JavaScript

Re: Page appearing way to wide.

3-Mar-17 21:14

Please do not expect people to go searching for details of your problem, post the details in your question.

ZurdoDev7-Mar-17 9:23

ZurdoDev

7-Mar-17 9:23

A forum member already solved it for you. But it is quite rude to only post a link and expect us to go hunt down what the issue is.

There are two kinds of people in the world: those who can extrapolate from incomplete data.

There are only 10 types of people in the world, those who understand binary and those who don't.

Setting SignalR URL in JQuery

Kevin Marois2-Mar-17 12:21

Kevin Marois

2-Mar-17 12:21

I'm following this example

There's a code block about halfway down where there is

JavaScript

<script type="text/javascript">
        $(function () {
            // Declare a proxy to reference the hub. 
            var chat = $.connection.chatHub;
            // Create a function that the hub can call to broadcast messages.
            chat.client.broadcastMessage = function (name, message) {
                // Html encode display name and message. 
                var encodedName = $('<div />').text(name).html();
                var encodedMsg = $('<div />').text(message).html();
                // Add the message to the page. 
                $('#discussion').append('<li>' + encodedName
                    + ':  ' + encodedMsg + '</li>');
            };
            // Get the user name and store it to prepend to messages.
            $('#displayname').val(prompt('Enter your name:', ''));
            // Set initial focus to message input box.<br />
            $('#message').focus();
            // Start the connection.
            $.connection.hub.start().done(function () {
                $('#sendmessage').click(function () {
                    // Call the Send method on the hub. 
                    chat.server.send($('#displayname').val(), $('#message').val());
                    // Clear text box and reset focus for next comment. 
                    $('#message').val('').focus();
                });
            });
        });
    </script>

I tried adding

JavaScript

$.connection.url = "myurl"

right above the $.connection, but I then get "Cannot read property '' of undefined"

I take this to mean that the ".connection' object is not set.

What's wrong here? How does the URL to the server get set?? Confused | :confused:

If it's not broken, fix it until it is.
Everything makes sense in someone's mind.
Ya can't fix stupid.

modified 2-Mar-17 18:33pm.

Re: Setting SignalR URL in JQuery

Nathan Minier3-Mar-17 1:26

Nathan Minier

3-Mar-17 1:26

You are correct, that means that $.connection is not defined at the point where you're trying to set the URL.

MS has a JavaScript API reference for SignalR at:
ASP.NET SignalR Hubs API Guide - JavaScript Client | Microsoft Docs

"There are three kinds of lies: lies, damned lies and statistics."
- Benjamin Disraeli

JavaScript Error - Internet Explorer - Object doesn't support property or method 'indexOf'

Kevin Marois2-Mar-17 10:54

Kevin Marois

2-Mar-17 10:54

Using IE11. I only get this error on IE. The app works fine in Chrome.

Here's the JS

JavaScript

<script src="/Scripts/jquery-3.1.1.min.js"></script>

<script src="/Scripts/jquery.signalR-2.1.2.js"></script>

<script src="/signalr/hubs"></script>

<script type="text/javascript">
    $(function () {

    var proxy = $.connection.DashboardHub;

    proxy.client.NotifyAllClientsOfChanges = function () {

        var searchUrl = "Home/GetData";

        $.ajax({
            url: searchUrl,
            type: "POST",
            success: function (data) {
                $("#divData").html(data);
            }
        });
    };
});

When I run the file jquery-3.1.1.min.js opens, and this is highligted:

h=a.indexOf(" ")

and I get the error "Object doesn't support property or method 'indexOf'"

Google results say that IE doesn't support this. Anyone heard of this? Any way to make this work?

If it's not broken, fix it until it is.
Everything makes sense in someone's mind.
Ya can't fix stupid.

(SOLVED) Call to Member function execute() on null...

samflex25-Feb-17 11:36

25-Feb-17 11:36

Even though this forum got me to where I am now on this project due largely the generous help of Richard Deeming, I have tried hard to avoid coming back here but I am having so much difficulty getting to the finish line which is this last bit.

First, on the screen that we use to preview data, I have this:

PHP

<body>
<form action='final.php' method = 'POST'>
<input type="hidden" name="employeename" value="<?php echo $employeename; ?>">
<input type="hidden" name="ttitle" value="<?php echo $ttitle; ?>">
<input type="hidden" name="email" value="<?php echo $email; ?>">
<?php foreach ($rowIDs as $id) { ?>
<input type="hidden" name="sourcename" value="<?php echo $_POST['sourcename' . $id]; ?>">
<input type="hidden" name="sourceaddress" value="<?php echo $_POST['sourceaddress' . $id]; ?>">
<input type="hidden" name="income" value="<?php echo $_POST['income' . $id]; ?>">
<?php }?>
<?php foreach ($row2IDs as $id) { ?>
<input type="hidden" name="spousename" value="<?php echo $_POST['spousename' . $id]; ?>">
<input type="hidden" name="spouseAddress" value="<?php echo $_POST['spouseAddress' . $id]; ?>">
<input type="hidden" name="spouseIncome" value="<?php echo $_POST['spouseIncome' . $id]; ?>">
<?php } ?>
<a href="javascript:history.go(-1)" data-icon="back" data-role="button" data-theme="b">Return to make changes</a>
<input type="submit" value="submit" />
</form>

There is more but I thought this is relevant part.

