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new to php - please help
Posted: Thu Apr 03, 2008 8:21 am
by jo_project08
Hi All,
I am rather new to php and am currently doing a project where I have set up a website along with mysql and php.
I am trying to do the following:
one one page I have a number of options with a sumbit button at the side:
<!--submit button -->
<td><form action="<?= $php_SELF ?>" method="POST">
</form><h5>navigation & search</h5></label>
<?php function trainee_course () {} ?>
<input name="trainee_course" type="submit" onclick="MM_goToURL('parent','individual_course_info2.php');return document.MM_returnValue" value="submit" />
then on the following page (which is linked) I am trying to display all training courses which are called 'navigation & search':
<!-- connect to the database -->
<?php
$con = mysql_connect("localhost","root","joanne");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("training", $con);
// perform the SQL query
function submit($query)
{}
function trainee_course()
{}
if(isset($trainee_course))
submit($query);
$trainee_course= "navigation & search";
$query = "select trainee_course, course_date from trainee_courses where trainee_course= '$trainee_course' ";
$result= mysql_query ($query)
or die("couldnt execute query.");
?>
everytime i run this it brings back my own error. Please help!!!
jo
Re: new to php - please help
Posted: Thu Apr 03, 2008 8:28 am
by it2051229
on the die('couldn't execute query').. would you mind changing it to "or die(mysql_error())" and then post the result here
Re: new to php - please help
Posted: Thu Apr 03, 2008 8:32 am
by jo_project08
Hi,
Thanks for replying, i get:
Unknown column 'trainee_course' in 'field list'
Cheers!
Re: new to php - please help
Posted: Thu Apr 03, 2008 8:34 am
by it2051229
check your database, seems like you did not create a COLUMN with that "field name" OR maybe you have misspelled it
Re: new to php - please help
Posted: Thu Apr 03, 2008 8:40 am
by jo_project08
I changed it and not getting any error now

thanks!
although I am still not getting the output.
I am wishing to display the course name and date.
cheers.
Re: new to php - please help
Posted: Thu Apr 03, 2008 8:43 am
by it2051229
Code: Select all
$query = "select trainee_course, course_date from trainee_courses where trainee_course= '$trainee_course' ";
for($i=0; $i<mysql_num_rows($query); $i++)
{
$traineeCourse = mysql_result($query,$i, "trainee_course");
$courseDate = mysql_result($query,$i, "course_date");
echo $traineeCourse."<---- Trainee Course";
echo $courseDate."<---- Course Date";
}
is that what you're looking for?? or something else
Re: new to php - please help
Posted: Thu Apr 03, 2008 8:52 am
by jo_project08
this is what I have :
<!-- connect to the database -->
<?php
$con = mysql_connect("localhost","root","joanne");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("training", $con);
// perform the SQL query
function submit($query)
{}
function course_name($course_name)
{}
if(isset($course_name))
submit($query);
$course_name= "navigation & search";
$query = "select course_name, course_date from trainee_courses where course_name= '$course_name' ";
for($i=0; $i<mysql_num_rows($query); $i++)
{
$course_name = mysql_result($query,$i, "course_name");
$course_date = mysql_result($query,$i, "course_date");
echo $traineeCourse."<---- Trainee Course";
echo $courseDate."<---- Course Date";
}
$result= mysql_query ($query)
or die(mysql_error());
?>
I am now getting the error:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\wamp\www\project\Admin\individual_course_info2.php on line 81
sorry for the hastle!
Re: new to php - please help
Posted: Thu Apr 03, 2008 8:57 am
by it2051229
wait you have done it wrong.... the QUERY should COME FIRST before the display of data something like
Code: Select all
$query = mysql_query("SELECT blah blah blah...") or die(mysql_error());
for($i=0; $i<mysql_num_rows($query); $i++)
{
echo "... blah blha blah";
}
Re: new to php - please help
Posted: Thu Apr 03, 2008 9:01 am
by it2051229
here.. i modified your code.. give it a try
Code: Select all
<!-- connect to the database -->
<?php
$con = mysql_connect("localhost","root","joanne");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("training", $con);
// perform the SQL query
function submit($query)
{}
function course_name($course_name)
{}
if(isset($course_name))
submit($query);
$course_name= "navigation & search";
$query = "select course_name, course_date from trainee_courses where course_name= '$course_name' ";
$result= mysql_query ($query)
or die(mysql_error());
for($i=0; $i<mysql_num_rows($result); $i++)
{
$course_name = mysql_result($result,$i, "course_name");
$course_date = mysql_result($result,$i, "course_date");
echo $traineeCourse."<---- Trainee Course";
echo $courseDate."<---- Course Date";
}
?>
Re: new to php - please help
Posted: Thu Apr 03, 2008 9:03 am
by jo_project08
yeah im getting the same error message as before along with the echoed line..any further suggestions???
Re: new to php - please help
Posted: Thu Apr 03, 2008 9:05 am
by it2051229
ok seems like there is no queried data..
to check try this..
Code: Select all
if(mysql_num_rows($result) > 0)
{
// place the display for loop code here
for($i=0; $i<mysql_num_rows($result); $i++)
{
/// blah blah blah
}
}
else
{
echo "SORRY THERE AIN'T NO DATA BOY!!";
}
Re: new to php - please help
Posted: Thu Apr 03, 2008 9:27 am
by jo_project08
no luck
$query = "select course_name, course_date from trainee_courses where course_name= '$course_name' ";
if(mysql_num_rows($query) > 0)
{
$course_name = mysql_result($query,$i, "course_name");
$course_date = mysql_result($query,$i, "course_date");
echo $course_name."<---- Trainee Course";
echo $course_date."<---- Course Date";
for($i=0; $i<mysql_num_rows($query); $i++)
else
echo "SORRY THERE AIN'T NO DATA BOY!!";
}
{
or die(mysql_error());
}
?>
its giving me an error for the else now!!!
Re: new to php - please help
Posted: Thu Apr 03, 2008 6:44 pm
by it2051229
lol..
.. try this
Code: Select all
$query = mysql_query("select course_name, course_date from trainee_courses where course_name= '$course_name' ") or die(mysql_error());
if(mysql_num_rows($query) > 0)
{
for($i=0; $i<mysql_num_rows($query); $i++)
{
$course_name = mysql_result($query,$i, "course_name");
$course_date = mysql_result($query,$i, "course_date");
echo $course_name."<---- Trainee Course";
echo $course_date."<---- Course Date";
}
}
else
{
echo "SORRY THERE AIN'T NO DATA BOY!!";
}
?>