RE: [PHP] in_array - what the...
-Original Message-
From: Gary . [mailto:php-gene...@garydjones.name]
Sent: 25 June 2010 08:18
To: PHP
Subject: [PHP] in_array - what the...
If I have an array that looks like
array(1) {
[mac_address]=
string(2) td
}
and I call
if (in_array($name, self::$aboveArray))
with $name as
string(11) mac_address
what should be the result?
FALSE -- in_array checks the *values*, not the keys, so would be looking at the
td for this element.
To do what you want to do, simply do an isset():
if (isset($array['mac_address'])):
// do stuff with $array['mac_address']
else:
// it doesn't exist
endif;
Cheers!
Mike
--
Mike Ford,
Electronic Information Developer, Libraries and Learning Innovation,
Leeds Metropolitan University, C507, Civic Quarter Campus,
Woodhouse Lane, LEEDS, LS1 3HE, United Kingdom
Email: m.f...@leedsmet.ac.uk
Tel: +44 113 812 4730
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Re: [PHP] in_array - what the...
Ford, Mike writes:
-Original Message-
If I have an array that looks like
array(1) {
[mac_address]=
string(2) td
}
and I call
if (in_array($name, self::$aboveArray))
with $name as
string(11) mac_address
what should be the result?
FALSE -- in_array checks the *values*, not the keys, so would be
looking at the td for this element.
Agh! So it does.
You know what's worse? I even looked at the documentation of the
function this morning wondering if that's what the problem was and
*still* didn't see it!
*slinks away in embarrassment*
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RE: [PHP] in_array - what the...
-Original Message-
From: Gary . [mailto:php-gene...@garydjones.name]
Sent: 25 June 2010 09:14
To: PHP
Subject: Re: [PHP] in_array - what the...
Ford, Mike writes:
-Original Message-
If I have an array that looks like
array(1) {
[mac_address]=
string(2) td
}
and I call
if (in_array($name, self::$aboveArray))
with $name as
string(11) mac_address
what should be the result?
FALSE -- in_array checks the *values*, not the keys, so would be
looking at the td for this element.
Agh! So it does.
You know what's worse? I even looked at the documentation of the
function this morning wondering if that's what the problem was and
*still* didn't see it!
*slinks away in embarrassment*
Not to worry -- happens to the best of us. (Been there, done that, got a
wardrobe full of T-shirts!)
Cheers!
Mike
--
Mike Ford,
Electronic Information Developer, Libraries and Learning Innovation,
Leeds Metropolitan University, C507, Civic Quarter Campus,
Woodhouse Lane, LEEDS, LS1 3HE, United Kingdom
Email: m.f...@leedsmet.ac.uk
Tel: +44 113 812 4730
To view the terms under which this email is distributed, please go to
http://disclaimer.leedsmet.ac.uk/email.htm
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RE: [PHP] in_array - what the...
-Original Message-
From: Gary . [mailto:php-gene...@garydjones.name]
Sent: Friday, June 25, 2010 1:14 AM
To: PHP
Subject: Re: [PHP] in_array - what the...
Ford, Mike writes:
-Original Message-
If I have an array that looks like
array(1) {
[mac_address]=
string(2) td
}
and I call
if (in_array($name, self::$aboveArray))
with $name as
string(11) mac_address
what should be the result?
FALSE -- in_array checks the *values*, not the keys, so would be
looking at the td for this element.
Agh! So it does.
You know what's worse? I even looked at the documentation of the
function this morning wondering if that's what the problem was and
*still* didn't see it!
*slinks away in embarrassment*
Why do this in_array() business??
Just do this...
if (self::$aboveArray[$name])
{
//something interesting here
}
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RE: [PHP] in_array - what the...
From: Daevid Vincent
Why do this in_array() business??
Just do this...
if (self::$aboveArray[$name])
{
//something interesting here
}
Does that gibberish actually do something? It doesn't make any sense to
me, while in_array() actually looks like what it does.
Bob McConnell
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Re: [PHP] in_array - what the...
On 25 June 2010 19:58, Bob McConnell r...@cbord.com wrote:
From: Daevid Vincent
Why do this in_array() business??
Just do this...
if (self::$aboveArray[$name])
{
//something interesting here
}
Does that gibberish actually do something? It doesn't make any sense to
me, while in_array() actually looks like what it does.
Gibberish?? Probably a good time to go look up some php tutorials.
Apart from that, it's rather bad form as a missing index will create a
notice at the least. The isset snippet is better, though if you care
about finding null values you need to go the route of array_keys().
Regards
Peter
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RE: [PHP] in_array - what the...
From: Peter Lind
On 25 June 2010 19:58, Bob McConnell r...@cbord.com wrote:
From: Daevid Vincent
Why do this in_array() business??
Just do this...
if (self::$aboveArray[$name])
{
//something interesting here
}
Does that gibberish actually do something? It doesn't make any sense to
me, while in_array() actually looks like what it does.
Gibberish?? Probably a good time to go look up some php tutorials.
No thanks. I tried to figure out that double colon nonsense over a decade ago
as part of an OOP development team. I still don't understand most of the code
written during those two years, even though I still maintain parts of it. All I
see is a lot of unnecessary overhead with no significant return on the
investment. I'll stick with the tried and true procedural notation, at least
until I retire next year.
Bob McConnell
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Re: [PHP] in_array breaks down for 0 as value
On 20 Nov 2008, at 23:09, Ashley Sheridan wrote:
On Thu, 2008-11-20 at 09:25 +, Stut wrote:
On 20 Nov 2008, at 06:55, Yashesh Bhatia wrote:
I wanted to use in_array to verify the results of a form submission
for a checkbox and found an interesting
behaviour.
