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>
> <title>Menus 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>
> > > <title>Menus 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]
> --------------------------------
>
>
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