Comments by Janardhanam Julapalli (Top 167 by date)

you are right. I did that mistakenly.

I am getting the json response properly in localhost but not in Server. Also the names like Abacavir 5’-β-D-Glucuronide Sodium Salt are coming as null. If I remove the single quote after the letter 5 it is displaying the name.

I hope it works. sorry If am wrong!

In the above code I need to replace the src with dynamic values coming from database(php).How can I do this? Quick help required!Thanks in Advance.

Any way thanks for the reply. Nice to have experts like u people for help.

Spot on @Richard Deeming! I already fixed the issue. I have noticed this array in the debugger and it is giving jsonData.data as undefined. I have changed my web service call like this:
return json_encode($project_ar_response);
instead of
return json_encode(array($project_ar_response));.

function CreateProperties(jsonData)
Here jsonData is having all the values like
0: Object
1: Object
2: Object
3: Object
4: Object
5: Object

When debugger goes to the for loop, it stops there only. It is not moving to the next step.

any one please help me out ASAP. Because of this I struck in the middle of the module.I have few more steps to do.

I have updated my question clearly.

just inside a div

This is my json data
[{"status":"1","error":"0","data":[{"project_id":"1","project_name":"Hoysala ACE Phase II","project_ar_url":"www.myfourwalls.in\/projects_data\/villa\/2BHK.wt3","project_ar_type":"Apartment","project_ar_title":"Exterior","project_marker":"www.myfourwalls.in\/projects_data\/marker\/villa.wtc"},{"project_id":"2","project_name":"WILASA Aikya","project_ar_url":"www.myfourwalls.in\/projects_data\/villa\/table.wt3","project_ar_type":"Plots","project_ar_title":"Exterior","project_marker":"www.myfourwalls.in\/projects_data\/marker\/villa.wtc"},{"project_id":"3","project_name":"Bluejay Ardley","project_ar_url":"www.myfourwalls.in\/projects_data\/villa\/villa.wt3","project_ar_type":"Villa","project_ar_title":"Interior","project_marker":"www.myfourwalls.in\/projects_data\/marker\/villa.wtc"},{"project_id":"4","project_name":"Hoysala Habitat","project_ar_url":"www.myfourwalls.in\/projects_data\/villa\/seater_sofa.wt3","project_ar_type":"Apartment","project_ar_title":"360 Model","project_marker":"www.myfourwalls.in\/projects_data\/marker\/villa.wtc"},{"project_id":"5","project_name":"AVS","project_ar_url":"www.myfourwalls.in\/projects_data\/villa\/bedroom.wt3","project_ar_type":"Villa","project_ar_title":"Exterior","project_marker":"www.myfourwalls.in\/projects_data\/marker\/villa.wtc"},{"project_id":"6","project_name":"JSONS","project_ar_url":"www.myfourwalls.in\/projects_data\/villa\/bedroom.wt3","project_ar_type":"Villa","project_ar_title":"Exterior","project_marker":"www.myfourwalls.in\/projects_data\/marker\/villa.wtc"}]}]

This is working for me

How can I compare this value with the value inside the json data. That is my main problem.That is reason y I posted this question.

am I doing it right?

function CreateProperties(jsonData) {
var items = jsonData.data;

jQuery(items.data).each(function () {
var CurrentProjId = jQuery(this)[0].project_id;
if (CurrentProjId = url) {
//$(this)[0] is the data you want
var project_name=items[i].project_name;
console.log(project_name);
}
});

@mgl chandra sekhar -- if(id==url)
{

var project_name=items[i].project_name;

console.log(project_name);

break;
}
this part is being skipped. I noticed in console.

please have a look at the question.I have updated it just now!

<script>
jQuery(document).ready(function()
{
jQuery.ajax(
{
type: 'GET',
url: 'projectNamesDropdown.php',
dataType: 'JSON',
success:function(data)
{
for(i=0;i<jquery(data).length;i++)
{
="" jquery('#project_name').append('&lt;option="" value="+ data[i].project_name+">' + data[i].project_name + '</option>');
}
}
});
});
</script>

y the value breaks down into pieces.How can I prevent that in jQuery ajax append.

<option name="project_name" value="Hoysala" ace="" phase="" ii="">Hoysala ACE Phase II</option>

Thank You for the reply.Thats fixed now.CORS helps.