$ php -v
PHP 5.2.5 (cli) (built: Jan 12 2008 14:54:37)
$
$ cat in_array2.php
?php
$node_review_types = array(
'page' = 'page',
'story' = 'story',
'nodereview' = 'abc',
);
if (in_array('page', $node_review_types)) {
print page found in node_review_types\n;
}
if (in_array('nodereview', $node_review_types)) {
print nodereview found in node_review_types\n;
}
?
$ php in_array2.php
page found in node_review_types
$
This works fine. but if i change the value of the key
'nodereview' to
0 it breaks down.
$ diff in_array2.php in_array3.php
6c6
'nodereview' = 'abc',
---
'nodereview' = 0,
$
$ php in_array3.php
page found in node_review_types
nodereview found in node_review_types
$
Any reason why in_array is returning TRUE when one has a 0 value on
the array ?
That's weird, 5.2.6 does the same thing. There's actually a comment
about this on the in_array manual page from james dot ellis at gmail
dot com...
quote
Be aware of oddities when dealing with 0 (zero) values in an array...
This script:
?php
$array = array('testing',0,'name');
var_dump($array);
//this will return true
var_dump(in_array('foo', $array));
//this will return false
var_dump(in_array('foo', $array, TRUE));
?
It seems in non strict mode, the 0 value in the array is evaluating
to
boolean FALSE and in_array returns TRUE. Use strict mode to work
around this peculiarity.
This only seems to occur when there is an integer 0 in the array. A
string '0' will return FALSE for the first test above (at least in
5.2.6).
/quote
So use strict mode and this problem will go away. Oh, and please read
the manual before asking a question in future.
-Stut
--
http://stut.net/
What about using the === and !== comparisons to compare and make sure
that 0 is not giving a false false.
That's effectively what using strict mode does. RTFM please.
-Stut
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Re: [PHP] in_array breaks down for 0 as value
On Fri, 2008-11-21 at 09:11 +, Stut wrote:
On 20 Nov 2008, at 23:09, Ashley Sheridan wrote:
On Thu, 2008-11-20 at 09:25 +, Stut wrote:
On 20 Nov 2008, at 06:55, Yashesh Bhatia wrote:
I wanted to use in_array to verify the results of a form submission
for a checkbox and found an interesting
behaviour.
$ php -v
PHP 5.2.5 (cli) (built: Jan 12 2008 14:54:37)
$
$ cat in_array2.php
?php
$node_review_types = array(
'page' = 'page',
'story' = 'story',
'nodereview' = 'abc',
);
if (in_array('page', $node_review_types)) {
print page found in node_review_types\n;
}
if (in_array('nodereview', $node_review_types)) {
print nodereview found in node_review_types\n;
}
?
$ php in_array2.php
page found in node_review_types
$
This works fine. but if i change the value of the key
'nodereview' to
0 it breaks down.
$ diff in_array2.php in_array3.php
6c6
'nodereview' = 'abc',
---
'nodereview' = 0,
$
$ php in_array3.php
page found in node_review_types
nodereview found in node_review_types
$
Any reason why in_array is returning TRUE when one has a 0 value on
the array ?
That's weird, 5.2.6 does the same thing. There's actually a comment
about this on the in_array manual page from james dot ellis at gmail
dot com...
quote
Be aware of oddities when dealing with 0 (zero) values in an array...
This script:
?php
$array = array('testing',0,'name');
var_dump($array);
//this will return true
var_dump(in_array('foo', $array));
//this will return false
var_dump(in_array('foo', $array, TRUE));
?
It seems in non strict mode, the 0 value in the array is evaluating
to
boolean FALSE and in_array returns TRUE. Use strict mode to work
around this peculiarity.
This only seems to occur when there is an integer 0 in the array. A
string '0' will return FALSE for the first test above (at least in
5.2.6).
/quote
So use strict mode and this problem will go away. Oh, and please read
the manual before asking a question in future.
-Stut
--
http://stut.net/
What about using the === and !== comparisons to compare and make sure
that 0 is not giving a false false.
That's effectively what using strict mode does. RTFM please.
-Stut
Hey, chill. If you offer advice, don't be so offensive to everyone.
Ash
www.ashleysheridan.co.uk
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Re: [PHP] in_array breaks down for 0 as value
On 22 Nov 2008, at 00:06, Ashley Sheridan wrote:
On Fri, 2008-11-21 at 09:11 +, Stut wrote:
On 20 Nov 2008, at 23:09, Ashley Sheridan wrote:
On Thu, 2008-11-20 at 09:25 +, Stut wrote:
On 20 Nov 2008, at 06:55, Yashesh Bhatia wrote:
I wanted to use in_array to verify the results of a form
submission
for a checkbox and found an interesting
behaviour.
$ php -v
PHP 5.2.5 (cli) (built: Jan 12 2008 14:54:37)
$
$ cat in_array2.php
?php
$node_review_types = array(
'page' = 'page',
'story' = 'story',
'nodereview' = 'abc',
);
if (in_array('page', $node_review_types)) {
print page found in node_review_types\n;
}
if (in_array('nodereview', $node_review_types)) {
print nodereview found in node_review_types\n;
}
?
$ php in_array2.php
page found in node_review_types
$
This works fine. but if i change the value of the key
'nodereview' to
0 it breaks down.
$ diff in_array2.php in_array3.php
6c6
'nodereview' = 'abc',
---
'nodereview' = 0,
$
$ php in_array3.php
page found in node_review_types
nodereview found in node_review_types
$
Any reason why in_array is returning TRUE when one has a 0 value
on
the array ?
That's weird, 5.2.6 does the same thing. There's actually a comment
about this on the in_array manual page from james dot ellis at
gmail
dot com...
quote
Be aware of oddities when dealing with 0 (zero) values in an
array...
This script:
?php
$array = array('testing',0,'name');
var_dump($array);
//this will return true
var_dump(in_array('foo', $array));
//this will return false
var_dump(in_array('foo', $array, TRUE));
?
