Hello, I have three courses, .net, j2ee, php with starting dates like 12-01-2007, 13-01-2007, 14-01,2007, the courses will be displayed in drop down box. the question is how to write the html code for this logic, if a user select .net then its starting date should be displayed i.e, 12-01-2007, if a user select j2ee then its starting should be displayed like 13-01-2007. Any help in this matter. Thanks.
you mean the html for it ?? <select name="course"> <option value="12-01-2007">.NET</option> <option value="13-01-2007">J2ee</option> <option value="14-01-2007">PHP</option> </select> HTML:
Courses should be displayed in one drop down list and immediately after it a text box should display the starting date, i.e when the user select .NET from the droop down list, its starting date should be displayed in the below text box i.e 12-01-07, how to write the code for this logic. Thanks.
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Untitled Document</title> </head> <body> <form name="mainform" method="post" action=""> <select name="course" onchange="document.mainform.date.value = this.value"> <option value="">Choose a course</option> <option value="12-01-2007">.NET</option> <option value="13-01-2007">J2ee</option> <option value="14-01-2007">PHP</option> </select> <input type="text" name="date" /> </form> </body> </html> HTML:
I have created this table in the database <?php $con=mysql_connect("localhost","root",""); if(!$con) { echo "Error". mysql_error(); } mysql_select_db("my_student7",$con); $sql="CREATE TABLE person(course varchar(30), date DATE)"; if(mysql_query($sql,$con)) { echo "Table created"; } else { echo "Error". mysql_error(); } mysql_close($con); ?> Code (markup): I am inserting the values in the database i.e., course and date values from the html form above in the database <?php if (!($con = mysql_connect("localhost","root",""))) die('Could not connect: ' . mysql_error()); else mysql_select_db( "my_student7", $con ); $course=stripslashes(trim($_POST['course'])); $date=stripslashes(trim($_POST['date'])); if( mysql_query( "INSERT INTO person (course, date) VALUES ('$course', '$date')",$con) ) { echo "Your details are added in the database and you can use this id, <b>$id</b>, for furture references"; } else { echo "Your details could not be inserted at this time, please contact support"; } mysql_close($con); ?> Code (markup): When i am trying to display the details i.e course and date, only date is displayed, ` <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("my_student7", $con);$result = mysql_query("SELECT * FROM person"); echo "<table border='1'>"; while($row = mysql_fetch_array($result, MYSQL_NUM)) { echo "<tr><td>$row[0]</td><td>$row[1]</td></tr>"; }echo "</table>"; mysql_close($con); ?> Code (markup): course is not displayed how to write the code for this logic. Thanks.