subject:"\[PHP\] Creating a javascript array from database data"

RE: [PHP] Creating a javascript array from database data

2001-08-16 Thread Tim Ward


I'm not sure if this is the problem but it is an important point. print
\$text\; does not output anything to screen, as with all prints and
echos it outputs it to html (in this case within a js function definition).
Try viewing the source produced.

Tim

--
From:  Neil Freeman [SMTP:[EMAIL PROTECTED]]
Sent:  15 August 2001 18:10
To:  Sheridan Saint-Michel
Cc:  php-general
Subject:  Re: [PHP] Creating a javascript array from database data

Thanks a lot for your help Sheridan,

This now appears to work ok :) but if I remove the:

print \$text\;

line (as it is not required), I receive errors complaining about a
syntax
error (related to what?) and that 'text' is undefined. As this line
simply
outputs the value to screen why should this cause a problem if it is
removed?
Or could it be some other problem?

Neil

PS: Here is my updated code:

--
html
head
titleMenus test/title

/head
body bgcolor=white

?php
$dbhost = 'localhost';
$dbuser = 'guest';
$dbpass = 'guest';
$dbname = 'IFE';
$dbtable = 'menus';

//-- DATABASE CONNECTION //
mysql_connect($dbhost,$dbuser,$dbpass)
   or die (Unable to connect to database);

mysql_select_db($dbname)
   or die (Unable to select database);


$sql= SELECT * FROM $dbtable;
$result = mysql_query($sql);

$number = mysql_numrows($result);

$i = 0;

if ($number == 0)
   print Error - No records found;
elseif ($number  0)
{
   echo Script Language=\JavaScript\\n;
   echo text = new Array(;
   while ($i  $number)
   {
  $text = mysql_result($result, $i, name);

   $i++;

   if ($i  $number)
print ,;
  else
print )\n;
  }
  echo /Script\n;
}

mysql_free_result($result);
mysql_close();
?

/body
/html
--

Sheridan Saint-Michel wrote:

 ***
  This message was virus checked with: SAVI 3.48
  last updated 14th August 2001
 ***

 The thing to ALWAYS remember when working with both PHP and
 JavaScript is that PHP is Server-Side and JavaScript is
Client-Side.

 What this mean is practical terms is that when going from
JavaScript
 to PHP you have to submit or redirect back to the server... and
when
 going from PHP to JavaScript (Like you are trying to do here) you
 have to make sure your PHP outputs JavaScript.

 So instead of printing $text you need to actually print the
JavaScript...

 So in your case change the middle of your script to something
like:

 $number = mysql_numrows($result);

  $i = 0;

  if ($number == 0)
 print Error - No records found;
 else
 {
   echo Script Language=\JavaScript\\n;
   echo text = new Array(;
   while ($i  $number)
   {
  $text = mysql_result($result, $i, name);
  print \$text\;
 $i++;
 if ($i  $number)
   print ,;
 else
   print )\n;
}
   echo /Script\n;
 }

 }

 Note:  This is my quick *untested* fix done by modifying
 your code as little as possible.  I would suggest you use
 mysql_fetch_row and use your loop to parse it rather than
 making several calls to mysql_result (especially if you are
 ever going to have more than a few entries in the DB).

 Sheridan Saint-Michel
 Website Administrator
 FoxJet, an ITW Company
 www.foxjet.com

 - Original Message -
 From: Neil Freeman [EMAIL PROTECTED]
 To: PHP General [EMAIL PROTECTED]
 Sent: Wednesday, August 15, 2001 11:16 AM
 Subject: [PHP] Creating a javascript array from database data

  Hi there,
 
  Well after a few hours roaming around various websites I am at a
loss.
  Here is what I am trying to do:
 
  1) Access a MySQL database which contains 1 table
  2) Read the records from this table
  3) Store the values returned from this table into javascript
array
  elements, ie, if I get the values dog, cat and cow back I
want
  these stored in an array as such:
  myArray[0] = valueReturned1
  myArray[1] = valueReturned2

Re: [PHP] Creating a javascript array from database data (Off-List)

2001-08-16 Thread Neil Freeman


Thanks for the clarification Sheridan - I think it must have been end of day
madness on my part :) I see what's going on now with the 'print' line.