It seems in non strict mode, the 0 value in the array is evaluating
to
boolean FALSE and in_array returns TRUE. Use strict mode to work
around this peculiarity.
This only seems to occur when there is an integer 0 in the array. A
string '0' will return FALSE for the first test above (at least in
5.2.6).
/quote
So use strict mode and this problem will go away. Oh, and please
read
the manual before asking a question in future.
-Stut
--
http://stut.net/
What about using the === and !== comparisons to compare and make
sure
that 0 is not giving a false false.
That's effectively what using strict mode does. RTFM please.
-Stut
Hey, chill. If you offer advice, don't be so offensive to everyone.
I don't believe I was being offensive, you're clearly a very delicate
little flower. The way I saw it you made a suggestion without
understanding how to use the function in question. In my opinion RTFM
is a perfectly reasonable response to that.
-Stut
--
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Re: [PHP] in_array breaks down for 0 as value
On 20 Nov 2008, at 06:55, Yashesh Bhatia wrote:
I wanted to use in_array to verify the results of a form submission
for a checkbox and found an interesting
behaviour.
$ php -v
PHP 5.2.5 (cli) (built: Jan 12 2008 14:54:37)
$
$ cat in_array2.php
?php
$node_review_types = array(
'page' = 'page',
'story' = 'story',
'nodereview' = 'abc',
);
if (in_array('page', $node_review_types)) {
print page found in node_review_types\n;
}
if (in_array('nodereview', $node_review_types)) {
print nodereview found in node_review_types\n;
}
?
$ php in_array2.php
page found in node_review_types
$
This works fine. but if i change the value of the key 'nodereview' to
0 it breaks down.
$ diff in_array2.php in_array3.php
6c6
'nodereview' = 'abc',
---
'nodereview' = 0,
$
$ php in_array3.php
page found in node_review_types
nodereview found in node_review_types
$
Any reason why in_array is returning TRUE when one has a 0 value on
the array ?
That's weird, 5.2.6 does the same thing. There's actually a comment
about this on the in_array manual page from james dot ellis at gmail
dot com...
quote
Be aware of oddities when dealing with 0 (zero) values in an array...
This script:
?php
$array = array('testing',0,'name');
var_dump($array);
//this will return true
var_dump(in_array('foo', $array));
//this will return false
var_dump(in_array('foo', $array, TRUE));
?
It seems in non strict mode, the 0 value in the array is evaluating to
boolean FALSE and in_array returns TRUE. Use strict mode to work
around this peculiarity.
This only seems to occur when there is an integer 0 in the array. A
string '0' will return FALSE for the first test above (at least in
5.2.6).
/quote
So use strict mode and this problem will go away. Oh, and please read
the manual before asking a question in future.
-Stut
--
http://stut.net/
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Re: [PHP] in_array breaks down for 0 as value
On Thu, 2008-11-20 at 09:25 +, Stut wrote:
On 20 Nov 2008, at 06:55, Yashesh Bhatia wrote:
I wanted to use in_array to verify the results of a form submission
for a checkbox and found an interesting
behaviour.
$ php -v
PHP 5.2.5 (cli) (built: Jan 12 2008 14:54:37)
$
$ cat in_array2.php
?php
$node_review_types = array(
'page' = 'page',
'story' = 'story',
'nodereview' = 'abc',
);
if (in_array('page', $node_review_types)) {
print page found in node_review_types\n;
}
if (in_array('nodereview', $node_review_types)) {
print nodereview found in node_review_types\n;
}
?
$ php in_array2.php
page found in node_review_types
$
This works fine. but if i change the value of the key 'nodereview' to
0 it breaks down.
$ diff in_array2.php in_array3.php
6c6
'nodereview' = 'abc',
---
'nodereview' = 0,
$
$ php in_array3.php
page found in node_review_types
nodereview found in node_review_types
$
Any reason why in_array is returning TRUE when one has a 0 value on
the array ?
That's weird, 5.2.6 does the same thing. There's actually a comment
about this on the in_array manual page from james dot ellis at gmail
dot com...
quote
Be aware of oddities when dealing with 0 (zero) values in an array...
This script:
?php
$array = array('testing',0,'name');
var_dump($array);
//this will return true
var_dump(in_array('foo', $array));
//this will return false
var_dump(in_array('foo', $array, TRUE));
?
It seems in non strict mode, the 0 value in the array is evaluating to
boolean FALSE and in_array returns TRUE. Use strict mode to work
around this peculiarity.
This only seems to occur when there is an integer 0 in the array. A
string '0' will return FALSE for the first test above (at least in
5.2.6).
/quote
So use strict mode and this problem will go away. Oh, and please read
the manual before asking a question in future.
-Stut
--
http://stut.net/
What about using the === and !== comparisons to compare and make sure
that 0 is not giving a false false.
Ash
www.ashleysheridan.co.uk
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Re: [PHP] in_array breaks down for 0 as value
2008/11/20 Stut [EMAIL PROTECTED]:
On 20 Nov 2008, at 06:55, Yashesh Bhatia wrote:
I wanted to use in_array to verify the results of a form submission
for a checkbox and found an interesting
behaviour.
$ php -v
PHP 5.2.5 (cli) (built: Jan 12 2008 14:54:37)
$
$ cat in_array2.php
?php
$node_review_types = array(
'page' = 'page',
'story' = 'story',
'nodereview' = 'abc',
);
if (in_array('page', $node_review_types)) {
print page found in node_review_types\n;
}
if (in_array('nodereview', $node_review_types)) {
print nodereview found in node_review_types\n;
}
?
$ php in_array2.php
page found in node_review_types
$
This works fine. but if i change the value of the key 'nodereview' to
0 it breaks down.
$ diff in_array2.php in_array3.php
6c6
'nodereview' = 'abc',
---
'nodereview' = 0,
$
$ php in_array3.php
page found in node_review_types
nodereview found in node_review_types
$
Any reason why in_array is returning TRUE when one has a 0 value on the
array ?
