Re: using PHP script to add address'es of all picture files in a dir to MYSQL DB
| From: | Robin Chen | Date: | Wed, 07 Nov 2001 10:49:44 +0000 |
| Subject: | Re: using PHP script to add address'es of all picture files in a dir to MYSQL DB | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-73716@lists.php.net to get a copy of this message | ||
If you are in an unix shell, you can use the find ./ -name '*.jpg' and
pipe it to awk '{ print "mysql -e 'insert into table (path_to_pict)
values ('" $0 "')" }' > execute.sh
Chown u+x the execute.sh and run it. I don't have the proper awk syntax
because of the ' and " but if you figure it out, it should be pretty
simple.
Robin
Daniel wrote:
>
> Hi,
>
> I want to make a simple database driven photo gallery for my site where
> users can search the picture descriptions to find the pictures they want,
> all fairly simple PHP/MySQL stuff which I know. My big problem is that I
> have like 500+ images I want to put into the MySQL DB and I am way too lazy
> to consider doing it the hard way. I was wondering if it would be possible
> to write a PHP script which would get a list of all the .jpg files in its
> current directory and then insert a reference for each file (basically just
> its location on the web server) into the MySQL PHOTO table? I was thinking
> about this on the train yesterday so heres the pseudo PHP/C code of the sort
> of script I was thinking about(I don't know the PHP file functions):
>
> //get the number of jpeg files in same dir as this script (I don't know code
> for this)
> $NumFiles = GetNumberOfJPGFilesInDirectory();
> While ($NumFiles > 0)
> {
> //get file location of current jpeg file (I don't know code for this)
> $FileName = GetFileNameAndLocation($NumFiles);
> // construct MySQL query
> $sql = "INSERT INTO PHOTO SET " . "FILE='$FileName ";
> //add file location of current jpeg and do error handling (easy stuff)
> if (mysql_query($sql))
> {
> echo("Photo " . $FileName ." has been added.<br>");
> }else
> {
> echo("Error adding photo: " . mysql_error() . " <br>");
> }
> $NumFiles = $NumFiles-1;
> }
>
> Would anyone be able to help me get the code for this script right? I can
> do basic PHP & MySQL stuff but I am not familiar with PHP file functions so
> I need someone to help me get it working...
>
> Cheers,
>
> Daniel
>
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