Cheers,

Neil

Sheridan Saint-Michel wrote:

 ***
  This message was virus checked with: SAVI 3.48
  last updated 14th August 2001
 ***

 Think in terms of output...  The JavaScript is literally going to
 try and run the output of the PHP (It doesn't even know PHP
 exists).  PHP is outputting the source code for the HTML and
 JavaScript.  So what you are trying to do is get PHP to take
 dog,cat,cow from MySQL and output:

 Script Language=JavaScript
 text = new Array(dog,cat,cow)
 /Script

 This block of code does exactly that, with the print\$text\;
 filling in dog cat and cow.  I've added comments so you can
 see how each line produces the above (I am using C-Style
 comments as the // gets mucked up in e-mail)

 The parts in [] are comments within comments to show flow
 control and are not output.

 else
 {
 /* Script Language=JavaScript */
   echo Script Language=\JavaScript\\n;
 /* text=new Array( */
   echo text = new Array(;
   while ($i  $number)

 $text = mysql_result($result, $i, name);
 /* dog [first time] */
 /* cat [second time] */
 /* cow [third time] */
 print \$text\;
 $i++;
 if ($i  $number)
 /* , [first time] */
 /* , [second time] */
print ,;
 else
 /* ) [third time] */
   print )\n;
 }
 /* /Script */
   echo /Script\n;
 }

 So if you can follow that (just walk all the way through
 the loop for each of the three passes) you can see that
 no part of the code (including the print \$text\;) is
 unnecessary

 Hope this helps

 Sheridan Saint-Michel
 Website Administrator
 FoxJet, an ITW Company
 www.foxjet.com

 - Original Message -
 From: Neil Freeman [EMAIL PROTECTED]
 To: Sheridan Saint-Michel [EMAIL PROTECTED]
 Cc: php-general [EMAIL PROTECTED]
 Sent: Wednesday, August 15, 2001 12:09 PM
 Subject: Re: [PHP] Creating a javascript array from database data

  Thanks a lot for your help Sheridan,
 
  This now appears to work ok :) but if I remove the:
 
  print \$text\;
 
  line (as it is not required), I receive errors complaining about a syntax
  error (related to what?) and that 'text' is undefined. As this line simply
  outputs the value to screen why should this cause a problem if it is
 removed?
  Or could it be some other problem?
 
  Neil
 
  PS: Here is my updated code:
 Snip

--

 Email:  [EMAIL PROTECTED]
 [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] Creating a javascript array from database data

2001-08-16 Thread Neil Freeman


Yeah thanks Tim. 5:30 in the afternoon does bad things to my logical thinking.
Not the best time to try and learn PHP :)

Neil

Tim Ward wrote:

 I'm not sure if this is the problem but it is an important point. print
 \$text\; does not output anything to screen, as with all prints and
 echos it outputs it to html (in this case within a js function definition).
 Try viewing the source produced.

 Tim

 --
 From:  Neil Freeman [SMTP:[EMAIL PROTECTED]]
 Sent:  15 August 2001 18:10
 To:  Sheridan Saint-Michel
 Cc:  php-general
 Subject:  Re: [PHP] Creating a javascript array from database data

 Thanks a lot for your help Sheridan,

 This now appears to work ok :) but if I remove the:

 print \$text\;

 line (as it is not required), I receive errors complaining about a
 syntax
 error (related to what?) and that 'text' is undefined. As this line
 simply
 outputs the value to screen why should this cause a problem if it is
 removed?
 Or could it be some other problem?

 Neil

 PS: Here is my updated code:

 --
 html
 head
 titleMenus test/title

 /head
 body bgcolor=white

 ?php
 $dbhost = 'localhost';
 $dbuser = 'guest';
 $dbpass = 'guest';
 $dbname = 'IFE';
 $dbtable = 'menus';

 //-- DATABASE CONNECTION //
 mysql_connect($dbhost,$dbuser,$dbpass)
or die (Unable to connect to database);

 mysql_select_db($dbname)
or die (Unable to select database);

 $sql= SELECT * FROM $dbtable;
 $result = mysql_query($sql);

 $number = mysql_numrows($result);

 $i = 0;

 if ($number == 0)
print Error - No records found;
 elseif ($number  0)
 {
echo Script Language=\JavaScript\\n;
echo text = new Array(;
while ($i  $number)
{
   $text = mysql_result($result, $i, name);

$i++;

if ($i  $number)
 print ,;
   else
 print )\n;
   }
   echo /Script\n;
 }

 mysql_free_result($result);
 mysql_close();
 ?