That's weird, 5.2.6 does the same thing. There's actually a comment about
this on the in_array manual page from james dot ellis at gmail dot com...
quote
Be aware of oddities when dealing with 0 (zero) values in an array...
This script:
?php
$array = array('testing',0,'name');
var_dump($array);
//this will return true
var_dump(in_array('foo', $array));
//this will return false
var_dump(in_array('foo', $array, TRUE));
?
It seems in non strict mode, the 0 value in the array is evaluating to
boolean FALSE and in_array returns TRUE. Use strict mode to work around this
peculiarity.
This only seems to occur when there is an integer 0 in the array. A string
'0' will return FALSE for the first test above (at least in 5.2.6).
/quote
So use strict mode and this problem will go away. Oh, and please read the
manual before asking a question in future.
-Stut
I wouldn't consider it weird; it's just how PHP handles loose type
comparisons. I would certainly agree that it's not terribly obvious
why it happens, though. :) That said, it's consistent with PHP
behaviour.
James Ellis almost got it right in his note. As I already noted, it's
not because of a conversion to boolean FALSE, but a conversion to
integer 0. You can test this by substituting FALSE for the 0 in the
array in the example and trying it.
Torben
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Re: [PHP] in_array breaks down for 0 as value
2008/11/19 Yashesh Bhatia [EMAIL PROTECTED]:
Hi.
I wanted to use in_array to verify the results of a form submission
for a checkbox and found an interesting
behaviour.
$ php -v
PHP 5.2.5 (cli) (built: Jan 12 2008 14:54:37)
$
$ cat in_array2.php
?php
$node_review_types = array(
'page' = 'page',
'story' = 'story',
'nodereview' = 'abc',
);
if (in_array('page', $node_review_types)) {
print page found in node_review_types\n;
}
if (in_array('nodereview', $node_review_types)) {
print nodereview found in node_review_types\n;
}
?
$ php in_array2.php
page found in node_review_types
$
This works fine. but if i change the value of the key 'nodereview' to
0 it breaks down.
$ diff in_array2.php in_array3.php
6c6
'nodereview' = 'abc',
---
'nodereview' = 0,
$
$ php in_array3.php
page found in node_review_types
nodereview found in node_review_types
$
Any reason why in_array is returning TRUE when one has a 0 value on the array
?
Thanks.
Hi Yasheed,
It looks like you've found the reason for the existence of the
optional third argument to in_array(): 'strict'. In your second
example (in_array3.php), what happens is that the value of
$node_review_types['nodereview'] is 0 (an integer), so it is compared
against the integer value of the first argument to in_array(), which
is also 0 (in PHP, the integer value of a string with no leading
numerals is 0). In other words, in_array() first looks at the first
element of $node_review_types and finds that it is a string, so it
compares that value as a string against the string value of its first
argument ('nodereview'). Same goes for the second element of
$node_review_types. However, when it comes time to check the third
element, in_array() sees that it is an integer (0) and thus compares
it against the integer value of 'nodereview' (also 0), and returns
true.
Make any sense? The problem goes away if you give a true value as the
third argument to in_array(): this tells it to check the elements of
the given array for type as well as value--i.e., it tells in_array()
to not automatically cast the value being searched for to the type of
the array element being checked.
Torben
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Re: [PHP] in_array() related problem
Hi,
Thanks for your reply. After a sleep overnight I found I said
something really stupid. Arrays are compared in deep, and also for
objects. I really forgot the old PHP4 way and thought PHP5 compares
object simply by address when using ==, which is not the real case. I
need to use === for comparing objects of the same instance. And thanks
Tom for pointing out to use the strict parameter.
On 11/4/06, Richard Lynch [EMAIL PROTECTED] wrote:
Try providing a custom comparison function.
Almost for sure, PHP is attempting to test the == by a deeper scan
than you think.
On Fri, November 3, 2006 10:56 am, tamcy wrote:
Hello all,
I'm new to this list. To not flooding the bug tracking system I hope
to clarify some of my understanding here.
I am referring to the (now bogus) bug report
http://bugs.php.net/bug.php?id=39356edit=2. This happens after my
upgrade to PHP 5.2, where the code shown produces a Fatal error:
Nesting level too deep - recursive dependency?. Same testing code
reproduced below:
?php
class A
{
public $b;
}
class B
{
public $a;
}
$a = new A;
$b = new B;
$b-a = $a;
$a-b = $b;
$test = array($a, $b);
var_dump(in_array($a, $test));
I think this is not rare for a child item to have knowledge about its
parent, forming a cross-reference.
This code runs with no problem in PHP5.1.6, but not in 5.2. Ilia
kindly points out that In php 5 objects are passed by reference, so
your code does in
fact create a circular dependency.. I know the passed by reference
rule. What I'm now puzzled is, why this should lead to an error.
To my knowledge, despite the type-casting issue and actual algorithm,
in_array() should actually do nothing more than:
function mimic_in_array($search, $list)
{
foreach ($list as $item)
if ($search == $item)
return true;
return false;
}
Which means:
1. in_array() isn't multi-dimensional.
2. in_array() doesn't care about the properties of any object.
That is, I don't expect in_array() to nest through all available inner
arrays for a match, not to mention those are object properties, not
arrays.
So here is the question: Why should in_array() throws such a Fatal
error: Nesting level too deep error? Why should it care? Is there any
behaviour I don't know?
Thanks all in advance.
Tamcy
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Re: [PHP] in_array() related problem
Try like this:
var_dump(in_array($a, $test, true));
Richard Lynch wrote:
Try providing a custom comparison function.
Almost for sure, PHP is attempting to test the == by a deeper scan
than you think.
On Fri, November 3, 2006 10:56 am, tamcy wrote:
Hello all,
I'm new to this list. To not flooding the bug tracking system I hope
to clarify some of my understanding here.
