"+row].value, //variable equal to drop down list, sent to external page
postrow: row}, //variable equal row being processed, sent to external page
function(output){
$('#training'+row).html(output).show(); //display external results in row div
//popupWindow = window.open('t4.php?variable1Name=Vicki&variable2Name=Maulline','popUpWindow','height=400,width=1000,left=10,top=10,resizable=yes,scrollbars=yes,toolbar=no,menubar=no,location=no,directories=no,status=yes')
});
}
</script>
Form data is
<tr>
<td>
<?PHP echo $i; ?>
</td>
<td>
<input type=text NAME="position<?PHP echo $i; ?>" id="position<?PHP echo $i; ?>" style="border: 1px solid #2608c3;color:red; width=200px" value="<? echo mysql_result($resultpositionjob,$r,0);?>">
</td>
<td>
<input type=text NAME="job<?PHP echo $i; ?>" id="job<?PHP echo $i; ?>" style="border: 1px solid #2608c3;color:red; width=200px" value="<? echo mysql_result($resultpositionjob,$r,1);?>">
</td>
<td>
<SELECT NAME="person<?PHP echo $i; ?>" id="person<?PHP echo $i; ?>" style="border: 1px solid #2608c3;color:red; width=200px" onchange="get(<? echo $i; ?>);">
<OPTION VALUE=0 >
<?=$optionpeople?>
</SELECT>
</td>
<td onclick="train(<? echo $i; ?>);" style="color:grey; cursor: pointer;">
<div id="training<?PHP echo $i; ?>"><font color=grey size=2></div>
</td>
<td>
</td>
</tr>
<?PHP
$i++;
$r++;
}
?>
The second page or page called by jquery, the output is:
echo
"
<table border=0 width=400>
<tr>
<td width=20>
</td>
<td width=150>
<b>Position<b>
</td>
<td>
<b>Job<b>
</td>
</tr>
";
while($line = mysql_fetch_array($result))
{
echo "
<tr>
<td>
";
?>
<input type=checkbox name="delete[]" id="delete[]" value="<?php echo $rows['id']; ?>">
<?PHP
echo "
</td>
<td>
";
echo $line['position'];
echo "
</td>
<td>
";
echo $line['job'];
echo "
</td>
</tr>
";
}
?>
<tr>
<td>
<input type=submit name="update" id="update" value="Update">
</td>
</tr>
</table>
<?PHP
}
To repeat I want the table checkbox element from the second page posted with the form data of the parent page.
Is this possible as my testing hasnt given me any luck.
Am I doing something wrong or do I need to modify the jquery code?
Thanks as always for the assistance.
There is no second page. Really - you're loading HTML content from the second page, but it's being inserted into the parent page. All content, from your browsers perspective, is in the same page. The fields should be included as long as they're inside the DOM inside the element. Use the developer tools for your browser or Firebug for Firefox to make sure that the content is placed within the form element in the DOM. The developer tools should also be able to show you exactly which variables are submitted to the server when the form is submitted.
The 'action' parameter of 'form' will indicate where the content gets submitted (and it seems you've left that part out of your HTML, so it's impossible to say if you've left out or just not included it in the paste.