 /body
 /html
 --

 Sheridan Saint-Michel wrote:

  ***
   This message was virus checked with: SAVI 3.48
   last updated 14th August 2001
  ***
 
  The thing to ALWAYS remember when working with both PHP and
  JavaScript is that PHP is Server-Side and JavaScript is
 Client-Side.
 
  What this mean is practical terms is that when going from
 JavaScript
  to PHP you have to submit or redirect back to the server... and
 when
  going from PHP to JavaScript (Like you are trying to do here) you
  have to make sure your PHP outputs JavaScript.
 
  So instead of printing $text you need to actually print the
 JavaScript...
 
  So in your case change the middle of your script to something
 like:
 
  $number = mysql_numrows($result);
 
   $i = 0;
 
   if ($number == 0)
  print Error - No records found;
  else
  {
echo Script Language=\JavaScript\\n;
echo text = new Array(;
while ($i  $number)
{
   $text = mysql_result($result, $i, name);
   print \$text\;
  $i++;
  if ($i  $number)
print ,;
  else
print )\n;
 }
echo /Script\n;
  }
 
  }
 
  Note:  This is my quick *untested* fix done by modifying
  your code as little as possible.  I would suggest you use
  mysql_fetch_row and use your loop to parse it rather than
  making several calls to mysql_result (especially if you are
  ever going to have more than a few entries in the DB).
 
  Sheridan Saint-Michel
  Website Administrator
  FoxJet, an ITW Company
  www.foxjet.com
 
  - Original Message -
  From: Neil Freeman [EMAIL PROTECTED]
  To: PHP General [EMAIL PROTECTED]
  Sent: Wednesday, August 15, 2001 11:16 AM
  Subject: [PHP] Creating a javascript array from database data
 
   Hi there,
  
   Well after a few hours roaming around various websites I am at a
 loss.
   Here is what I am trying to do:
  
   1) Access a MySQL database which contains 1 table
   2) Read the records from

[PHP] Creating a javascript array from database data

2001-08-15 Thread Neil Freeman


Hi there,

Well after a few hours roaming around various websites I am at a loss.
Here is what I am trying to do:

1) Access a MySQL database which contains 1 table
2) Read the records from this table
3) Store the values returned from this table into javascript array
elements, ie, if I get the values dog, cat and cow back I want
these stored in an array as such:
myArray[0] = valueReturned1
myArray[1] = valueReturned2
myArray[2] = valueReturned3

You get the idea.

Problem being that I cannot work out how to implement the javascript
section of this. At the moment my php script writes the values returned
from the database to screen but I require these to be stored in a
javascript array. Please can someone help me before I go mad :)

Here is my current .php script:

--
html
head
titleMenus test/title
/head
body bgcolor=white

?php
$dbhost = 'localhost';
$dbuser = 'guest';
$dbpass = 'guest';
$dbname = 'IFE';
$dbtable = 'menus';

//-- DATABASE CONNECTION //
mysql_connect($dbhost,$dbuser,$dbpass)
   or die (Unable to connect to database);

mysql_select_db($dbname)
   or die (Unable to select database);


$sql= SELECT * FROM $dbtable;
$result = mysql_query($sql);

$number = mysql_numrows($result);

$i = 0;

if ($number == 0)
   print Error - No records found;
elseif ($number  0)
{
   while ($i  $number)
   {
  $text = mysql_result($result, $i, name);
  print $text;

   $i++;
   }
}

mysql_free_result($result);
mysql_close();
?

/body
/html
--

Thanks, Neil


 Email:  [EMAIL PROTECTED]
 [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] Creating a javascript array from database data

2001-08-15 Thread Jon Farmer




--
while you are looping through the date jump in and out of php like this to
populate the javascript array


script language=Javascript

function DataArray()
{
a = new Array()

?
while ($i  $number)
{
$text = mysql_result($result, $i, name);
?
a[?=$i?] = ?=$text?
?

