Re: [PHP] Finding next recored in a array
At 4:14 PM -0400 9/19/07, brian wrote:
tedd wrote:
At 11:52 AM -0400 9/17/07, brian wrote:
tedd wrote:
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next.
What I do not understand is if I know the last number was say 5
how do I tell the script that that is the current number so I
can select the next record
What the next record?
Try:
$array = array('0','1','3','5','8','15','25');
$val = 5;
echo($array[array_search($val, $array)+1]);
Cheers,
tedd
Not quite:
$array = array('0','1','3','5','8','15','25');
$val = 25;
echo($array[array_search($val, $array)+1]);
Notice: Undefined offset: 7 ...
brian
Duh?
You program for that -- you want me to write the entire code?
~sigh~
Grasshopper, the point i was trying to make is that your example
displays what *not* to do with array_search(). Though a wonderful
teaching aid in itself, it falls somewhat short of being a
reasonable solution for the OP.
brian
~sigh~
Sorry to disappoint you master. But I was using array_search() to
search an array -- it seemed like a reasonable thing to do.
So why do you say it's an example of what *not* to do with array_search()?
Please be specific.
Cheers,
tedd
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Re: [PHP] Finding next recored in a array
tedd wrote:
At 4:14 PM -0400 9/19/07, brian wrote:
tedd wrote:
Duh?
You program for that -- you want me to write the entire code?
~sigh~
Grasshopper, the point i was trying to make is that your example
displays what *not* to do with array_search(). Though a wonderful
teaching aid in itself, it falls somewhat short of being a reasonable
solution for the OP.
brian
~sigh~
Sorry to disappoint you master. But I was using array_search() to search
an array -- it seemed like a reasonable thing to do.
OK, so my joke left a bitter taste; sorry about that.
What i'm trying to get across is that this list is here so that we may
post our various problems in the hopes that someone else might see the
solution we are missing. Sometimes, when we post what we *think* is the
correct path to scripting Nirvana, another of us may point out a flaw in
our logic.
It happens to all of us.
The responses to this list should not be taken to be either complete or
fully-tested solutions. But, neither should potential bugs be ignored if
noticed by any other reader. While your response certainly does *seem*
to work just fine, it contains a bug that *will* bite at some point.
So why do you say it's an example of what *not* to do with array_search()?
Please be specific.
Do you really expect me to write the entire code? Perhaps you could
just meditate on this a little longer:
Not quite:
$array = array('0','1','3','5','8','15','25');
$val = 25;
echo($array[array_search($val, $array)+1]);
Notice: Undefined offset: 7 ...
brian
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Re: [PHP] Finding next recored in a array
At 11:52 AM -0400 9/17/07, brian wrote:
tedd wrote:
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next.
What I do not understand is if I know the last number was say 5
how do I tell the script that that is the current number so I can
select the next record
What the next record?
Try:
$array = array('0','1','3','5','8','15','25');
$val = 5;
echo($array[array_search($val, $array)+1]);
Cheers,
tedd
Not quite:
$array = array('0','1','3','5','8','15','25');
$val = 25;
echo($array[array_search($val, $array)+1]);
Notice: Undefined offset: 7 ...
brian
Duh?
You program for that -- you want me to write the entire code?
Cheers,
tedd
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Re: [PHP] Finding next recored in a array
tedd wrote:
At 11:52 AM -0400 9/17/07, brian wrote:
tedd wrote:
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next. What
I do not understand is if I know the last number was say 5 how do I
tell the script that that is the current number so I can select the
next record
What the next record?
Try:
$array = array('0','1','3','5','8','15','25');
$val = 5;
echo($array[array_search($val, $array)+1]);
Cheers,
tedd
Not quite:
$array = array('0','1','3','5','8','15','25');
$val = 25;
echo($array[array_search($val, $array)+1]);
Notice: Undefined offset: 7 ...
brian
Duh?
You program for that -- you want me to write the entire code?
