This is a discussion on Re: putting variables in a variable within the PHP General forums, part of the PHP Programming Forums category; Hulf wrote: > Hi, > > I am making and HTML email. I have 3 images to put in. Currently ...
|
|||||||
| FAQ | Members List | Calendar | Search | Today's Posts | Mark Forums Read |
|
|||
|
Hulf wrote:
> Hi, > > I am making and HTML email. I have 3 images to put in. Currently I have > > $body .=" > <table> > <tr> > <td><img src=\"image1.jpg\"></td> > </tr> > > <tr> > <td></td> > </tr> > </table> > "; > > > ideally I would like to have > > $myimage1 = "image1.jpg"; > $myimage2 = "image2.jpg"; > $myimage3 = "image3.jpg"; > > > and put them into the HTML body variable. I have tried escaping them in > every way i can think of, dots and slashes and the rest. Any ideas? > > > Ross My first idea is to ask you what you WANT exactly and what's wrong with what you have. Currently you're showing 1 string, and 3 variables. Nothing else... what do you expect to happen that is not hapenning? Or what do you expect not to happen that IS hapenning ? |
|
|||
|
At 4:47 PM +0100 3/28/08, M. Sokolewicz wrote:
>Hulf wrote: >>Hi, >> >>I am making and HTML email. I have 3 images to put in. Currently I have >> >>$body .=" >><table> >> <tr> >> <td><img src=\"image1.jpg\"></td> >> </tr> >> >> <tr> >> <td></td> >> </tr> >></table> >>"; >> >> >>ideally I would like to have >> >>$myimage1 = "image1.jpg"; >>$myimage2 = "image2.jpg"; >>$myimage3 = "image3.jpg"; >> >> >>and put them into the HTML body variable. I have tried escaping >>them in every way i can think of, dots and slashes and the rest. >>Any ideas? >> >> >>Ross >My first idea is to ask you what you WANT exactly and what's wrong >with what you have. Currently you're showing 1 string, and 3 >variables. Nothing else... what do you expect to happen that is not >hapenning? Or what do you expect not to happen that IS hapenning ? I think it's fuzzy thinking. He probably wants: $body .=" <table> <tr> <td><img src=\"$image.jpg\"></td> </tr> <tr> <td></td> </tr> </table> "; Where $image.jpg is "image1.jpg" or "image2.jpg" or "image3.jpg" Cheers, tedd -- ------- http://sperling.com http://ancientstones.com http://earthstones.com |