Then on the next page (final.php) after user clicks the submit button, I have this:

PHP

<?php

// Database connection
$conn = mysqli_connect("localhost","myusername","mypass","myDB");
if(!$conn) {
die('Problem in database connection: ' . mysql_error());
}

// Data insertion into database

$query = 'INSERT INTO `mydb`.`wp_mytable` ( `employeename`, `ttitle`, `email` )'
     . ' VALUES ( ? , ? , ? )';
if( $sth = mysqli_prepare($conn,$query) ) {
   mysqli_stmt_bind_param($sth,'sss'
      ,$_POST["employeename"]
      ,$_POST["ttitle"]
      ,$_POST["email"]
   );
   if( mysqli_stmt_execute($sth) ) {
     echo 'Successfully created employee record;';
   } else {
      printf("Error: %s\n",mysqli_stmt_error($sth));
   }
} else {
   printf("Error: %s\n",mysqli_connect_error($conn));
}
$last_id = mysqli_insert_id($conn);


$sql = 'INSERT INTO `myDB`.`wp_myTable` ( `employeeID`'
     . ', `sourcename`, `sourceaddress`, `income`,`spousename`,`spouseAddress`,`spouseincome` )'
     . ' VALUES ( ? , ? , ? , ? , ? , ? , ? )';

if( $sth = mysqli_prepare($conn,$sql) ) {
for($i=0; $i<count($_POST['sourcename']); $i++)
{
$stmt->execute(array($last_id,
$_POST['sourcename'][$i],
$_POST['sourceaddress'][$i],
$_POST['income'][$i]),
$_POST['spousename'][$i],
$_POST['spouseAddress'][$i],
$_POST['spouseIncome'][$i]));
 }

   if( mysqli_stmt_execute($sth) ) {
     echo '<h1>Success!</h1><br>Your record successfully submitted <br><br><h1>Thank You!</h1';
   } else {
      printf("Error: %s\n",mysqli_stmt_error($sth));
   }
} else {
   printf("Error: %s\n",mysqli_connect_error($conn));
}
?>

If I can get records to submit correctly, I am done with this project.

However, I am stuck here with the following error messages:

Notice: Undefined variable: stmt in C:\xampp\htdocs\closures\forms\final.php on line 37

Fatal error: Call to a member function execute() on null in C:\xampp\htdocs\closures\forms\final.php on line 37

This error starts on this line:

$stmt->execute(array($last_id,...

I have spent my entire weekend on this without any progress. I will be truly grateful once again for solution.

Thanks

modified 28-Feb-17 11:38am.

Richard MacCutchan25-Feb-17 20:47

25-Feb-17 20:47

samflex wrote:
Notice: Undefined variable: stmt in C:\xampp\htdocs\closures\forms\final.php on line 37

You have never declared the variable stmt but you are trying to use it in the statement at line 37. You probably meant to write

PHP

$sth->execute(array($last_id,

Using more descriptive variable names would help to find such errors more easily.

samflex25-Feb-17 21:25

25-Feb-17 21:25

Thank you very much Richard.

Resolved those issues but created two new ones:

Warning: mysqli_stmt::execute() expects exactly 0 parameters, 4 given in C:\xampp\htdocs\closures\forms\final.php on line 43
Error: No data supplied for parameters in prepared statement

Any ideas?

Richard MacCutchan25-Feb-17 21:29

25-Feb-17 21:29

samflex wrote:
Any ideas?

The best idea is to make use of the documentation: PHP: mysqli_stmt::execute - Manual[^].

samflex25-Feb-17 21:37

25-Feb-17 21:37

I think the issue is that I have 7 parameters and need to bound them with the $sth as in

PHP

mysqli_stmt_bind_param($sth,'sssssss'($last_id, ...

but how do I integrate this with array?

Richard MacCutchan25-Feb-17 21:42

25-Feb-17 21:42

See PHP: mysqli_stmt::bind_param - Manual[^].

samflex25-Feb-17 21:47

25-Feb-17 21:47

Thanks Richard but I know how to do that though.

You can see I have that on one of the INSERT statements in my original post.

I just can't use with array.

That's where the struggle comes in.

Richard MacCutchan26-Feb-17 21:06

26-Feb-17 21:06

samflex wrote:
I just can't use with array.

Why not, it is exactly the same issue?

samflex27-Feb-17 0:34

27-Feb-17 0:34

Sorry sir but why can't I get a sample from you if you know how?