May be because of cross-domain.How can I solve it.

may I know y this error is coming in console.
XMLHttpRequest cannot load http://myfourwalls.in/test/ar_property_details.php?url=undefined. No 'Access-Control-Allow-Origin' header is present on the requested resource. Origin 'http://localhost' is therefore not allowed access.

Sorry that I disturbed u and stolen some valuable time. I was doing a thing that is absolutely not needed. My superiors misdirected me. They need only the path and they are telling that now.

I fixed those.Now I am getting a bulk of encoded chars as output when I use fopen or fread or file_get_contents

please have a look at this link!
http://myfourwalls.in/test/file_reader.php

I run the above code, It shows errors like,

Warning: filesize() [function.filesize]: stat failed for http://myfourwalls.in/test/ar_images/models/Interior/2BHK.wt3 in /home/wwwmyfou/public_html/test/file_reader.php on line 3

Warning: fread() [function.fread]: Length parameter must be greater than 0 in /home/wwwmyfou/public_html/test/file_reader.php on line 3

simply I don't know how to do it.

I cant able to read that file at all in php.

I got the path from DB.Is this step is right?

Please help me out whether I am doing it in right way!

sorry for that! I did not noticed at all.

If u don't mind, please send me piece of code to do that. I am very new with PHP. Anyway I need to fix this issue by today end of the day.

I am getting all the image paths and .wt3 files from my DB and files are stored in server.

Done Sir!

Alright.What mistake exactly I have done here?

Thanks for your time and patience.I am grateful to u!

$link = mysql_connect("localhost", "root", "");
mysql_select_db("rohan_iksha", $link);

$test = "total_4bhk";
$result = mysql_query("select $test from project_master", $link);
$num_rows = mysql_num_rows($result);

while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
$row_array['total_4bhk'] = $row['total_4bhk'];

echo $row['total_4bhk'];
}
It is displaying the required values for me.

s

It is returning "53 Rows" as result

No sir.I am using the same db name before itself.

$conn = mysql_connect("localhost", "root", "myfourwalls");
mysql_select_db("rohan_iksha", $conn);
$num_rows = mysql_num_rows($conn);
echo $num_rows;

It is not returning anything for this.
$result = mysql_num_rows($conn);
echo $result;

53 rows

Showing rows 0 - 24 (53 total, Query took 0.0010 seconds.)

$towerValue_4bhk = "total_4bhk";
$floor_names_query = MYSQL_QUERY("select ".$towerValue_4bhk." from project_master");
This is what I am doing.Not getting anything.

$towerValue_4bhk = "total_4bhk"; Even I tried with hard coded value.

Can anyone please explain what exactly the query will do!

It is giving Resource id #4.I am not sure what it will be.

I tried it.But no success.

sorry! MYSQL_QUERY("select $towerValue_4bhk from project_master") is my requirement

In this script I already have the value called towerValue_1bhk.The same value I need to use it in the script that I have posted in the question.Is it possible?

jQuery(document).ready(function()
{
jQuery('.floor_plan_image_hotspot').hide();
var floor_count;
var body = jQuery("body");
body.delegate('.towers_for_1bhk', 'click', function()
{
var towerValue_1bhk = jQuery(this).find('span').html();
//alert(towerValue_3bhk);
jQuery.ajax(
{
url:"floors_for_1bhk.php?towerValue_1bhk=" + towerValue_1bhk,
type:"GET",
dataType:'JSON',
success:function(data)
{
//alert(towerValue_1bhk)
floor_count="";
for(i=0;i<jquery(data).length;i++)
{
="" if(floor_count="=" ""){
=""
="" floor_count="<div>Pick a Floor</div><hr><a href="#"><div name="floor_value" class="floors_for_1bhk" id="" +data[i].floor_name+'"=""><span name="floorValue_1bhk" class="floor_value">' + data[i].floor_label + '</span><div class="flats_available_for_1bhk">' + data[i].total_1bhk + '</div>' + '</div></div>';
}
else{
floor_count = floor_count + '<div name="floor_value" class="floors_for_1bhk" id="'+data[i].floor_name+'"><span name="floorValue_1bhk" class="floor_value">' + data[i].floor_label + '</span><div class="flats_available_for_1bhk">' + data[i].total_1bhk + '</div>' + '</div>';
}
}
if(floor_count!="" & floor_count!="undefined")
{
jQuery('#floor_count').html(floor_count);
}
}
});
});
});

I have this value in my another script.How can I use the same in this script

Thats not my requirement

Please help me out!