I am referring to the (now bogus) bug report
http://bugs.php.net/bug.php?id=39356edit=2. This happens after my
upgrade to PHP 5.2, where the code shown produces a Fatal error:
Nesting level too deep - recursive dependency?. Same testing code
reproduced below:
?php
class A
{
public $b;
}
class B
{
public $a;
}
$a = new A;
$b = new B;
$b-a = $a;
$a-b = $b;
$test = array($a, $b);
var_dump(in_array($a, $test));
I think this is not rare for a child item to have knowledge about its
parent, forming a cross-reference.
This code runs with no problem in PHP5.1.6, but not in 5.2. Ilia
kindly points out that In php 5 objects are passed by reference, so
your code does in
fact create a circular dependency.. I know the passed by reference
rule. What I'm now puzzled is, why this should lead to an error.
To my knowledge, despite the type-casting issue and actual algorithm,
in_array() should actually do nothing more than:
function mimic_in_array($search, $list)
{
foreach ($list as $item)
if ($search == $item)
return true;
return false;
}
Which means:
1. in_array() isn't multi-dimensional.
2. in_array() doesn't care about the properties of any object.
That is, I don't expect in_array() to nest through all available inner
arrays for a match, not to mention those are object properties, not
arrays.
So here is the question: Why should in_array() throws such a Fatal
error: Nesting level too deep error? Why should it care? Is there any
behaviour I don't know?
Thanks all in advance.
Tamcy
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Re: [PHP] in_array() related problem
Try providing a custom comparison function.
Almost for sure, PHP is attempting to test the == by a deeper scan
than you think.
On Fri, November 3, 2006 10:56 am, tamcy wrote:
Hello all,
I'm new to this list. To not flooding the bug tracking system I hope
to clarify some of my understanding here.
I am referring to the (now bogus) bug report
http://bugs.php.net/bug.php?id=39356edit=2. This happens after my
upgrade to PHP 5.2, where the code shown produces a Fatal error:
Nesting level too deep - recursive dependency?. Same testing code
reproduced below:
?php
class A
{
public $b;
}
class B
{
public $a;
}
$a = new A;
$b = new B;
$b-a = $a;
$a-b = $b;
$test = array($a, $b);
var_dump(in_array($a, $test));
I think this is not rare for a child item to have knowledge about its
parent, forming a cross-reference.
This code runs with no problem in PHP5.1.6, but not in 5.2. Ilia
kindly points out that In php 5 objects are passed by reference, so
your code does in
fact create a circular dependency.. I know the passed by reference
rule. What I'm now puzzled is, why this should lead to an error.
To my knowledge, despite the type-casting issue and actual algorithm,
in_array() should actually do nothing more than:
function mimic_in_array($search, $list)
{
foreach ($list as $item)
if ($search == $item)
return true;
return false;
}
Which means:
1. in_array() isn't multi-dimensional.
2. in_array() doesn't care about the properties of any object.
That is, I don't expect in_array() to nest through all available inner
arrays for a match, not to mention those are object properties, not
arrays.
So here is the question: Why should in_array() throws such a Fatal
error: Nesting level too deep error? Why should it care? Is there any
behaviour I don't know?
Thanks all in advance.
Tamcy
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Re: [PHP] in_array w/statement
reason it seems to always be true, ... something i'm doing wrong? btw, i
cannot add the in_array to the statement because if the $buddylist is empty
it will generate errors because of the empty implode.
you could add the is_array() check.
$buddylist = preg_split('/( )+/', trim($userinfo['buddylist']), -1,
PREG_SPLIT_NO_EMPTY);
if($buddylist)
{
$buddy = in_array($uname['uid'], array(implode(',', $buddylist)));
}
not sure I understand this, implode returns a string, then you are
putting that string into the array() function. I dont know what kind
of data you would get back from that. $buddylist should already be an
array.
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Re: [PHP] in_array w/statement
yeah, you're right though.. i had to use explode not implode
eg, if(in_array($uname['uid'], explode(' ', trim($userinfo['buddylist']
so i ditched the preg_split()
cheers.
- Original Message -
From: Jason Wong [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, December 16, 2004 3:39 PM
Subject: Re: [PHP] in_array w/statement
On Friday 17 December 2004 02:33, Sebastian wrote:
I cannot solve this problem,. sorry if this looks confusing,.
It is ...
i have a form and don't want to set the variable if the in_array is
true..
the code works, up until i add the last !$buddy in the statement, for
some
reason it seems to always be true, ... something i'm doing wrong? btw, i
cannot add the in_array to the statement because if the $buddylist is
empty
it will generate errors because of the empty implode.
$buddylist = preg_split('/( )+/', trim($userinfo['buddylist']), -1,
PREG_SPLIT_NO_EMPTY);
OK, it looks like $userinfo['buddylist'] is a string containing buddies
separated by some whitespace:
'buddy1 buddy2'
After the above statement $buddylist becomes an array.
if($buddylist)
{
$buddy = in_array($uname['uid'], array(implode(',', $buddylist)));
}
Here implode() returns a string containing 'buddy1,buddy2', then you stick
that into an array() which is effectively:
array('buddy1,buddy2'); // note that there is only *1* element
Now unless your $uname['uid'] really is literally 'buddy1,buddy2' then
$buddy
will be false.
Hope that's enough to get you going.
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Re: [PHP] in_array w/statement
On Friday 17 December 2004 02:33, Sebastian wrote:
I cannot solve this problem,. sorry if this looks confusing,.
It is ...
i have a form and don't want to set the variable if the in_array is true..
the code works, up until i add the last !$buddy in the statement, for some
reason it seems to always be true, ... something i'm doing wrong? btw, i
cannot add the in_array to the statement because if the $buddylist is empty
it will generate errors because of the empty implode.