$i++;
}
?
}

HTH
Jon


Jon Farmer
Systems Programmer, Entanet www.enta.net
Tel 01952 428969 Mob 07968 524175
PGP Key available, send blank email to [EMAIL PROTECTED]

-Original Message-
From: Neil Freeman [mailto:[EMAIL PROTECTED]]
Sent: 15 August 2001 17:16
To: PHP General
Subject: [PHP] Creating a javascript array from database data


Hi there,

Well after a few hours roaming around various websites I am at a loss.
Here is what I am trying to do:

1) Access a MySQL database which contains 1 table
2) Read the records from this table
3) Store the values returned from this table into javascript array
elements, ie, if I get the values dog, cat and cow back I want
these stored in an array as such:
myArray[0] = valueReturned1
myArray[1] = valueReturned2
myArray[2] = valueReturned3

You get the idea.

Problem being that I cannot work out how to implement the javascript
section of this. At the moment my php script writes the values returned
from the database to screen but I require these to be stored in a
javascript array. Please can someone help me before I go mad :)

Here is my current .php script:

--
html
head
titleMenus test/title
/head
body bgcolor=white

?php
$dbhost = 'localhost';
$dbuser = 'guest';
$dbpass = 'guest';
$dbname = 'IFE';
$dbtable = 'menus';

//-- DATABASE CONNECTION //
mysql_connect($dbhost,$dbuser,$dbpass)
   or die (Unable to connect to database);

mysql_select_db($dbname)
   or die (Unable to select database);


$sql= SELECT * FROM $dbtable;
$result = mysql_query($sql);

$number = mysql_numrows($result);

$i = 0;

if ($number == 0)
   print Error - No records found;
elseif ($number  0)
{
   while ($i  $number)
   {
  $text = mysql_result($result, $i, name);
  print $text;

   $i++;
   }
}

mysql_free_result($result);
mysql_close();
?

/body
/html
--

Thanks, Neil


 Email:  [EMAIL PROTECTED]
 [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]


-- 
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] Creating a javascript array from database data

2001-08-15 Thread Sheridan Saint-Michel


The thing to ALWAYS remember when working with both PHP and
JavaScript is that PHP is Server-Side and JavaScript is Client-Side.

What this mean is practical terms is that when going from JavaScript
to PHP you have to submit or redirect back to the server... and when
going from PHP to JavaScript (Like you are trying to do here) you
have to make sure your PHP outputs JavaScript.

So instead of printing $text you need to actually print the JavaScript...

So in your case change the middle of your script to something like:

$number = mysql_numrows($result);
 
 $i = 0;
 
 if ($number == 0)
print Error - No records found;
else 
{
  echo Script Language=\JavaScript\\n;
  echo text = new Array(;
  while ($i  $number)
  {
 $text = mysql_result($result, $i, name);
 print \$text\;
$i++;
if ($i  $number)
  print ,;
else
  print )\n;
   }
  echo /Script\n;
}

}

Note:  This is my quick *untested* fix done by modifying
your code as little as possible.  I would suggest you use 
mysql_fetch_row and use your loop to parse it rather than
making several calls to mysql_result (especially if you are
ever going to have more than a few entries in the DB).

Sheridan Saint-Michel
Website Administrator
FoxJet, an ITW Company
www.foxjet.com


- Original Message - 
From: Neil Freeman [EMAIL PROTECTED]
To: PHP General [EMAIL PROTECTED]
Sent: Wednesday, August 15, 2001 11:16 AM
Subject: [PHP] Creating a javascript array from database data


 Hi there,
 
 Well after a few hours roaming around various websites I am at a loss.
 Here is what I am trying to do:
 
 1) Access a MySQL database which contains 1 table
 2) Read the records from this table
 3) Store the values returned from this table into javascript array
 elements, ie, if I get the values dog, cat and cow back I want
 these stored in an array as such:
 myArray[0] = valueReturned1
 myArray[1] = valueReturned2
 myArray[2] = valueReturned3
 
 You get the idea.
 