~sigh~
Grasshopper, the point i was trying to make is that your example
displays what *not* to do with array_search(). Though a wonderful
teaching aid in itself, it falls somewhat short of being a reasonable
solution for the OP.
brian
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Re: [PHP] Finding next recored in a array
Rick Pasotto wrote:
On Sun, Sep 16, 2007 at 06:04:45PM -0700, Richard Kurth wrote:
Richard Kurth wrote:
Rick Pasotto wrote:
On Sun, Sep 16, 2007 at 07:09:02PM -0400, brian wrote:
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next. What I
do not understand is if I know the last number was say 5 how do I tell
the script that that is the current number so I can select the next
record
||
I think you'll need your own function for this.
Nope. Just use array_search().
$k = array_search('5',$Campaign_array);
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
I tried this and it gives me nothing back. It should give me a 8
$Campaign_array= array('0','1','3','5','8','15','25');
$val=5;
$k = array_search($val,$Campaign_array);
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
I figured out way it was not working $k + 1 count needed to be $k + 1
count
Yup. Sorry 'bout that.
But now it works perfect
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
is very over-the-top. What's wrong with
if (isset($Campaign_array[$k+1]) { echo $Campaign_array[$k + 1]; }
Not to mention I don't really like the whole way this has been handled,
I'm sure there are better ways, but for this we'd need to know more
about what you're doing and what you're trying to achieve, which we
obviously don't.
- Tul
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Re: [PHP] Finding next recored in a array
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next. What
I do not understand is if I know the last number was say 5 how do I
tell the script that that is the current number so I can select the
next record
What the next record?
Try:
$array = array('0','1','3','5','8','15','25');
$val = 5;
echo($array[array_search($val, $array)+1]);
Cheers,
tedd
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Re: [PHP] Finding next recored in a array
tedd wrote:
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next. What I
do not understand is if I know the last number was say 5 how do I tell
the script that that is the current number so I can select the next
record
What the next record?
Try:
$array = array('0','1','3','5','8','15','25');
$val = 5;
echo($array[array_search($val, $array)+1]);
Cheers,
tedd
Not quite:
$array = array('0','1','3','5','8','15','25');
$val = 25;
echo($array[array_search($val, $array)+1]);
Notice: Undefined offset: 7 ...
brian
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Re: [PHP] Finding next recored in a array
Richard Kurth wrote:
$query = SELECT day
FROMemailcampaign
where campaign_id = '$emailcampaign'
AND member_id = '$members_id'
;
$DB_Change_Campaign_Results = safe_query($query);
while ( $row = mysql_fetch_array($DB_Change_Campaign_Results) ) {
$Campaign_array[$row['day']] = $row;
}
# At this point you have arrays as values for your $Campaign_array
# So, unless $k is an array that matches a sub array of $Campaign_array
# you're never going to get a match
$k = array_search($val,$Campaign_array);
# What is $k at this point? An int (1, 2, 3, etc...) , string (Sunday, Monday,
etc...)
# Before I go any further I will need to know the above information.
if ( ($k + 1) count($Campaign_array) ) {
echo $Campaign_array[$k + 1];
}
--
Jim Lucas
Some men are born to greatness, some achieve greatness,
and some have greatness thrust upon them.
Twelfth Night, Act II, Scene V
by William Shakespeare
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Re: [PHP] Finding next recored in a array
M. Sokolewicz wrote:
Rick Pasotto wrote:
On Sun, Sep 16, 2007 at 06:04:45PM -0700, Richard Kurth wrote:
Richard Kurth wrote:
Rick Pasotto wrote:
On Sun, Sep 16, 2007 at 07:09:02PM -0400, brian wrote:
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next.
What I do not understand is if I know the last number was say 5
how do I tell the script that that is the current number so I
can select the next record
||
I think you'll need your own function for this.
Nope. Just use array_search().
