php-general Digest 5 May 2012 22:35:42 -0000 Issue 7802
php-general Digest 5 May 2012 22:35:42 - Issue 7802
Topics (messages 317793 through 317804):
Re: Calculating driving distance between UK postcodes
317793 by: tamouse mailing lists
Re: function
317794 by: tamouse mailing lists
317797 by: Jim Giner
317801 by: tamouse mailing lists
317802 by: tamouse mailing lists
Re: PHP Emacs
317795 by: tamouse mailing lists
Re: Retrieve pages from an ASP driven site
317796 by: tamouse mailing lists
Re: Running through an enormous SQL file
317798 by: Brian Dunning
317800 by: tamouse mailing lists
Re: get content rss feed
317799 by: tamouse mailing lists
Re: code deployment through php
317803 by: tamouse mailing lists
Re: PHP Database Problems -- Code Snippets
317804 by: Matijn Woudt
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--
---BeginMessage---
On Fri, May 4, 2012 at 9:18 AM, Terry Ally (Gmail) terrya...@gmail.com wrote:
Google works in Javascript extensively - not a language with which I
have in-depth experience hence my reason for asking for PHP solution.
For example the following will get me a JSON output with the distance in
Kms and time. I don't know how to get PHP to read this information and
extract just the distance. I need the distance so that I can calculate cost
of a trip.
form id=google action=
http://maps.googleapis.com/maps/api/distancematrix/json; method=get
input type=text name=origins value= /
input type=text name=destinations value= /
input type=hidden name=sensor value=false
input type=hidden name=submitted value=1
bra type=submit
onClick=document.getElementById('google').submit()strongstrongGet
Distance/strong/strong/a
/form
Using Google Maps API is pretty straight-forward. You don't need to
set up a form or a use a POST to get the info. This page should
describes how to use a standard GET query to get the info you want:
https://developers.google.com/maps/documentation/distancematrix/
Setting up the proper URL to call, you can activate it using
file_get_contents provided you have allow_url_fopen set to true in
php.ini. (Do make sure to check for possible errors returned.)
You can get the response back as either JSON or XML, both of which PHP
can parse into useful data structures:
http://us.php.net/manual/en/function.json-decode.php
http://us.php.net/manual/en/book.simplexml.php
---End Message---
---BeginMessage---
On Thu, May 3, 2012 at 9:12 PM, Ron Piggott
ron.pigg...@actsministries.org wrote:
I need to access a FUNCTION I programmed within a different FUNCTION. Are
these able to be passed like a variable? Or are they able to become like a
$_SESSION variable in nature? How am I able to do this?
I am essentially programming:
===
function name( $flag1, $flag2 ) {
# some PHP
echo name_of_a_different_function( $flag1 , $flag2 );
}
===
The error I am receiving is “Call to undefined function
name_of_a_different_function”
Where is name_of_a_different_function defined? If it is somewhere in
the same file as name, that shouldn't be a problem, provided it is
defined in the same namespace/scope as name. If it is defined in a
different file, you need to include that file before you make the echo
statement.
For example:
function func1 ($flag1, $flag2) {
# blah blah
echo func2($flag1, $flag2);
}
function func2 ($flag1, $flag2) {
#blah blah
return some string value;
}
in the same file should be just fine. It doesn't really matter what
order func1 and func2 are declared in.
However, if func2 is defined in some_other_file.php, you need to
include it in this_file.php (where func1 is defined) first:
this_file.php:
include('some_other_file.php');
function func1 ($flag1, $flag2) {
#blah blah
echo func2 ($flag1, $flag2);
}
some_other_file.php:
function func2 ($flag1, $flag2) {
#blah blah
return some string value;
}
If func2 is a method for an object/class, you'll have to access it
that way in func1:
this_file.php:
include('MyClass.php');
function func1 ($flag1, $flag2) {
# blah blah, instantiate object?
$myobj = new MyClass();
echo $myobj-func2 ($flag1, $flag2);
}
MyClass.php:
class MyClass
{
function func2 ($flag1, $flag2) {
#blah blah
return some string value;
}
}
---End Message---
---BeginMessage---
But the OP says function is defined inside a different function. Your
theories to a solution don't fit that problem.
tamouse mailing lists tamouse.li...@gmail.com wrote in message
news:cahuc_t-416_-lpcn3mo8qqxwrh4pnq5fmwouhwpdk+hmkgh...@mail.gmail.com...