 Problem being that I cannot work out how to implement the javascript
 section of this. At the moment my php script writes the values returned
 from the database to screen but I require these to be stored in a
 javascript array. Please can someone help me before I go mad :)
 
 Here is my current .php script:
 
 --
 html
 head
 titleMenus test/title
 /head
 body bgcolor=white
 
 ?php
 $dbhost = 'localhost';
 $dbuser = 'guest';
 $dbpass = 'guest';
 $dbname = 'IFE';
 $dbtable = 'menus';
 
 file://-- DATABASE CONNECTION //
 mysql_connect($dbhost,$dbuser,$dbpass)
or die (Unable to connect to database);
 
 mysql_select_db($dbname)
or die (Unable to select database);
 
 
 $sql= SELECT * FROM $dbtable;
 $result = mysql_query($sql);
 
 $number = mysql_numrows($result);
 
 $i = 0;
 
 if ($number == 0)
print Error - No records found;
 elseif ($number  0)
 {
while ($i  $number)
{
   $text = mysql_result($result, $i, name);
   print $text;
 
$i++;
}
 }
 
 mysql_free_result($result);
 mysql_close();
 ?
 
 /body
 /html
 --
 
 Thanks, Neil



-- 
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] Creating a javascript array from database data

2001-08-15 Thread Neil Freeman


Thanks a lot for your help Sheridan,

This now appears to work ok :) but if I remove the:

print \$text\;

line (as it is not required), I receive errors complaining about a syntax
error (related to what?) and that 'text' is undefined. As this line simply
outputs the value to screen why should this cause a problem if it is removed?
Or could it be some other problem?

Neil

PS: Here is my updated code:

--
html
head
titleMenus test/title

/head
body bgcolor=white

?php
$dbhost = 'localhost';
$dbuser = 'guest';
$dbpass = 'guest';
$dbname = 'IFE';
$dbtable = 'menus';

//-- DATABASE CONNECTION //
mysql_connect($dbhost,$dbuser,$dbpass)
   or die (Unable to connect to database);

mysql_select_db($dbname)
   or die (Unable to select database);


$sql= SELECT * FROM $dbtable;
$result = mysql_query($sql);

$number = mysql_numrows($result);

$i = 0;

if ($number == 0)
   print Error - No records found;
elseif ($number  0)
{
   echo Script Language=\JavaScript\\n;
   echo text = new Array(;
   while ($i  $number)
   {
  $text = mysql_result($result, $i, name);

   $i++;

   if ($i  $number)
print ,;
  else
print )\n;
  }
  echo /Script\n;
}

mysql_free_result($result);
mysql_close();
?

/body
/html
--

Sheridan Saint-Michel wrote:

 ***
  This message was virus checked with: SAVI 3.48
  last updated 14th August 2001
 ***

 The thing to ALWAYS remember when working with both PHP and
 JavaScript is that PHP is Server-Side and JavaScript is Client-Side.

 What this mean is practical terms is that when going from JavaScript
 to PHP you have to submit or redirect back to the server... and when
 going from PHP to JavaScript (Like you are trying to do here) you
 have to make sure your PHP outputs JavaScript.

 So instead of printing $text you need to actually print the JavaScript...

 So in your case change the middle of your script to something like:

 $number = mysql_numrows($result);

  $i = 0;

  if ($number == 0)
 print Error - No records found;
 else
 {
   echo Script Language=\JavaScript\\n;
   echo text = new Array(;
   while ($i  $number)
   {
  $text = mysql_result($result, $i, name);
  print \$text\;
 $i++;
 if ($i  $number)
   print ,;
 else
   print )\n;
}
   echo /Script\n;
 }

 }

 Note:  This is my quick *untested* fix done by modifying
 your code as little as possible.  I would suggest you use
 mysql_fetch_row and use your loop to parse it rather than
 making several calls to mysql_result (especially if you are
 ever going to have more than a few entries in the DB).

 Sheridan Saint-Michel
 Website Administrator
 FoxJet, an ITW Company
 www.foxjet.com

 - Original Message -
 From: Neil Freeman [EMAIL PROTECTED]
 To: PHP General [EMAIL PROTECTED]
 Sent: Wednesday, August 15, 2001 11:16 AM
 Subject: [PHP] Creating a javascript array from database data

  Hi there,
 
  Well after a few hours roaming around various websites I am at a loss.
  Here is what I am trying to do:
 
  1) Access a MySQL database which contains 1 table
  2) Read the records from this table
  3) Store the values returned from this table into javascript array
  elements, ie, if I get the values dog, cat and cow back I want
  these stored in an array as such:
  myArray[0] = valueReturned1
  myArray[1] = valueReturned2
  myArray[2] = valueReturned3
 
  You get the idea.
 