$k = array_search('5',$Campaign_array);
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k +
1]; }
I tried this and it gives me nothing back. It should give me a 8
$Campaign_array= array('0','1','3','5','8','15','25');
$val=5;
$k = array_search($val,$Campaign_array);
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
I figured out way it was not working $k + 1 count needed to be
$k + 1 count
Yup. Sorry 'bout that.
But now it works perfect
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
is very over-the-top. What's wrong with
if (isset($Campaign_array[$k+1]) { echo $Campaign_array[$k + 1]; }
Not to mention I don't really like the whole way this has been
handled, I'm sure there are better ways, but for this we'd need to
know more about what you're doing and what you're trying to achieve,
which we obviously don't.
What I am trying to is get all the days from a table where email
campaign = number. and then look at the last day that was sent and find
it in the list and get the next day that follows that day. At this point
the script below is not working. So if you have any Ideas that would be
a better way of doing this please let me know.
$query = SELECT day FROM emailcampaign where campaign_id =
'$emailcampaign' AND member_id = '$members_id';
$DB_Change_Campaign_Results = safe_query($query);
for ($i=0; $i mysql_num_rows($DB_Change_Campaign_Results); $i++) {
$Change_Campaign_row = mysql_fetch_array($DB_Change_Campaign_Results);
$Campaign = $Change_Campaign_row['day'];
$Campaign_array[$Campaign] = $Change_Campaign_row;
}
$k = array_search($val,$Campaign_array);
if ($k + 1 count($Campaign_array)){ echo $Campaign_array[$k + 1]; }
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Re: [PHP] Finding next recored in a array
Richard Kurth wrote:
What I am trying to is get all the days from a table where email
campaign = number. and then look at the last day that was sent and find
it in the list and get the next day that follows that day. At this point
the script below is not working. So if you have any Ideas that would be
a better way of doing this please let me know.
$query = SELECT day FROM emailcampaign where campaign_id =
'$emailcampaign' AND member_id = '$members_id';
$DB_Change_Campaign_Results = safe_query($query);
for ($i=0; $i mysql_num_rows($DB_Change_Campaign_Results); $i++) {
$Change_Campaign_row = mysql_fetch_array($DB_Change_Campaign_Results);
$Campaign = $Change_Campaign_row['day'];
$Campaign_array[$Campaign] = $Change_Campaign_row;
}
$k = array_search($val,$Campaign_array);
if ($k + 1 count($Campaign_array)){ echo $Campaign_array[$k + 1]; }
You can simplify this greatly:
$Campaign_array = Array();
$DB_Change_Campaign_Results = safe_query($query);
while ($row = mysql_fetch_array($DB_Change_Campaign_Results))
{
array_push($Campaign_array, $row['day']);
}
Because you're only selecting day from the table, your row data consists
of only one column. You're repeating stuff unnecessarily by doing this:
$Campaign = $Change_Campaign_row['day'];
$Campaign_array[$Campaign] = $Change_Campaign_row;
IOW, for the row that contains '5', you'd end up with $Campaign_array[5]
== an array that essentially just points to the value '5'.
I also noticed that your dump shows that the days begin with 0. If these
are calendar days you're adding a further layer of complexity.
brian
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Re: [PHP] Finding next recored in a array
brian wrote:
Richard Kurth wrote:
What I am trying to is get all the days from a table where email
campaign = number. and then look at the last day that was sent and
find it in the list and get the next day that follows that day. At
this point the script below is not working. So if you have any Ideas
that would be a better way of doing this please let me know.