On Thu, May 3, 2012 at 9:12 PM, Ron Piggott
ron.pigg...@actsministries.org wrote:
Where is
Re: [PHP] PHP Database Problems -- Code Snippets
On Thu, May 3, 2012 at 4:20 PM, Ethan Rosenberg eth...@earthlink.net wrote:
At 06:47 PM 5/2/2012, Matijn Woudt wrote:
On Wed, May 2, 2012 at 11:43 PM, Ethan Rosenberg eth...@earthlink.net
wrote: Dear list - Sorry for the attachment. Â Here are code snippets
--- Ethan, I don't want to sound rude, but it appears to me you don't have
any understanding of what you're doing. It might help if you understand what
the code is doing... Let me explain. GET THE DATA FROM INTAKE3: Â Â
function handle_data() Â Â { Â Â Â global $cxn; Â Â Â $query =
select * from Intake3 where  1;   Â
if(isset($_Request['Sex']) trim($_POST['Sex']) != '' ) $_Request does not
exists, you're looking for $_REQUEST. And why are you mixing $_REQUEST and
$_POST here? Â Â Â { Â Â Â Â Â Â if ($_REQUEST['Sex'] === 0)
      {        $sex = 'Male';      Â
}       else       {        $sex =
'Female'; Â Â Â Â Â Â } Â Â Â } Â Â } What is the point of
the handle_data function above? It doesn't do anything. Â Â
$allowed_fields = array    (  'Site' =$_POST['Site'], 'MedRec' =
$_POST['MedRec'], 'Fname' = $_POST['Fname'], 'Lname' = $_POST['Lname'] ,
      'Phone' = $_POST['Phone'] , 'Sex' = $_POST['Sex']  ,
'Height' = $_POST['Height'] Â ); Â Â if(empty($allowed_fields)) Â
 {      echo ouch;   }   $query = select * from
Intake3  where  1 ;   foreach ( $allowed_fields as $key = $val )
  {    if ( (($val != '')) )   {    $query .=
AND ($key  = '$val') ;   }    $result1 = mysqli_query($cxn,
$query); Â Â } First, this will allow SQL injections, because you insert
the values directly from the browser. Second, you should move the last line
($result1=...), outside of the foreach loop, now you're executing the query
multiple times. Third, you should check if $result1 === FALSE, in case the
query fails   $num = mysqli_num_rows($result1);   if(($num =
mysqli_num_rows($result1)) == 0) Doing the same thing twice? Â Â { ?
  br /br /centerbp style=color: red; font-size:14pt; No
Records Retrieved #1/center/b/style/p ?php   exit(); Â
 } DISPLAY THE INPUT3 DATA: THIS SEEMS TO BE THE ROUTINE THAT
IS FAILINGÂ Â centerbSearch Results/b/centerbr / Â
 centertable border=4 cellpadding=5 cellspacing=55  rules=all
 frame=box   tr class=\heading\   thSite/th  Â
thMedical Record/th   thFirst Name/th   thLast Name/th
  thPhone/td   thHeight/td   thSex/td  Â
thHistory/td   /tr ?php    while ($row1 =
mysqli_fetch_array($result1, MYSQLI_BOTH)) Â Â Â { Â Â Â Â Â Â
print_r($_POST); Doesn't really make sense to print $_POST here.. Â Â Â
    global $MDRcheck;        $n1++;     Â
  echo br /n1 br /;echo $n1;       {     Â
  if (($n1 2) ($MDRcheck == $row1[1]))        {
          echo 2==  ;          Â
echo $MDRcheck; Â Â Â Â Â Â Â Â Â Â echo td $row1[0] /td\n;
          echo td $row1[1] /td\n;     Â
     echo td $row1[2] /td\n;          Â
echo td $row1[3] /td\n; Â Â Â Â Â Â Â Â Â Â echo td
$row1[4] /td\n; Â Â Â Â Â Â Â Â Â Â echo td $row1[5]
/td\n; Â Â Â Â Â Â Â Â Â Â echo td $row1[6] /td\n; Â
         echo td $row1[7] /td\n;      Â
    echo /tr\n;        }       Â
elseif (($n1 2) ($MDRcheck != $row1[1])) Â Â Â Â Â Â Â { Â
         echo 2!=  ;          Â
echo $MDRcheck;Â Â Â Â Â Â Â Â Â Â continue; continue
doesn't do anything here. Â Â Â Â Â Â Â } Â Â Â Â Â Â Â
elseif ($n1 == 2) Â Â Â Â Â Â Â { Â Â Â Â Â Â Â Â Â Â
define( MDR , Â $row1[1]); Â Â Â Â Â Â Â Â Â Â echo br /row1
br;echo $row1[1]; Â Â Â Â Â Â Â Â Â Â echo tr\n; Â Â
        $_GLOBALS['mdr']= $row1[1];        Â
  $_POST['MedRec'] = $row1[1]; You're not supposed to set variables in
$_POST... Â Â Â Â Â Â Â Â Â Â $MDRold = $_GLOBALS['mdr']; It
appears you want the old value of mdr, if so, then you should do this before
you set it again 2 lines above.. Â Â Â Â Â Â Â Â Â Â echo td
$row1[0] /td\n; Â Â Â Â Â Â Â Â Â Â echo td $row1[1]
/td\n; Â Â Â Â Â Â Â Â Â Â echo td $row1[2] /td\n; Â
         echo td $row1[3] /td\n;      Â
    echo td $row1[4] /td\n;          Â
echo td $row1[5] /td\n; Â Â Â Â Â Â Â Â Â Â echo td
$row1[6] /td\n; Â Â Â Â Â Â Â Â Â Â echo td $row1[7]
/td\n; Â Â Â Â Â Â Â Â Â
[PHP] I'm missing something
I have a discrepancy in the number of elements in my arrays and can't see
why.
