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Posted by Richard Heinz on 05/15/06 22:27
Never mind I had the path messed up and now the images display.
Dick
"Richard Heinz" <richardkheinz@bellsouth.net> wrote in message
news:I64ag.48305$MM6.5815@bignews3.bellsouth.net...
>I inserted the code as you stated and I know I can access the image file
>becuase if I change the name the get imagesize reports an error message.
>The problem I have is it does not display the image but it displays the alt
>info.
> So what is wrong? It displays a a square with an 'x' and reserves room
> for the image.
>
> Dick
>
> "Martin Jay" <martin@spam-free.org.uk> wrote in message
> news:mmnFdhEiJKaEFw9z@spam-free.org.uk...
>> In message <CP%9g.48253$MM6.35416@bignews3.bellsouth.net>, Richard Heinz
>> <richardkheinz@bellsouth.net> writes
>>>I have a directory of pictures of individuals and a database with the
>>>individual information. One filed or column in the database has the
>>>location (directory/filename) of picture. I display the personal
>>>information form the database in a table amd now want to display the
>>>corresponding picture with the individual information in the table.
>>>
>>>This is the page with the indiviual info without the pictures.
>>>
>>>http://sadcma.com/Missions/DistrictMissionary.php
>>
>> Try inserting the following where you want the picture to be display:
>>
>>
>> <?php
>> $image = ; // image location and filename
>> $name = ; // name of person
>>
>> $size=getimagesize($image); // set the dimensions of image
>> ?>
>>
>> <img src="<?php echo $image; ?>" alt="<?php echo $name;?>" <?php echo
>> $size[3]; ?>">
>>
>>
>> $image is the directory/filename from your database, and $name is the
>> name of the person.
>>
>> And finally $size=getimagesize($image); returns an array of information
>> about the image. $size[3]; should contain it's dimensions in a format
>> such as:
>>
>> width="" height=""
>>
>> --
>> Martin Jay
>
>
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