$query = SELECT day FROM emailcampaign where campaign_id =
'$emailcampaign' AND member_id = '$members_id';
$DB_Change_Campaign_Results = safe_query($query);
for ($i=0; $i mysql_num_rows($DB_Change_Campaign_Results); $i++) {
$Change_Campaign_row = mysql_fetch_array($DB_Change_Campaign_Results);
$Campaign = $Change_Campaign_row['day'];
$Campaign_array[$Campaign] = $Change_Campaign_row;
}
$k = array_search($val,$Campaign_array);
if ($k + 1 count($Campaign_array)){ echo $Campaign_array[$k + 1]; }
You can simplify this greatly:
$Campaign_array = Array();
$DB_Change_Campaign_Results = safe_query($query);
while ($row = mysql_fetch_array($DB_Change_Campaign_Results))
{
array_push($Campaign_array, $row['day']);
}
Because you're only selecting day from the table, your row data
consists of only one column. You're repeating stuff unnecessarily by
doing this:
$Campaign = $Change_Campaign_row['day'];
$Campaign_array[$Campaign] = $Change_Campaign_row;
IOW, for the row that contains '5', you'd end up with
$Campaign_array[5] == an array that essentially just points to the
value '5'.
I also noticed that your dump shows that the days begin with 0. If
these are calendar days you're adding a further layer of complexity.
0 represents the email campain to send 0 mens send the one that gois out
the day the campain starts. the next number in the list is the next day
a email is sent out would represent 5 day after the starting day
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Re: [PHP] Finding next recored in a array
REVISED
Try this
?php
$query = SELECT day
FROMemailcampaign
WHERE campaign_id = '$emailcampaign'
AND member_id = '$members_id'
;
$DB_Change_Campaign_Results = safe_query($query);
##
## NOTICE: changed from array to assoc
##
$Campaign_array = array();
while ( $row = mysql_fetch_assoc($DB_Change_Campaign_Results) ) {
## Switched to an indexed array with the day as the value instead of the key
$Campaign_array[] = $row['day'];
}
# At this point you have arrays as values for your $Campaign_array
# So, unless $k is an array that matches a sub array of $Campaign_array
# you're never going to get a match
$k = array_search($val,$Campaign_array);
# What is $k at this point? An int (1, 2, 3, etc...) , string (Sunday, Monday,
etc...)
# Before I go any further I will need to know the above information.
if ( isset($Campaign_array[($k + 1)])) {
echo $Campaign_array[($k + 1)];
} else {
echo 'Not found';
}
--
Jim Lucas
Some men are born to greatness, some achieve greatness,
and some have greatness thrust upon them.
Twelfth Night, Act II, Scene V
by William Shakespeare
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Re: [PHP] Finding next recored in a array
Richard Kurth wrote:
include (includes/location.php);
$query = SELECTday
FROMemailcampaign
wherecampaign_id = '1'
ANDmember_id = '8'
;
$DB_Change_Campaign_Results = safe_query($query);
$Campaign_array = array();
while ( $row = mysql_fetch_assoc($DB_Change_Campaign_Results) ) {
$Campaign_array[] = $row['day'];
}
if ( isset($Campaign_array[($k + 1)])) {
echo $Campaign_array[($k + 1)];
} else {
echo 'Not found';
}
var_dump($Campaign_array);
This is what I get now when I run this
1
*array*
0 = string '0' /(length=1)/
1 = string '1' /(length=1)/
2 = string '3' /(length=1)/
3 = string '6' /(length=1)/
4 = string '9' /(length=1)/
5 = string '12' /(length=2)/
6 = string '15' /(length=2)/
7 = string '20' /(length=2)/
8 = string '25' /(length=2)/
9 = string '30' /(length=2)/
Are there going to be wholes in the date range?
if so, you will have to do that last bit like this.
?php
include (includes/location.php);
#
# Setting $k
# Make sure that $k is an integer, not a string.
# hence, no quotes
$k = 5;
$query = SELECT day
FROMemailcampaign
WHERE campaign_id = '1'
AND member_id = '8'
;
$DB_Change_Campaign_Results = safe_query($query);
$Campaign_array = array();
while ( $row = mysql_fetch_assoc($DB_Change_Campaign_Results) ) {
$Campaign_array[] = $row['day'];
}
sort($Campaign_array);
foreach ( $Campaign_array AS $day ) {
if ( $day = $k ) {
$day = next($Campaign_array);
break;
}
}
echo $day;
var_dump($Campaign_array);
?