Here is some code:
Note the lines with the ***
*** $plyrs = 0;
unset($plyrnames_ar);
unset($js_names);
unset($js_seeds);
*** $rows = mysql_num_rows($qrslts);
***echo in mysql there are $rows rowsbr;
while ($row = mysql_fetch_array($qrslts))
{ // build the name value here
$mi = ($row['MI']=='') ? '' : .$row['MI'];
$nm = $row['LastName']., .$row['FirstName'].$mi;
if ($row['srtdbls']=='X')
{
$mi = ($row['partMI']=='') ? '' : .$row['partMI'];
$nm .= /.$row['partLN']., .$row['partFN'].$mi;
}
*** $plyrs++;
$plyrnames_ar[$nm] = $row['Draw_pos'];
}
*** echo before sort there are .count($plyrnames_ar). in
plyrsnames_arbr;
ksort($plyrnames_ar);
foreach ($plyrnames_ar as $nm=$sd)
{
$js_names[] = $nm;
$js_seeds[] = $sd;
}
*** echo plyrs is $plyrs and there are .count($js_names). entries in
js_rnames and .count
***($plyrnames_ar). entries in plyrnames_ar and .count($js_seeds). in
js_seeds;
exit();
The problem is that my $plyrs field comes up 18, while in truth there are
only 17 rows of data in my database.
All of the other array-size counts echo out as being only 17, but my $rows
and $plyrs fields come up as 18. I was having problems with my javascript
showing an undefined array element and took a bit o time to determine what
it was and where it was happening.
Questions - does mysql_num_rows return a extra row that somehow doesn't get
processed in the while loop? And if it does how come the counter ($plyrs)
that I put into the loop comes up higher than it should.?
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Re: [PHP] I'm missing something
On Sun, May 6, 2012 at 12:38 AM, Jim Giner jim.gi...@albanyhandball.com wrote:
I have a discrepancy in the number of elements in my arrays and can't see
why.
Here is some code:
Note the lines with the ***
*** $plyrs = 0;
unset($plyrnames_ar);
unset($js_names);
unset($js_seeds);
*** $rows = mysql_num_rows($qrslts);
***echo in mysql there are $rows rowsbr;
while ($row = mysql_fetch_array($qrslts))
{ // build the name value here
$mi = ($row['MI']=='') ? '' : .$row['MI'];
$nm = $row['LastName']., .$row['FirstName'].$mi;
if ($row['srtdbls']=='X')
{
$mi = ($row['partMI']=='') ? '' : .$row['partMI'];
$nm .= /.$row['partLN']., .$row['partFN'].$mi;
}
*** $plyrs++;
$plyrnames_ar[$nm] = $row['Draw_pos'];
}
*** echo before sort there are .count($plyrnames_ar). in
plyrsnames_arbr;
ksort($plyrnames_ar);
foreach ($plyrnames_ar as $nm=$sd)
{
$js_names[] = $nm;
$js_seeds[] = $sd;
}
*** echo plyrs is $plyrs and there are .count($js_names). entries in
js_rnames and .count
***($plyrnames_ar). entries in plyrnames_ar and .count($js_seeds). in
js_seeds;
exit();
The problem is that my $plyrs field comes up 18, while in truth there are
only 17 rows of data in my database.
All of the other array-size counts echo out as being only 17, but my $rows
and $plyrs fields come up as 18. I was having problems with my javascript
showing an undefined array element and took a bit o time to determine what
it was and where it was happening.
Questions - does mysql_num_rows return a extra row that somehow doesn't get
processed in the while loop? And if it does how come the counter ($plyrs)
that I put into the loop comes up higher than it should.?
My guess would be that you end up with 2 rows having the same $nm,
overwriting the value that's already in $plyrnames_ar.
- Matijn
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Re: [PHP] I'm missing something
My guess would be that you end up with 2 rows having the same $nm, overwriting the value that's already in $plyrnames_ar. - Matijn Genius at work! Thanks - I'll look into that. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] I'm missing something
Yup that was it! Something I knew would happen during my design, but forgot to code for now. Jim Giner jim.gi...@albanyhandball.com wrote in message news:e2.dc.30075.c6ea5...@pb1.pair.com... My guess would be that you end up with 2 rows having the same $nm, overwriting the value that's already in $plyrnames_ar. - Matijn Genius at work! Thanks - I'll look into that. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] I'm missing something
On Sun, May 6, 2012 at 12:53 AM, Jim Giner jim.gi...@albanyhandball.com wrote: Yup that was it! Something I knew would happen during my design, but forgot to code for now. Jim Giner jim.gi...@albanyhandball.com wrote in message news:e2.dc.30075.c6ea5...@pb1.pair.com... My guess would be that you end up with 2 rows having the same $nm, overwriting the value that's already in $plyrnames_ar. - Matijn Genius at work! Thanks - I'll look into that. Jim, Two things: 1) You should bottom post on this (probably any) mailing list. 2) You can avoid these errors if you define unique fields in your database. - Matijn -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
