Posts by nav33n

You can use array_multisort. More on array_multisort here .

How do you want to split it ?

When the form is submitted, get the value of the select box. $selectboxvalue = $_POST;
Then update the table depending upon the value of $selectboxvalue !

Explode the value on +. Then add the required fields.
Eg.

$value1 = $_POST['field1'];
//consider $value1 has 100+xyz
$value2 = $_POST['field2'];
//$value2 = 300+abc
$newval1 = explode("+",$value1);
$newval2 = explode("+",$value2);
//$newval1 and $newval2 will be an array where $newval1[0] will have 100, $newval1[1] will have xyz and $newval2[0] will have 300, $newval2[1] will have the value abc. 
$total = $newval1[0] + $newval2[0];
echo $total; // 100+300 = 400
?>

Hope that helps :)

Welcome :)

Welcome Babs!

Welcome to Daniweb xa2009 !

Welcome!

Welcome to Daniweb!

Use <?php instead of <? (very bad habit).

Matti Ressler
Suomedia

I don't think thats the problem. Because, If short tags are disabled on his computer, it will just print the 'code' as text.

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Welcome to Daniweb Johnny :)

Hi Sodium1! Welcome to Daniweb!

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Thanks nav33n.. I think it's working now.. I change the the function query not to return anything since it's a delete query. did that caused the error? "Warning:mysql_num_rows(): supplied argument is not a valid mysql result resource in .... line 42

Ah! You are right ! :)

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When the user clicks apply for the first time, get his username (and password ?) from the table. Add it to the session variable. Then check if the session variable exists. If it does, then don't ask him for username and password. If session variable doesn't exist, ask him for username and password and add it to the session.

Welcome !

Welcome !

Welcome to Daniweb! :)

Welcome to Daniweb :)

Welcome to Daniweb!

Welcome to Daniweb :)

Welcome to Daniweb :)

Hmm.. Well, how bout this. When you click on a button, call a javascript function, which opens a new window, window.reload the parent window ?
Maybe someone has an answer for this.

Cheers,
Naveen

Dude, you have to be specific on the errors you get. What do you mean by 'error on mysql_num_row' ? What is the error ?

Try this.

unction exequery($query){
if (!$this->conn) {
die('Could not connect: ' . mysql_error());
}
echo $query;
exit;
$results = mysql_query($query) or die(mysql_error());
//mysql_close($this->conn);
return $results;
}

Check if its printing the query. Also in the class, check if database connection is established.

Well, AFAIK, If you want the parent window to be refreshed, you can't do it using php. That can only be done using javascript.
You can declare a php variable in javascript. But you can't use it in javascript. But this works.

<script type='text/javascript'>
function test() {
<?php 
$x=1000;
?>
....
}
</script>

Return some value on execution of the delete query. Check if its returning the right value.

I dont see an "action" in the html form ! When is your php script getting called ? And why do you have different forms for different form variables ?!?

but its not doing it

Which part of the script isn't working ? Executing the delete query or establishing the connection ?

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Yep. Javascript is the only way where you can have a popup. Well, check this thread. http://www.daniweb.com/forums/thread114307.html The OP wanted something same.

textbox inside <?php tags ? Yep, you have to echo it. But this way. echo "<input type=\"text\" name=\"name\" value=\"$firstname\">";

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int
aqua
ant

Discussion

http://www.w3schools.com/php/php_mysql_select.asp Check this. Instead of echoing $row and $row in the example, assign it to a variable and use it as the 'value' of the textbox. <input type="text" name="firstname" value="<?php echo $firstname; ?>">

In the php script, where is $username coming from ? Have a hidden variable in the upload form and pass the username. Check if that user already have a folder. If yes, copy the image, else, create a folder and copy it. What error are you getting ?

General idea is... I click a button, goes through php code, opens a new page and then refresh orginal page.

You mean the 'new page' is a popup and 'original page' is the parent window ?

count is not a reserved word AFAIK. hmmm..

What do you mean 'is not working' ? What error are you getting ?

Welcome to Daniweb! :)

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Umm.. Update or insert ? Print out your query to see if all the required values are passed. Print_r($checkbox_name) will print the values of those checkbox which were selected.

:) You are welcome!

Can you post your code ?

Use a variable with another name.. not $weekday ! :)

Huh! thats strange. Before foreach loop, assign null to $weekday and then try again !

Edit: Or, use another variablename $weekday1. Check this example.

<?php
echo $id;
?>
<html>
<body>
<form method="post" action='test.php'>
<input type=text name=id[]><br>
<input type=submit name=submit value=submit>
</form>
</body>
</html>

Form variables are global by default. (I think).

Thats because, while inserting, you insert this way. insert into table (col1) values ('value'); But since your value already has a ', its considered as the end of value, throwing an error. If you are using php, you can escape ' or " by using mysql_real_escape_string or addslashes.

Basically, its like this. You have for example 3 pages. homepage.php, courses.php and students.php.
homepage.php is a login script, where the user can login. You should check if the user is a valid user. You shouldn't let the user to go to page2 or page3 (ie., courses.php and students.php) if the user hasn't logged in. For this, you can use sessions. Once the user has logged in, add his username to the session. ($_SESSION=$username; ). Then on page2 and page3, check if $_SESSION is not null. If its null, then he isn't a valid user, so redirect him back to homepage.php. If he's a valid user, show him page2. That can be done as shown by petr.pavel in post2.

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