--
Jim Lucas
Some men are born to greatness, some achieve greatness,
and some have greatness thrust upon them.
Twelfth Night, Act II, Scene V
by William Shakespeare
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Re: [PHP] Finding next recored in a array
Richard Kurth wrote:
What I am trying to is get all the days from a table where email
campaign = number. and then look at the last day that was sent and find
it in the list and get the next day that follows that day. At this point
the script below is not working. So if you have any Ideas that would be
a better way of doing this please let me know.
$query = SELECT day FROM emailcampaign where campaign_id =
'$emailcampaign' AND member_id = '$members_id';
$DB_Change_Campaign_Results = safe_query($query);
for ($i=0; $i mysql_num_rows($DB_Change_Campaign_Results); $i++) {
$Change_Campaign_row = mysql_fetch_array($DB_Change_Campaign_Results);
$Campaign = $Change_Campaign_row['day'];
$Campaign_array[$Campaign] = $Change_Campaign_row;
}
$k = array_search($val,$Campaign_array);
if ($k + 1 count($Campaign_array)){ echo $Campaign_array[$k + 1]; }
Ok, so just to recap (because I'm very confused now, even more after the
many tries on the list since):
You have table in your database called 'emailcampaign'. This table has a
column called 'day'.
Now, your table holds 1 row per campaign and the 'day' column
signifies when the last mail has been sent.
-- am I correct on this so far ? --
Next, you have an array which contains a FIXED collection of days on
which an email should be sent, which looks like so:
$days_to_send_email_on = array('0','1','3','5','8','15','25');
-- am I correct on this so far [2] ? --
You know the 'day' the last mail was sent, and you wish to check your
FIXED array to find out when the next email should be sent.
-- am I correct on this so far [3] ? --
If so, then there are a few things I'm wondering:
1. How did you come up with that fixed array? Is it the result of some
kind of calculation?
1.1 If so, it'd be easier to solve it the mathematical way.
1.2 If #1 is a random result (ie. user-input, or different), then
how about storing it in a table and doing:
SELECT day WHERE day $last_day_it_was_sent_on ORDER BY day ASC LIMIT 1
2. Actually, that's all I'm wondering :)
- Tul
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Re: [PHP] Finding next recored in a array
Jim Lucas wrote:
Richard Kurth wrote:
$query = SELECTday
FROMemailcampaign
wherecampaign_id = '$emailcampaign'
ANDmember_id = '$members_id'
;
$DB_Change_Campaign_Results = safe_query($query);
while ( $row = mysql_fetch_array($DB_Change_Campaign_Results) ) {
$Campaign_array[$row['day']] = $row;
}
# At this point you have arrays as values for your $Campaign_array
# So, unless $k is an array that matches a sub array of $Campaign_array
# you're never going to get a match
$k = array_search($val,$Campaign_array);
# What is $k at this point? An int (1, 2, 3, etc...) , string
(Sunday, Monday, etc...)
# Before I go any further I will need to know the above information.
if ( ($k + 1) count($Campaign_array) ) {
echo $Campaign_array[$k + 1];
}
This is what I get if I run the above script.