  Problem being that I cannot work out how to implement the javascript
  section of this. At the moment my php script writes the values returned
  from the database to screen but I require these to be stored in a
  javascript array. Please can someone help me before I go mad :)
 
  Here is my current .php script:
 
  --
  html
  head
  titleMenus test/title
  /head
  body bgcolor=white
 
  ?php
  $dbhost = 'localhost';
  $dbuser = 'guest';
  $dbpass = 'guest';
  $dbname = 'IFE';
  $dbtable = 'menus';
 
  file://-- DATABASE CONNECTION //
  mysql_connect($dbhost,$dbuser,$dbpass)
 or die (Unable to connect to database);
 
  mysql_select_db($dbname)
 or die (Unable to select database);
 
 
  $sql= SELECT * FROM $dbtable;
  $result = mysql_query($sql);
 
  $number = mysql_numrows($result);
 
  $i = 0;
 
  if ($number == 0)
 print Error - No records found;
  elseif ($number  0)
  {
 while ($i  $number)
 {
$text = mysql_result($result, $i, name);
print $text;
 
 $i++;
 }
  }
 
  mysql_free_result($result);
  mysql_close();
  ?
 
  /body
  /html
  --
 
  Thanks, Neil

--

 Email:  [EMAIL PROTECTED]
 [EMAIL PROTECTED]




-- 
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED

Re: [PHP] Creating a javascript array from database data (Off-List)

2001-08-15 Thread Sheridan Saint-Michel


Think in terms of output...  The JavaScript is literally going to
try and run the output of the PHP (It doesn't even know PHP
exists).  PHP is outputting the source code for the HTML and
JavaScript.  So what you are trying to do is get PHP to take
dog,cat,cow from MySQL and output:

Script Language=JavaScript
text = new Array(dog,cat,cow)
/Script

This block of code does exactly that, with the print\$text\;
filling in dog cat and cow.  I've added comments so you can
see how each line produces the above (I am using C-Style
comments as the // gets mucked up in e-mail)

The parts in [] are comments within comments to show flow
control and are not output.

else
{
/* Script Language=JavaScript */
  echo Script Language=\JavaScript\\n;
/* text=new Array( */
  echo text = new Array(;
  while ($i  $number)


$text = mysql_result($result, $i, name);
/* dog [first time] */
/* cat [second time] */
/* cow [third time] */
print \$text\;
$i++;
if ($i  $number)
/* , [first time] */
/* , [second time] */
   print ,;
else
/* ) [third time] */
  print )\n;
}
/* /Script */
  echo /Script\n;
}

So if you can follow that (just walk all the way through
the loop for each of the three passes) you can see that
no part of the code (including the print \$text\;) is
unnecessary

Hope this helps

Sheridan Saint-Michel
Website Administrator
FoxJet, an ITW Company
www.foxjet.com


- Original Message -
From: Neil Freeman [EMAIL PROTECTED]
To: Sheridan Saint-Michel [EMAIL PROTECTED]
Cc: php-general [EMAIL PROTECTED]
Sent: Wednesday, August 15, 2001 12:09 PM
Subject: Re: [PHP] Creating a javascript array from database data


 Thanks a lot for your help Sheridan,

 This now appears to work ok :) but if I remove the:

 print \$text\;

 line (as it is not required), I receive errors complaining about a syntax
 error (related to what?) and that 'text' is undefined. As this line simply
 outputs the value to screen why should this cause a problem if it is
removed?
 Or could it be some other problem?

 Neil

 PS: Here is my updated code:
Snip


-- 
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] Creating a javascript array from database data

Re: [PHP] Creating a javascript array from database data (Off-List)

Re: [PHP] Creating a javascript array from database data

[PHP] Creating a javascript array from database data

RE: [PHP] Creating a javascript array from database data

Re: [PHP] Creating a javascript array from database data

Re: [PHP] Creating a javascript array from database data

Re: [PHP] Creating a javascript array from database data (Off-List)

8 matches

Site Navigation

Mail list logo

Footer information