I would have solved it by now if I know the answer.

I have been playing around with it since your last response and can't get it to work.

Richard MacCutchan27-Feb-17 0:42

27-Feb-17 0:42

samflex wrote:
can't get it to work.

Sorry but I do not understand what your problem is. Whether you have individual items, or items in an array, you can still provide their references in your call to mysqli_stmt_bind_param

samflex27-Feb-17 3:22

27-Feb-17 3:22

Ok thanks for your help.

Hopefully, someone else will be able to help.

Here is my INSERT statement:

PHP

$sql = 'INSERT INTO `disclosures`.`wp_disclosures` ( `employeeID`'
     . ', `sourcename`, `sourceaddress`, `income`,`spousename`,`spouseAddress`,`spouseincome` )'
     . ' VALUES ( ? , ? , ? , ? , ? , ? , ? )';

if( $sth = mysqli_prepare($conn,$sql) ) {
for($i=0; $i<count($last_id); $i++)
{
$sth->execute(array($last_id,
$_POST['sourcename'][$i],
$_POST['sourceaddress'][$i],
$_POST['income'][$i]),
$_POST['spousename'][$i],
$_POST['spouseAddress'][$i],
$_POST['spouseIncome'][$i]);
 }

Just given the way the INSERT statement is written, I am getting the following error:

rning: mysqli_stmt::execute() expects exactly 0 parameters, 4 given in C:\xampp\htdocs\closures\forms\final.php on line 43
Error: No data supplied for parameters in prepared statement

But I believe that I am getting those errors because I could not bind the parameters like I did with the other INSERT statement like this:

PHP

( $sth = mysqli_prepare($conn,$query) ) {
   mysqli_stmt_bind_param($sth,'sss'
      ,$_POST["employeename"]
      ,$_POST["ttitle"]
      ,$_POST["email"]
   );

Notice this line:

sqli_stmt_bind_param($sth,'sss'

I need to find a way to add something like this:

sqli_stmt_bind_param($sth,'ssssssss' to my INSERT statement to get resolve the problem.

Just can't figure out how to integrate that with the mysqli_prepare(...) statement or the $sth->execute(array($last_id,... array statement.

Richard MacCutchan27-Feb-17 3:33

27-Feb-17 3:33

You are just repeating your original question, and I already explained the answer. What exactly is the problem in adding the parameter references to the mysqli_stmt_bind_param statement? All you need is:

PHP

mysqli_stmt_bind_param($sth,'ssssssss',
        parameter1,
        parameter2,
        parameter3,
        parameter4,
        parameter5,
        parameter6,
        parameter7,
        parameter8
      );

samflex27-Feb-17 4:04

27-Feb-17 4:04

Sorry sir but I keep getting half-baked answers that confuse me more.

I tried what you just gave.

Here it is again. I am very sorry for keep dumping code.

PHP

if( $sth = mysqli_prepare($conn,$sql) ) {
for($i=0; $i<count($last_id); $i++)
{
$sth->execute(array(
mysqli_stmt_bind_param($sth,'sssssss',
$last_id,
$_POST['sourcename'][$i],
$_POST['sourceaddress'][$i],
$_POST['income'][$i]),
$_POST['spousename'][$i],
$_POST['spouseAddress'][$i],
$_POST['spouseIncome'][$i]));
 }

This now gives me the same error I got before when I tried it like that:

Fatal error: Only variables can be passed by reference in C:\xampp\htdocs\closures\forms\final.php on line 40

I am just trying to show what I have tried.

Again, integrating binding the parameters with the array is an issue that I am struggling with.

This is just the las phase of this annoying project.

Richard MacCutchan27-Feb-17 4:21

27-Feb-17 4:21

PHP

$sth->execute(array(

What is the array expression supposed to mean here?

samflex27-Feb-17 4:31

27-Feb-17 4:31

I may be using it incorrectly but without the array, I am getting an error that says I am trying to insert string when I should be inserting array of records.

Richard MacCutchan27-Feb-17 4:38

27-Feb-17 4:38

I do not think it helps that you are trying to use compound statements, which make it more difficult to find the error. Try something like:

PHP

if( $sth = mysqli_prepare($conn,$sql) ) {
    for($i=0; $i<count($last_id); $i++)
    {
        mysqli_stmt_bind_param($sth,'sssssss',
        $last_id,
        $_POST['sourcename'][$i],
        $_POST['sourceaddress'][$i],
        $_POST['income'][$i]),
        $_POST['spousename'][$i],
        $_POST['spouseAddress'][$i],
        $_POST['spouseIncome'][$i]));
        $sth->execute();
    }
}

samflex27-Feb-17 5:35

27-Feb-17 5:35

still getting the same error - Only variables can be passed by reference.

Richard MacCutchan27-Feb-17 6:49