From a var_dump($Campaign_array); I get
*array*
0 =
*array*
0 = string '0' /(length=1)/
'day' = string '0' /(length=1)/
1 =
*array*
0 = string '1' /(length=1)/
'day' = string '1' /(length=1)/
3 =
*array*
0 = string '3' /(length=1)/
'day' = string '3' /(length=1)/
6 =
*array*
0 = string '6' /(length=1)/
'day' = string '6' /(length=1)/
9 =
*array*
0 = string '9' /(length=1)/
'day' = string '9' /(length=1)/
12 =
*array*
0 = string '12' /(length=2)/
'day' = string '12' /(length=2)/
15 =
*array*
0 = string '15' /(length=2)/
'day' = string '15' /(length=2)/
20 =
*array*
0 = string '20' /(length=2)/
'day' = string '20' /(length=2)/
25 =
*array*
0 = string '25' /(length=2)/
'day' = string '25' /(length=2)/
30 =
*array*
0 = string '30' /(length=2)/
'day' = string '30' /(length=2)/
From a $val=5;
$k = array_search($val,$Campaign_array);
var_dump($k); I get
boolean false
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Re: [PHP] Finding next recored in a array
Jim Lucas wrote:
REVISED
Try this
?php
$query = SELECTday
FROMemailcampaign
WHEREcampaign_id = '$emailcampaign'
ANDmember_id = '$members_id'
;
$DB_Change_Campaign_Results = safe_query($query);
##
## NOTICE: changed from array to assoc
##
$Campaign_array = array();
while ( $row = mysql_fetch_assoc($DB_Change_Campaign_Results) ) {
## Switched to an indexed array with the day as the value instead of
the key
$Campaign_array[] = $row['day'];
}
# At this point you have arrays as values for your $Campaign_array
# So, unless $k is an array that matches a sub array of $Campaign_array
# you're never going to get a match
$k = array_search($val,$Campaign_array);
# What is $k at this point? An int (1, 2, 3, etc...) , string
(Sunday, Monday, etc...)
# Before I go any further I will need to know the above information.
if ( isset($Campaign_array[($k + 1)])) {
echo $Campaign_array[($k + 1)];
} else {
echo 'Not found';
}
include (includes/location.php);
$query = SELECTday
FROMemailcampaign
wherecampaign_id = '1'
ANDmember_id = '8'
;
$DB_Change_Campaign_Results = safe_query($query);
$Campaign_array = array();
while ( $row = mysql_fetch_assoc($DB_Change_Campaign_Results) ) {
$Campaign_array[] = $row['day'];
}
if ( isset($Campaign_array[($k + 1)])) {
echo $Campaign_array[($k + 1)];
} else {
echo 'Not found';
}
var_dump($Campaign_array);
This is what I get now when I run this
1
*array*
0 = string '0' /(length=1)/
1 = string '1' /(length=1)/
2 = string '3' /(length=1)/
3 = string '6' /(length=1)/
4 = string '9' /(length=1)/
5 = string '12' /(length=2)/
6 = string '15' /(length=2)/
7 = string '20' /(length=2)/
8 = string '25' /(length=2)/
9 = string '30' /(length=2)/
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Re: [PHP] Finding next recored in a array
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next. What I do
not understand is if I know the last number was say 5 how do I tell the
script that that is the current number so I can select the next record
||
I think you'll need your own function for this. Pass in the array and
loop through it until you find the key, increment that, ensure that
there is another value with that key, and return the key (or the value).
(untested)
function nextInArray($arr, $val)
{
$next_key = NULL;
for ($i = 0; $i sizeof($arr);$i++)
{
if ($arr[$i] == $val)
{
$next_key = ++$i;
break;
}
}
// return the key:
return (array_key_exists($next_key) ? $next_key : NULL);
// or the value:
return (array_key_exists($next_key) ? $arr[$next_key] : NULL);
}
However, in your example, you're searching for the key that points to
the value '5'. What if the value '5' occurs more than once?
brian
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Re: [PHP] Finding next recored in a array
brian wrote:
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next. What I
do not understand is if I know the last number was say 5 how do I
tell the script that that is the current number so I can select the
next record
||
I think you'll need your own function for this. Pass in the array and
loop through it until you find the key, increment that, ensure that
there is another value with that key, and return the key (or the value).