Thank You so much.

<div class="display_details"><div class="flat_no" id="'+data[i].flat_no+'">'+data[i].flat_no+'</div><div class="flat_area" class="flat_area" id="'+data[i].area+'">'+data[i].area+'</div><div class="flat_price" id="'+data[i].price+'">'+data[i].price+'</div><div class="flat_details"><input type="button" class="view_flat_details_3bhk" value="View details"></div></div>

I got it!My problem is with div structure.In my actual code,the static div is closing only after all the data is loaded inside of it.Now I have created a div dynamically and closing the div after each row of data.Now its working fine.

You are right!It should work fine.But for some reason it is not working for me.I don't know y?

should I remove h3 here?

When I click I am getting the required number.But I have 'n' number of such details.When I click on each item I should get that particular no.I am getting the same no. as alert here with this.

can I replace the "id" as "data-id" when I am generating it dynamically in jQuery

please have a look at my fiddle http://jsfiddle.net/6o7rahmt/

This has worked for me!Thanks a Lot buddy!

I need only the first value like G 1 2 3

I am updating my question with my jquery

But your code is working fine when I am doing it in a separate html file

I used something like this

jQuery(document).ready(function()
{
var body = $("body");
body.delegate('#groundFloor', 'click', function() {
var bhk1_floor0_value = jQuery('#floor_count').find('.floors_for_1bhk span:eq(0)').html();
console.log(bhk1_floor0_value);
});
});

When I add click event like this, click is not generating anything.Its idle.

http://jsfiddle.net/ke5y6q19/

please have a look at my jsfiddle

All the code inside the <div id="floor_count">is generated dynamically from jQuery with the values fetched from DB(php).

its not working!

I have created a folder called "projects_data" inside www.When I submit my form,I am sending the value called project_name and an image with variables called $project and file upload.I want to create a folder inside "projects_data" whose name should be the $project.Inside this folder I need to store the image.

$name is already creating the new folder but only the image is not stored in that path.

will it create a new folder in the name of $name

when I submit this, It is creating a folder with by $name and echo is producing with full path but the image is not stored in the required folder.

I am using above ajax code to send the images to php but I know that it will not work.I don,t know how to proceed.Just I need to store the selected image files to be stored in a folder via AJAX.

<script>
jQuery(document).ready(function()
{
jQuery('#submit').click(function()
{
var featured_image = jQuery('#featured_image')[0].files;
var no_of_files = jQuery('#featured_image')[0].files.length;

var form_data = new FormData();
form_data.append('myfile', featured_image)
$.ajax(
{
url: 'uploader.php', // point to server-side PHP script
dataType: 'text', // what to expect back from the PHP script, if anything
cache: false,
contentType: false,
processData: false,
type: 'post',
success: function(php_script_response)
{
alert(php_script_response); //display response from the PHP script, if any
}
});
});
});
</script>

I should not use any plugins.I should do it by ajax.

still cant do it but looks better than before.Form is few pixels higher than the adminpanel.It works fine in Firefox, but it is creating a vertical scroll bar for the form.

My problem is with #sidebar-left and #content

mycss file:

.navbar-header{
background-color:#088A85;
}

.logout a{
text-decoration:none;
float:right;
margin: 5px 10px;
}

#sidebar-left{
background-color:#088A85 ;
float: left;
width:150px;
height:2050px;

}

#content{
background-color:#BDBDBD;
height:2050px;
padding-left:150px;


}


.dropdown-pages, .dropdown-pages a , .sub-menu-pages a, .dropdown-media a{
text-decoration:none;
padding-top:20px;
font-size:20;
}

.sub-menu-pages, .sub-menu-media{
display:none;
width:100px;
}
#menu a, .dropdown-pages, .dropdown-media{
text-decoration:none;
padding-left: 10px;
padding-top:20px;
font-size:20;
width:140px; float:left; overflow: hidden;

}
.test{

overflow: auto;
}

When I click on Add New, it is appending a new page inside the required div.But the size of the div is much larger than both content and sidebar.Also it is changing from system to system.I am manually adjusting the width in each and every system that I run this program.How van I solve this?

I am getting 500 Internal Server Error for this!Is there any errors in this code or any other reason for that.When I give some other .php file in action it is working fine.

ok.I will check.

But the thing is I have nearly 15 input fields to be sent to .php file. SO I need to send everything through jQuery.ajax.I am Updating my whole form data in my Question.

