RE: [PHP-DB] How to..? From: Richard Allsebrook (richard.allsebrook <email protected>)
Date: 07/13/00

  $myArray=array("one","two","three");
  $Temp='"'.implode('";"',$a).'"';
  $myArray=explode(";",$d);

Any good?

BTW: this will only work for arrays of strings which do not contain the
semicolon ( ; ) character.

Pick another separator if this is a problem

  _____

~ Richard Allsebrook ~
Applications Developer and Webmaster
Easysoft Limited, No3 The Embankment, Leeds, LS1 4BJ
http://www.easysoft.com <http://www.easysoft.com>
The Home of Data Access Middleware
"A computer lets you make more mistakes faster than any invention in human
history-with the possible exceptions of handguns and tequila."

  _____

-----Original Message-----
From: Vojtech Dvorak [mailto:vojtech_dvorak <email protected>]
Sent: Thursday, July 13, 2000 2:15 PM
To: Wesley Choate
Cc: php-db <email protected>
Subject: Re: [PHP-DB] How to..?

Well my example is enough, try yourself

<?PHP
$customer="No One"
$customer='"'.$customer.'"' ;
echo $customer
?>
this will give you: "No One" and that is exactly what I need.
But it not answer my question.
How can I make this for an array of variables?

Vojtech

> You need to put the escape character in front of the quotes if you want
them
> in the string.
>
> Example
> $test = ' /" ' . $var . ' /" /;
>
> I think this should work.
>
> Wesley Choate
>
> > -----Original Message-----
> > From: Vojtech Dvorak [SMTP:vojtech_dvorak <email protected>]
> > Sent: Thursday, July 13, 2000 7:58 AM
> > To: php-db <email protected>
> > Subject: [PHP-DB] How to..?
> >
> > . << File: ATT00001.txt; charset = iso-8859-2 >>
>
>

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