$buddylist = preg_split('/( )+/', trim($userinfo['buddylist']), -1,
PREG_SPLIT_NO_EMPTY);
OK, it looks like $userinfo['buddylist'] is a string containing buddies
separated by some whitespace:
'buddy1 buddy2'
After the above statement $buddylist becomes an array.
if($buddylist)
{
$buddy = in_array($uname['uid'], array(implode(',', $buddylist)));
}
Here implode() returns a string containing 'buddy1,buddy2', then you stick
that into an array() which is effectively:
array('buddy1,buddy2'); // note that there is only *1* element
Now unless your $uname['uid'] really is literally 'buddy1,buddy2' then $buddy
will be false.
Hope that's enough to get you going.
--
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Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
--
Search the list archives before you post
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Re: [PHP] in_array not operating as 'expected'
Ing. Ivo F.A.C. Fokkema wrote:
Hi guys and gals,
I'm not screaming Bug! Bug! but this _does_ look 'illogical' to me. I've
searched the archives, but found no earlier conversation. Sorry if
I missed it. Consider the following code:
var_dump(in_array('test', array(0)));
What does this return? I expect bool(false), but it returns bool(true).
After some searching the web, I bumped into this:
http://www.phpdiscuss.com/article.php?id=67763group=php.bugs
Basically, it is said by derick [at] php.net that this behavior is
expected. The following code :
var_dump('test' == 0);
'test' and 0 are not of the same type (string vs int), so string is
converted to int, 0. In the examples below no type casting is needed.
also returns bool(true). But my logic tells me, that if 'test' == 0, then :
if (0) {
...
}
should do the same as
if ('test') {
...
}
but it doesn't! The first if-statement is _not_ executed, the latter is.
In my opinion, this is not correct. Any thoughts on this? Am I not seeing
the logic here?
Thanks for your thoughts.
Ivo Fokkema
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Re: [PHP] in_array()/finding page Problem
You solution is quite resource expensive. I would do:
[find $maxpages]
SELECT COUNT(*)/12 FROM table WHERE id = $real_id;
list($maxpages) = fetch_row()
In result.php use ORDER BY id
Ben G. McCullough wrote:
I think I have a flaw of logic in trying to find the correct page in a
mutli-page sql result.
Goal - find the correct page [results.php?page=x] when linking from
another page.
Current method - loop through a pagination function looking for the
matching $id in an array [simplified for illustration]:
[find $maxpages]
$page =1;
$catch=array();
while($page = $maxpages) {
[set start #]
$result = [get sql results - limit $startnumber, 12]
while($record = mysql_fetch_array($result)) {
extract($record);
$catch[] = $found_id;
}
if(in_array($real_id, $catch, TRUE)) {
$here = $page;
}
$page++;
}
echo results.php?page=$here;
I never seem to get a true return on in_array(), and in testing, I
don't seem to be getting a full result set with my look up.
I feel that I am missing something in the logic - or I am approaching
this from the wrong direction.
I originally wanted to do the while test as if in_array is FALSE,
continue loop but I couldn't get that to work at all.
Thank you for any help you can give this newbie.
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Re: [PHP] in_array()/finding page Problem
Ben G. McCullough wrote:
I agree - my solution is VERY resource intensive, but I think I may have
over simplified my question.
Each item is listed by two categories, $medium and $period. Users can
get to the item via a 'browse' of either category. I want to user to be
able to get back to the browse page they came from, or go to the other
listing.
Pass current url to the next page, or only the relevant variables, then
on the next page build the url from suplied variables.
To compound the issue - items are ordered by $sku, not $id.
So the select statement would look more like this:
SELECT COUNT(*)/12 FROM table WHERE medium = $medium ORDER BY $sku
You need to find out $sku of the current row and use
SELECT COUNT(*)/12 FROM table WHERE medium = $medium AND $sku =
$current_sku
But how would I find the page?
I had figured the only way was to actually go through each loop of pages
looking for the $id, which is not working, and takes a lot of resources.
While this is only a test project, this issue is based on a real - world
problem.
You solution is quite resource expensive. I would do:
[find $maxpages]
SELECT COUNT(*)/12 FROM table WHERE id = $real_id;
list($maxpages) = fetch_row()
In result.php use ORDER BY id
Ben G. McCullough wrote:
I think I have a flaw of logic in trying to find the correct page in
a mutli-page sql result.
Goal - find the correct page [results.php?page=x] when linking from
another page.
Current method - loop through a pagination function looking for the
matching $id in an array [simplified for illustration]:
[find $maxpages]
$page =1;
$catch=array();
while($page = $maxpages) {
[set start #]
$result = [get sql results - limit $startnumber, 12]
while($record = mysql_fetch_array($result)) {
extract($record);
$catch[] = $found_id;
}
if(in_array($real_id, $catch, TRUE)) {
$here = $page;
}
$page++;
}
echo results.php?page=$here;
I never seem to get a true return on in_array(), and in testing, I
don't seem to be getting a full result set with my look up.
I feel that I am missing something in the logic - or I am approaching
this from the wrong direction.
I originally wanted to do the while test as if in_array is FALSE,
continue loop but I couldn't get that to work at all.
Thank you for any help you can give this newbie.
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Re: [PHP] in_array()
Hi Riccardo,
if(mysql_num_rows($rty-resu)) { //result
$rec = mysql_fetch_array($rty-resu);
if(!isset($_SESSION[bkmks]) ||
When you save something for the first time, the element of $_SESSION[bkmks] is a
string and you can compare the array $rec with this string.
!in_array($rec, $_SESSION[bkmks])) {
Than you convert the element into an array
$_SESSION[bkmks][] = $rec;
and AFAIK in_array() can´t compare two arrays, correct me if i am wrong...
