[Syntax error - creating a table in PHPmyadmin]

You havent specified the list of values for the enum type field.

[Daily updating table?]

any code?

Starting from scratch or if I can find one online

I am pretty sure I am going to make it 2 tables

One table will have all the info for each day

the other table will be the view and it will update each day, connected to the first table

 

so what does this have to do with mysql?

[Daily updating table?]

any code?

[Help with my new project]

So you don't want to use an RDBMS and you want to somehow use cookies to keep track of user uploads.

Sounds terribly inefficient and I don't see it working.

How will you separate each users pictures if each user does not have a unique id?

Where will you store the images?

[echo php in a variable with html]

You are already inside php tags, you don't need them again.'

You need to concatenate the array value since the single index quotes will break the string.

 

$ext = ' <form action="/members/clubs/view-activity.php" method="post" name="venuefrm">
            <button type="submit" id="close" class="link"><span>View Activity</span></button>
            <input name="activity" type="hidden" value="' . $activites['activityID'] . '" />
            </form>'; 

[PHP Image upload help: copy() [function.copy]: Filename cannot be empty in]

$HTTP_POST_FILES is deprecated and should no longer be used.

Use the superglobal $_FILES array instead.

You should be checking for the files existence before attempting to copy it.

Also, use move_uploaded_file

[Run php in a table cell]

Then you really have two options, you can integrate the login.php script into the page that you want to call it from, and simply toggle its visibility onclick (I think xyph was after this solution).

Or you can use jquery and the .load() function (using the links I provided) to load the script into a specified element onclick.

To use jquery you will need to have the library on your server or link to an external repository (google and several others host the lib).

[Run php in a table cell]

Oh, you want to simply toggle visibility of a cell?

I was under the impression that you wanted to inject a page onclick.

My mistake. Disregard my post.

If you want the element hidden without it taking up it designated space, use display: none;

[loop adds an extra entry to db]

Well, there are a few misleading and incorrect posts in this thread.

My first reply was to make sure that you were using array names in the inputs.

If they were not in array format then the code you posted would have been flat out incorrect.

Since they are in array format, only your loop needed changing, as PFM and myself suspected.

You can either change <= to <, or set i = 1;

When looping arrays, I myself like to set the index to 0 as you have done.

[Run php in a table cell]

http://api.jquery.com/category/ajax/

 

particularly, http://api.jquery.com/load/

[loop adds an extra entry to db]

As I suspected, the names are in array format.

Which means if you make the change that PFM suggested, it will add the correct amount of rows.

If it doesn't, something else is wrong and you should post the updated relevant code.

 

[Warning: mysql_select_db() expects parameter 2 to be resource]

No where in the code do I see the variable $default being defined.

Judging from the error, I would say that it isn't at all, unless it's coming from the required page.

[PHP query always returns a result as interger]

Well my friend, the error is coming from the code YOU gave me.  So I am asking you: why this error?

 

It has nothing to do with the code that i gave you, it has to do with where you put it.

[loop adds an extra entry to db]

Do you have several fields with the same name? That doesn't work. Each field is unique and needs its own name that will then be posted.

 

Hence, you should neither loop on POST variables. They are atomic strings.

 

Neither of these statements are true.

you can have input names in array format, though it's typically avoided.

[loop adds an extra entry to db]

This code makes little to no sense.

Why are you treating $_POST values as arrays by using count and sizeof, when they should be strings.

Why are you using a loop at all? unless you have multiple <inputs>s will array names.

But if you are only expecting one row to be inserted, I'd say that's not the case.

[changing click event from h2 to a span]

In the click function, change children() back to siblings()

[changing click event from h2 to a span]

Oh, i wasn't sure which span you wanted it attached to.

 

$('.dragbox')
.each(function(){
	$(this).hover(function(){
		$(this).find('h2').addClass('collapse');
	}, function(){
	$(this).find('h2').removeClass('collapse');
	})
	.find('h2').hover(function(){
		$(this).find('.configure').css('visibility', 'visible');
	}, function(){
		$(this).find('.configure').css('visibility', 'hidden');
	})
                .end()
	.find('span.handle')
                .click(function(){
		$(this).children('.dragbox-content').toggle();
		updateWidgetData();
	})
	.end()
	.find('.configure').css('visibility', 'hidden');
});;

[changing click event from h2 to a span]

$('.dragbox')
.each(function(){
	$(this).hover(function(){
		$(this).find('h2').addClass('collapse');
	}, function(){
	$(this).find('h2').removeClass('collapse');
	})
	.find('h2').hover(function(){
		$(this).find('.configure').css('visibility', 'visible');
	}, function(){
		$(this).find('.configure').css('visibility', 'hidden');
	})
                .end()
	.find('div.dragbox-content span')
                .click(function(){
		$(this).siblings('.dragbox-content').toggle();
		updateWidgetData();
	})
	.end()
	.find('.configure').css('visibility', 'hidden');
});;

[PHP query always returns a result as interger]

Error:  Notice: Undefined variable: strCaptainName in E:\Web Server\baltimorebeach_com\htdocs\registration\verify.php on line 34

 

However the username now did not have the integer.

 

this error is self explanatory..

[PHP query always returns a result as interger]

I am really not trying to be a jerk here. 

 

The query itself is $strCaptainSearch = "SELECT COUNT(*) FROM captain WHERE username LIKE '%s%%'";

 

However I did not feel that this part of the code was revelant.  My apologies.  Does this help now?

 

It is absolutely relevant, using a format modifier inside of a LIKE statement will often produce unexpected results.

Try this:

 

$sqlCaptainSearch = sprintf
    (
        "SELECT COUNT(*) FROM captain WHERE username LIKE '%s'",
        "%" . $strCaptainName . "%"
    );

 

I'm actually surprised that your original query didn't throw an error.

[new to form submission - getting error]

i'm talking about most of the document.

I want to see the <form> and the js used to submit it.

[Issue with LEFT JOIN query]

`timeStart` is either a DATE or a DATETIME type field. Which one is it?

Also, make sure that `fkEquipID` and `timeStart` are not ambiguous.

Also, use mysql_error to debug the query, assuming your RDBMS is mysql.

[Table Styling Question]

Thanks AyKay47.

I do plan on using an external style sheet in my project.

But my real question is can I style the <td> in the stylesheet or should it be done inline?

 

sure, there are different ways to go about this, it depends on how many td elements you want to style.

You can give each <td> a class and use the external style sheet to style them. e.g

 

external css:

 

td.class
{
height: 20px;
background-color: red;
}

 

html:

 

<td class='class'>something</td>

 

just incorporate it into the php loop.

If you want to style only one td element, you can substitute class for id.

[new to form submission - getting error]

if you are sure that frm is correct, then the variable is either getting overwritten or is not being injected correctly into the document.forms[] call.

Post the entire relevant code.

[new to form submission - getting error]

have you checked to make sure that frm is what you expect it to be?