Jump to content

Search the Community

Showing results for tags 'if statement'.



More search options

  • Search By Tags

    Type tags separated by commas.
  • Search By Author

Content Type


Forums

  • W3Schools
    • General
    • Suggestions
    • Critiques
  • HTML Forums
    • HTML/XHTML
    • CSS
  • Browser Scripting
    • JavaScript
    • VBScript
  • Server Scripting
    • Web Servers
    • Version Control
    • SQL
    • ASP
    • PHP
    • .NET
    • ColdFusion
    • Java/JSP/J2EE
    • CGI
  • XML Forums
    • XML
    • XSLT/XSL-FO
    • Schema
    • Web Services
  • Multimedia
    • Multimedia
    • FLASH

Calendars

  • Community Calendar

Group


AIM


MSN


Website URL


ICQ


Yahoo


Jabber


Skype


Location


Interests


Languages

Found 4 results

  1. Hi everyone, I need a little help with trying to figure out what I need to do. I am writing an IF statement that allowed only certain users can see. So what I am trying to do, in my DB i have it created that if you are going to have certain access, the value in the DB is set to '1' for true. What I need help with is seeing how I can write and IF statement that reads that value and allows the user to see it. This is the current code I have: <?php if (isset($_SESSION['id'])) { echo "Welcome"; } else { echo "You not logged in or don't have access."; } ?> With this code, anyone that logs in will be able to see it. What I want is another line of code that allows those with the value of '1' in the DB to see it. I have tried many different ways of seeing if I can get it to work but it never does. It either always returned the value TRUE and reads the code or the value is FALSE; even if when I am login, the value turn FALSE. So if anyone can give me just a quick moment of their time, that would be great. NOTE: My DB Connection is in a header.
  2. WHILE LOOP with IF STATEMENT

    I would like to create a stored routine for MySQL that figures out the number of business or working days for a month (Working Days are Monday thru Friday). I have not created the procedure portion of it yet as I want to ensure that the code works first. Please note that there is a function called first_day. This is a function provided by my hosting provider. I also have another function for this that will figure out the first day of the month if I didn't have it from the hosting company. It is not currently used and can be removed for testing purposes. I am testing this through phpmyadmin SQL Query. phpmyadmin uses the mysql extension and not the mysqli extension processing if that makes a difference. (The hosting company refuses to update it). It's a syntax error however I don't know what the syntax error is. All it tells me is: 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHILE(@daycount < @totaldays) DO IF (WEEKDAY(@checkweekday) < 6) THEN ' at line 2 My Syntax Error is in the following: WHILE(@daycount < @totaldays) DO IF (WEEKDAY(@checkweekday) < 6) THEN My Code: SELECT MONTH(CURDATE()) INTO @curmonth; SELECT MONTHNAME(CURDATE()) INTO @curmonthname; SELECT DAY(LAST_DAY(CURDATE())) INTO @totaldays; SELECT FIRST_DAY(CURDATE()) INTO @checkweekday; SELECT DAY(@checkweekday) INTO @checkday; SET @daycount = 0; SET @workdays = 0; BEGIN WHILE(@daycount < @totaldays) DO IF (WEEKDAY(@checkweekday) < 6) THEN SET @workdays = @workdays+1; END IF; SET @daycount = @daycount+1; SELECT ADDDATE('@checkweekday', INTERVAL 1 DAY) INTO @checkweekday; END WHILE; END; SELECT @workdays; I receive the same error with the following bit of code so it probably has something to do with this: SET @workdays = 0; IF (WEEKDAY('2013-06-13') < 6) THEN SET @workdays = @workdays+1; END IF; SELECT @workdays; Is someone able to assist?
  3. If statement doesn't work?

    I've created some if / else statements to get a download when a user hit click jf he fields name and email but doesn"t work for my site http://my-easy-woodw...ng-projects.com because it is do download without field name and email contact.php is <?php$field_name = $_POST['cf_name'];$field_email = $_POST['cf_email'];$field_message = $_POST['cf_message'];$mail_to = 'dedalusonline@yahoo.com';$subject = 'Message from a site visitor '.$field_name;$body_message = 'From: '.$field_name."\n";$body_message .= 'E-mail: '.$field_email."\n";$body_message .= 'Message: '.$field_message;$headers = 'From: '.$field_email."\r\n";$headers .= 'Reply-To: '.$field_email."\r\n";$mail_status = mail($mail_to, $subject, $body_message, $headers);if ($mail_status) { echo "<script>\n"; echo 'var str = \"download\"' . "\n"; echo "document.write(str.link(\"http://www.myshedplans.com/12BY8SHED.pdf\"));\n"; echo "</script>\n";}else { echo "<script language=\"javascript\" type=\"text/javascript\">"; echo "alert('Message failed. Please, send an email to gordon@template-help.com');"; echo "window.location = 'contact_page.html';"; echo "</script>";}?>
  4. MySQL If statement

    Morning/Afternoon All, I'm trying to create a simple if statement in SQLThis has been casuing me so much trouble since yesterdayHere's what i'm trying to achieve. IF((SELECT COUNT(*) FROM computers WHERE name='PC-01' AND domain='mydomain.local')==1)SELECT ComputerID FROM computers WHERE name='PC-01' AND domain='mydomain.local'ELSE IF((SELECT COUNT(*) FROM computers WHERE name='PC-01' AND domain='mydomain')==1)SELECT ComputerID FROM computers WHERE name='PC-01' AND domain='mydomain'ELSE IF(SELECT COUNT(*) FROM computers WHERE name='PC-01')==1)SELECT ComputerID FROM computers WHERE name='PC-01' I have tried: IF((SELECT COUNT(*) FROM computers WHERE name='PC-01' AND domain='mydomain.local')==1)BEGINSELECT ComputerID FROM computers WHERE name='PC-01' AND domain='mydomain.local'ENDELSE IF((SELECT COUNT(*) FROM computers WHERE name='PC-01' AND domain='mydomain')==1)BEGINSELECT ComputerID FROM computers WHERE name='PC-01' AND domain='mydomain'ENDELSE IF(SELECT COUNT(*) FROM computers WHERE name='PC-01')==1)BEGINSELECT ComputerID FROM computers WHERE name='PC-01'END Along with other suggests which google returned. to no luck. How can a simple IF statement be so complicated? (Side Note: using phpmyadmin i cannot do print/echo/prompt commands is that normal?)
×