Greetings
Jochen
*** REPLY SEPARATOR ***
On 22.08.02 at 12:06 Riccardo Sepe wrote:
Hi every1 I got this script that works fine on my local windows pc but
on the remote server (FreeBSD)
I get this message:
Warning: Wrong datatype for first argument in call to in_array in
/usr/local/..
the script should bookmark an user choice storing it in the
$_SESSION[bkmks] array.
this is the code:
$rty-mark($id,$tab); // call to the method
that perform a standard query on the db
if(mysql_num_rows($rty-resu)) { //result
$rec = mysql_fetch_array($rty-resu);
if(!isset($_SESSION[bkmks]) ||
!in_array($rec, $_SESSION[bkmks])) {
$_SESSION[bkmks][] = $rec;
when I store for the first time no problem ... When I try to store
another item or the same item again
I got that awful error
thanks in advance !
Ricky
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RE: [PHP] in_array problems (another pair of eyes?)
Unless you are using PHP version 4.2 or higher, the first argument can't be
an array.
Kirk
-Original Message-
From: Jas [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, May 21, 2002 11:46 AM
To: [EMAIL PROTECTED]
Subject: [PHP] in_array problems (another pair of eyes?)
I don't think I am using the syntax correctly, I have been
looking at this
function on php.net and everything I have seen says my code should be
working.
A form allows the user to upload a file:
form name=img1 method=post action=upload_done.php
enctype=multipart/form-data
input type=file name=img1 size=25
input type=submit name=Submit value=save
input type=reset name=reset value=reset
/form
Resulting file (upload_done.php):
?php
$types = array(.gif,
.jpg,
.jpeg,
.htm,
.pdf); //place file type into array
if (in_array(array ('.jpg', '.jpeg'), $types)) { //this is
the error line
(line 7)
print jpg file; }
?
And here is my error:
Warning: Wrong datatype for first argument in call to in_array in
upload_done.php on line 7
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Re: [PHP] in_array problems (another pair of eyes?)
On Wednesday 22 May 2002 01:45, Jas wrote:
I don't think I am using the syntax correctly, I have been looking at this
function on php.net and everything I have seen says my code should be
working.
What version of php are you using? In PHP versions before 4.2.0 needle was not
allowed to be an array.
if (in_array(array ('.jpg', '.jpeg'), $types)) { //this is the error line
(line 7)
print jpg file; }
?
And here is my error:
Warning: Wrong datatype for first argument in call to in_array in
upload_done.php on line 7
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Re: [PHP] in_array algorithm
On Wed, 2002-02-06 at 08:26, John Fulton wrote: Does anyone know which algorithm in_array() uses? For example, if I say in_array(foo, $arr) Does in_array() do an unordered sequential serach of $arr for foo which takes up to n comparisons [where n = count($arr)], or does it do a binary search which takes about lg(n) comparisons? Is it up to me to maintain a sorted array in the later case? Thanks, John Well, the source for the currect version of that function (as of 4.2.0-dev) is here: http://cvs.php.net/co.php/php4/ext/standard/array.c?r=1.156 Search down the page for 'php_search_array'--that's the function which actually does the searching. Looks like a simple sequential search to me. Torben -- Torben Wilson [EMAIL PROTECTED] http://www.thebuttlesschaps.com http://www.hybrid17.com http://www.inflatableeye.com +1.604.709.0506 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] in_array error
Steve,
What version of PHP are you running. in_array is = 4.0. is_array was in 3.0
so this may be an issue for you.
Gerard
-Original Message-
From: Steve Osborne [mailto:[EMAIL PROTECTED]]
Sent: Thursday, November 29, 2001 3:08 PM
To: PHP-General (E-mail)
Subject: [PHP] in_array error
Can anyone explain why I am getting the following error?
Fatal error: Call to unsupported or undefined function in_array() in
includes/chinlib21stCentury.inc on line 3131
Code:
if( (is_array($List)) AND (is_array($RemoveList)) )
{
$ListItems = count($List);
sort($List);
for($ListItem=0; $ListItem $ListItems; $ListItem++)
{
//printf(brList value: $List[$ListItem]br\n);
if(!(in_array($List[$ListItem],$RemoveList)) AND (trim($List[$ListItem])
) ) // Line 3131
$diff[] = $List[$ListItem];
}
}elseif($debugit){
echo In function ListDiff List and RemoveList are NOT arraysBR;
}
return ($diff);
Thanks,
Steve Osborne
Database Programmer
Chinook Multimedia Inc.
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Re: [PHP] in_array error
I'm using php 4 on my machine, however the server that I am testing on only
supports php3
- Original Message -
From: Gerard Onorato [EMAIL PROTECTED]
To: Steve Osborne [EMAIL PROTECTED]; PHP-General
(E-mail) [EMAIL PROTECTED]
Sent: Thursday, November 29, 2001 11:29 AM
Subject: RE: [PHP] in_array error
Steve,
What version of PHP are you running. in_array is = 4.0. is_array was in
3.0
so this may be an issue for you.
Gerard
-Original Message-
From: Steve Osborne [mailto:[EMAIL PROTECTED]]
Sent: Thursday, November 29, 2001 3:08 PM
To: PHP-General (E-mail)
Subject: [PHP] in_array error
Can anyone explain why I am getting the following error?
Fatal error: Call to unsupported or undefined function in_array() in
includes/chinlib21stCentury.inc on line 3131
Code:
if( (is_array($List)) AND (is_array($RemoveList)) )
{
$ListItems = count($List);
sort($List);
for($ListItem=0; $ListItem $ListItems; $ListItem++)
{
//printf(brList value: $List[$ListItem]br\n);
if(!(in_array($List[$ListItem],$RemoveList)) AND
(trim($List[$ListItem])
) ) // Line 3131
$diff[] = $List[$ListItem];
}
}elseif($debugit){
echo In function ListDiff List and RemoveList are NOT arraysBR;
}
return ($diff);
Thanks,
Steve Osborne
Database Programmer
Chinook Multimedia Inc.
