Posted by Captain Paralytic on 11/06/06 14:02

strawberry wrote:
> Hi,
>
> Here's my (latest) attempt at building MySQL statements from a
> multidimensional array. It doesn't quite work, of course, but I
> can't figure out why. I suspect there's lots of reasons. Any
> pointers would be helpful!
>
> function array_to_mysql($array) //1 define the function
> {
> if(is_array($array)){ //3 if $array is an array
> foreach($array as $key=>$value){ //4 foreach key/value pair
> within the array
> if(is_array($value)){ //5 either the current value is itself
> an array
> array_to_mysql($value); //6 (so call this function
> again to process that array)
> } else { //7 or it isn't
> if($key == 'name'){ //8 (in which case, if the key paired
> with that value = 'name'
> $table = $value; //9 then assign that value to $table)
> } elseif //10 or else
> ($key == 'ID'){ //11 (if the key paired with that value =
> 'ID'
> $parent = $value;} //12 then asign that value to $parent
> //13 and also issue the following statement
> echo "INSERT INTO $table ($key) VALUES ('$value') ON DUPLICATE KEY
> UPDATE <BR>\n";
> if(!is_null($parent)){ //Finally, if $parent is assigned issue
> the following
> echo "INSERT INTO $table (ID) VALUES ('$parent') ON DUPLICATE KEY
> UPDATE<BR>\n";
> }
> }
> }
> }
> }
>
> The $table variable is supposed to be assigned whenever the function
> encounters a key called 'name' (it should be case sensitive). It should
> retain that key's associated value until it encounters another key
> called 'name'. But it doesn't. :-(
>
> So here's some example output:
>
> INSERT INTO PROJECT (name) VALUES ('PROJECT') UPDATE ON DUPLICATE KEY
> INSERT INTO (COMPANY) VALUES ('Myself LLC') UPDATE ON DUPLICATE KEY
> INSERT INTO (WEBLINK) VALUES ('www.myselfllc.net') UPDATE ON DUPLICATE
> KEY
> INSERT INTO (VIEW-DATE) VALUES ('2005-12-26') UPDATE ON DUPLICATE KEY
>
> and this is how it should look:
>
> INSERT INTO PROJECT (COMPANY) VALUES ('Myself LLC') UPDATE ON DUPLICATE
> KEY
> INSERT INTO PROJECT (WEBLINK) VALUES ('www.myselfllc.net') UPDATE ON
> DUPLICATE KEY
> INSERT INTO PROJECT (VIEW-DATE) VALUES ('2005-12-26') UPDATE ON
> DUPLICATE KEY
>
> Note the first INSERT statement should not have happened. It should
> have realised that PROJECT was the 'name' of the table and assigned
> $table accordingly. $table did get assigned but the assignment was
> apparently lost by the time it reached COMPANY.
>
> TIA

One thing that ocurs to me is that the element called 'name' may not be
the first element in the array. So when you hit the first field/value
pair, there may not yet be a name.

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