Okay!Thanks for Your time and patience!Have a Great Day!

Not in c# and .NET.Is it possible in Javascript?

I am good enough to differentiate between class and ID.Even It was not working.When I click the menu, the php page was just appearing and disappearing.It looked strange to me yesterday!

Also Please share any links which helps me in making text boxes and textareas as Rich Texboxes.

I Have used jQuery('#test').load('load.php').May I know why it does not work?

Thanks for the Explanation!

Can u please tell me what this line of code will do
$('#sidebar-left').unbind('click').click(function(e){

Thank You very much dude!It worked for me and You started my morning with some success.

Both files are in the same directory only.I am totally upset that I cant crack this.Yesterday It loaded in the content div and I don't know what happened after that.Please anyone help me!

http://jsfiddle.net/Janardhanam/sbot73L9/
look at this fiddle please

is there a facility to share a screen shot here?

even I tried with only add_new.php.

now is it okay to understand?

I am trying my best to align the code but cant do it!

Thank You.I will check.

Done with it!Thank u so much for the Help.Everyday I am learning atleast one or two things from this forum.This site has been a great help for me since I started learning PHP.

I will check.

should I need to edit this examples/gmail.phps file

previously I was getting "mail sending failed" as echo.Now I am not getting anything.When I see in console(network) it is showing 500-Internal server error.

I have placed the PHPMailer-master folder in the same directory where my code is present.Now I need to include this require 'PHPMailerAutoload.php'; in my .php file right?

Do i need to download this library and place it in my required server location

how can i do that

no

Thank u all for ur replies.I am seeing this forum as very useful one.

Very Impressive dude@Jorgen Andersson

@pikoh-This is working, but there is a small alias error.This is my updated code
SELECT * FROM
(SELECT *
FROM user_details
ORDER BY salary DESC
LIMIT 2)
user_details ORDER BY salary
LIMIT 1

@George Jonsson-This is working fine.

This is working fine.@member60

Here I need to append the values from database to UI in a structure where user can edit those values and can Re-submit their values.

Anyone think all programmers(may be Javascript) know this answer?

Thanks for the explanation.I was thinking that some obvious reason might be there.But it is so simple.

hi i have posted a question yesterday.No replies so far. Please help me out if possible.

Anyone help me out on this

thanks man its working

thanks man its working

I have posted a question

can i ask u one more query?

Fractions are not displayed properly.The above fractions are improper ones.I need those fractions to be displayed in correct format.

I have achieved my functionality in this way.
var category = jQuery('#parentCategory').val();
var subcategory = jQuery('#subCategory').val();
and appended these values as
url: 'result.php?category='+category+'&subcategory='+subcategory,

I am clicking on checkbox, selecting a category value from DD1, selecting subcategory value from DD2 and then save button.
Ex:sports(cat)->cricket(subcat).I want the selected values to be stored in DB in the same table where i have saved the video details only once for the whole loop.

where should I place the $_POST['category']?inside foreach or outside?

In the same example i want to insert the drop down selected values into the Database by using same result.php file or the new one.How can i do that?

Sure sure.Thank You very much.

Got Success.Its working.Thank you very much for your valuable time.If i have any queries in future how can i contact you.

Still not working.The problem is that the data is not coming to result.php through json_decode.

I can't format the whole code.Can i send u email?

I have added the same question as a new question

This is for youtube API.Very large code.I will paste the code here.

The url is hitting the result.php. But there is no response coming from the front.I checked it in REST Client in Firefox.

k k.That i can do?

No success.Can u tell me any other option with my existing code.I can able to insert one row atleast. It will be a great help for me.

very disgusting.you are my only hope.

previously i was getting data:"success" message in console.Now i am getting " "

when i click save button, nothing is happening. No error messages also.

yes i have done.No success.

I have done this already above the AJAX call.
var jsonString = JSON.stringify(videoDetails);

same error again

In console i can see the values but not inserted into DB.

its not working.An error occurred! Internal Server Error.This is the error.

The above code is working fine.
Can i use foreach like this: foreach($videoDetails as $value)
and where should i use it?Please help me out

where should i use foreach?

i used foreach, but no success for me.I am very new for PHP.

When i click one check box and submit, i am getting an array containing 5 values from POST.Also i can able to insert those values into DB.But when i click two check boxes, i will receive two arrays. how can i insert those values into DB.

please check it now

what should i do?really i don't know.

this is my php code

please can anyone help me out?