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Re: [PHP] in_array error
You will not be able to use this and many other functions on a php3 only
machine. For your convenience, all of the function references in the manual
state which versions of php support them and which do not.
Fred
Steve Osborne [EMAIL PROTECTED] wrote in message
008401c17916$dd87f200$[EMAIL PROTECTED]">news:008401c17916$dd87f200$[EMAIL PROTECTED]...
I'm using php 4 on my machine, however the server that I am testing on
only
supports php3
- Original Message -
From: Gerard Onorato [EMAIL PROTECTED]
To: Steve Osborne [EMAIL PROTECTED]; PHP-General
(E-mail) [EMAIL PROTECTED]
Sent: Thursday, November 29, 2001 11:29 AM
Subject: RE: [PHP] in_array error
Steve,
What version of PHP are you running. in_array is = 4.0. is_array was in
3.0
so this may be an issue for you.
Gerard
-Original Message-
From: Steve Osborne [mailto:[EMAIL PROTECTED]]
Sent: Thursday, November 29, 2001 3:08 PM
To: PHP-General (E-mail)
Subject: [PHP] in_array error
Can anyone explain why I am getting the following error?
Fatal error: Call to unsupported or undefined function in_array() in
includes/chinlib21stCentury.inc on line 3131
Code:
if( (is_array($List)) AND (is_array($RemoveList)) )
{
$ListItems = count($List);
sort($List);
for($ListItem=0; $ListItem $ListItems; $ListItem++)
{
file://printf(brList value: $List[$ListItem]br\n);
if(!(in_array($List[$ListItem],$RemoveList)) AND
(trim($List[$ListItem])
) ) // Line 3131
$diff[] = $List[$ListItem];
}
}elseif($debugit){
echo In function ListDiff List and RemoveList are NOT arraysBR;
}
return ($diff);
Thanks,
Steve Osborne
Database Programmer
Chinook Multimedia Inc.
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Re: [PHP] in_array
On Thu, 13 Sep 2001 14:17:12 +0300, you wrote:
i wrote php scripts with php 4. but my server's php version is php 3. i
used in_array function while i was writing the scripts. i used that
function to check posted variables is available or not.
is there an another way to check this posted variables or another one
likes in_array function?
Straight from the manual:
http://www.php.net/manual/en/function.in-array.php
function in_array($needle,$haystack) {
for($i=0; $icount($haystack) $haystack[$i] !=$needle; $i++);
return ($i!=count($haystack));
}
djo
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Re: [PHP] in_array() with associate array?
in_array is for testing that a value exists in an array. What you want is if ( isset( $some_array[some_key] ) ) print HAS KEY; else print DOESN'T HAVE KEY; At 14:23 31/07/2001 -0400, Jaxon wrote: hi, in_array is confusing me :) can someone show me an example of how to test if $some_array[some_key] actually has a value, without outputting the value? tia, jaxon -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] - Brian White Step Two Designs Pty Ltd - SGML, XML HTML Consultancy Phone: +612-93197901 Web: http://www.steptwo.com.au/ Email: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP] in_array() function not supported on my server... anything else?
yeah.. the loop.
see php.net/arrays
most of the coolest array functions were introduced in PHP4
if you are trying to make a dynamic portable application you might want to
look into phpversion();
this will tell you which version it runs, and based on that will select the
best way to find the results on any server.
Sincerely,
Maxim Maletsky
Founder, Chief Developer
PHPBeginner.com (Where PHP Begins)
[EMAIL PROTECTED]
www.phpbeginner.com
-Original Message-
From: Richard [mailto:[EMAIL PROTECTED]]
Sent: Friday, April 27, 2001 6:23 AM
To: [EMAIL PROTECTED]
Subject: [PHP] in_array() function not supported on my server...
anything else?
Greetings..
This is how I currently check for instances of words and other:
for ($i=0; $i $total_lines; $i++){
$line_array = explode(|,$line[$i]);
$swhat=strtolower($txtLinkname);
$xos =
array(strtolower($line_array[1]),strtolower($line_array[2]),strtolower($line
_array[3]),strtolower($line_array[4]),strtolower($line_array[5]));
if (in_array($swhat,$xos,true)) {
$found++;
Problem is, that the server where I have my files don't support in_array. It
is claimed that the servers PHP server is on version 4.0, how can this be
when in_array is a function that came first with PHP4 (according to my
documents) ??
Is there something else I could use?
Thanks
- Richard
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Re: [PHP] in_array() with multidimensional array
recusrion is your friend.
?php
$test[0][0] = 'a';
$test[0][1] = 'b';
$test[0][2] = 'c';
$test[1][0] = 'd';
$test[1][1] = 'e';
$test[1][2] = 'f';
$test[2][0] = 'g';
$test[2][1] = 'h';
$test[2][2] = 'i';
function in_multi_array($needle, $haystack)
{
foreach($haystack as $pos = $val)
{
if (is_array($val))
{
if (in_multi_array($needle, $val))
return 1;
} else
if ($val == $needle)
return 1;
}
}
if (in_multi_array('d', $test))
echo "TRUE br\n";
else
echo "FALSE br\n";
?
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Chris Lee
Mediawaveonline.com
ph. 250.377.1095
ph. 250.376.2690
fx. 250.554.1120
[EMAIL PROTECTED]
""Christian Dechery"" [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
How can I check if a value is in a multidimensional array?
like I have
?php
$i=0;
while($idt = mssql_fetch_row($query_result))
{
//echo $idt[0]."br";
$produtos_sem_tracking[$i]['cod']=$idt[0];
$produtos_sem_tracking[$i]['idt']=$idt[1];
$produtos_sem_tracking[$i]['gen']=$idt[2];
}
?
how can I check for an existing $produtos_sem_tracking['cod'] value for
example?
. [ Christian Dechery ]
. Webdeveloper @ T Na Mesa!
. Listmaster @ Gaita-L
. http://www.tanamesa.com.br
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