(untested)
function nextInArray($arr, $val)
{
$next_key = NULL;
for ($i = 0; $i sizeof($arr);$i++)
{
if ($arr[$i] == $val)
{
$next_key = ++$i;
break;
}
}
// return the key:
return (array_key_exists($next_key) ? $next_key : NULL);
// or the value:
return (array_key_exists($next_key) ? $arr[$next_key] : NULL);
}
However, in your example, you're searching for the key that points to
the value '5'. What if the value '5' occurs more than once?
brian
In my script the value of 5 will not reoccur the numbers are number of
days from 0 up to 30 days.
Thanks for the function
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Re: [PHP] Finding next recored in a array
On Sun, Sep 16, 2007 at 07:09:02PM -0400, brian wrote:
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next. What I do
not understand is if I know the last number was say 5 how do I tell the
script that that is the current number so I can select the next record
||
I think you'll need your own function for this.
Nope. Just use array_search().
$k = array_search('5',$Campaign_array);
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
Pass in the array and loop
through it until you find the key, increment that, ensure that there is
another value with that key, and return the key (or the value).
(untested)
function nextInArray($arr, $val)
{
$next_key = NULL;
for ($i = 0; $i sizeof($arr);$i++)
{
if ($arr[$i] == $val)
{
$next_key = ++$i;
break;
}
}
// return the key:
return (array_key_exists($next_key) ? $next_key : NULL);
// or the value:
return (array_key_exists($next_key) ? $arr[$next_key] : NULL);
}
However, in your example, you're searching for the key that points to the
value '5'. What if the value '5' occurs more than once?
From the docs:
If needle is found in haystack more than once, the first matching key
is returned. To return the keys for all matching values, use
array_keys() with the optional search_value parameter instead.
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Rick Pasotto[EMAIL PROTECTED]http://www.niof.net
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Re: [PHP] Finding next recored in a array
Rick Pasotto wrote:
On Sun, Sep 16, 2007 at 07:09:02PM -0400, brian wrote:
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next. What I do
not understand is if I know the last number was say 5 how do I tell the
script that that is the current number so I can select the next record
||
I think you'll need your own function for this.
Nope. Just use array_search().
$k = array_search('5',$Campaign_array);
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
I tried this and it gives me nothing back. It should give me a 8
$Campaign_array= array('0','1','3','5','8','15','25');
$val=5;
$k = array_search($val,$Campaign_array);
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
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Re: [PHP] Finding next recored in a array
Richard Kurth wrote:
Rick Pasotto wrote:
On Sun, Sep 16, 2007 at 07:09:02PM -0400, brian wrote:
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next. What
I do not understand is if I know the last number was say 5 how do I
tell the script that that is the current number so I can select the
next record
||
I think you'll need your own function for this.
Nope. Just use array_search().
$k = array_search('5',$Campaign_array);
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
I tried this and it gives me nothing back. It should give me a 8
$Campaign_array= array('0','1','3','5','8','15','25');
$val=5;
$k = array_search($val,$Campaign_array);
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
I figured out way it was not working $k + 1 count needed to be $k +
1 count
But now it works perfect
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Re: [PHP] Finding next recored in a array
On Sun, Sep 16, 2007 at 06:04:45PM -0700, Richard Kurth wrote:
Richard Kurth wrote:
Rick Pasotto wrote:
On Sun, Sep 16, 2007 at 07:09:02PM -0400, brian wrote:
Richard Kurth wrote:
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next. What I
do not understand is if I know the last number was say 5 how do I tell
the script that that is the current number so I can select the next
record
||
I think you'll need your own function for this.
Nope. Just use array_search().
$k = array_search('5',$Campaign_array);
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
I tried this and it gives me nothing back. It should give me a 8
$Campaign_array= array('0','1','3','5','8','15','25');
$val=5;
$k = array_search($val,$Campaign_array);
if ($k + 1 count($Campaign_array)) { echo $Campaign_array[$k + 1]; }
I figured out way it was not working $k + 1 count needed to be $k + 1
count
Yup. Sorry 'bout that.
But now it works perfect
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resentment. -- Dale Carnegie
Rick Pasotto[EMAIL PROTECTED]http://www.niof.net
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