Question

I have a xcode project in which i need to enter data on button click and then display it in the TableView.

Example: Twitter (follow/unfollow).

1.I click on friend’s icon, a list of my friends appear.

2.When I click on any of my friends, their friends list appear in tableview with follow buttons corresponding to them (mutual friends have unfollow buttons).

3.Now, when I click on the follow button and go back to check my friend list, that particular friend has been added twice.

What I have done so far:

Checked if the sql db exist somewhere else as well.

Checked if the webservice is being called twice.

Everything’s fine. I have iMac and Macbook Pro as well and both of them have same issues.
I have reinstalled Xampp twice but still the same result.

For your reference, following is the php i am using for my Xcode project.

<?php
 $id =  $_REQUEST['id'];
 $fid = $_REQUEST['fId'];
 $firstName = $_REQUEST['firstName'];
 $lastName = $_REQUEST['lastName'];

 $con = mysql_connect("localhost","root","");
 if(!$con)
 die("Connection failed"."<br/>");

 $db = mysql_select_db("db",$con);
 if(!$db) die("Database not found!"."<br/>");

 $query = "INSERT INTO $id(firstName,lastName,folks_id)     VALUES('$firstName','$lastName','$fid')";
 print  $query;
 mysql_query($query,$con);
 mysql_close($con);

?>

EDIT: Xcode part for reference

-(IBAction)followButtonClicked:(UIButton *)sender
    {

  NSMutableString *postString = [NSMutableString stringWithString:kFollowURL];
    [postString appendString: [NSString stringWithFormat:@"?%@=%@", kId, [user objectForKey:@"id"] ]];
    [postString appendString: [NSString stringWithFormat:@"&%@=%@", kfId, [fId objectForKey:@"fID"] ]];
    [postString appendString: [NSString stringWithFormat:@"&%@=%@", kfirstName, [firstNameDict objectForKey:@"firstName"] ]];
    [postString appendString: [NSString stringWithFormat:@"&%@=%@", klastName, [lastNameDict objectForKey:@"lastName"] ]];
    [postString appendString: [NSString stringWithFormat:@"&%@=%@", kprofileType, [profileTypeDict objectForKey:@"profileType"] ]];

    [postString setString: [postString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
    NSLog(@"post string = %@", postString);

    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:postString]];

    [request setHTTPMethod:@"POST"];

    followConnection = [[NSURLConnection alloc] initWithRequest:request delegate:self startImmediately:YES];
    NSLog(@"postconnection: %@", followConnection);

    //Get JSON Response from server

    NSData *dataURL =  [NSData dataWithContentsOfURL: [ NSURL URLWithString: postString ]];    
    NSString *serverOutput = [[NSString alloc] initWithData:dataURL encoding: NSASCIIStringEncoding];

    NSLog(@"serverOutput = %@", serverOutput);

 } 

Answer 1

[SOLVED] The thing which did the trick was that i made fid as unique… so now whenever i click on the follow button, it is entered one and not twice.

Thanx to all of you guys for the help and support.