Hi Jim,

echo cannot display the content of an array or object. try print_r() or
var_dump()

- Frank

> Ok,
> 
>  
> 
> I'm a PHP newbie and I'm having a problem.  All I want to do is parse
> some information out of a database and send someone and email stating
> that their posting has been approved.  
> 
>  
> 
> I've verified that my SQL is indeed returning the correct value for
> $key, but when I go to fetch the array for the row so that I can
extract
> the pertinent tidbits of information from that row, I get nada.  My
> troubleshooting echo statements give me the following out put.  I've
> included the actual snippet of code below.  
> 
>  
> 
> 54
> SELECT * FROM jobs WHERE pk_job = '54'
> Resource id #4
> Array
> 
> 0
> 
>  
> 
>  
> 
> //use the date to find the record we're looking for approved but not
yet
> posted
> 
> $findkey = "SELECT pk_job from jobs WHERE (display = 'Yes') AND
> (approval_stamp = '$matchnow') AND (posted IS NULL)";
> 
> list($getkey) = mysql_fetch_row(mysql_query ($findkey));
> 
> $key = ($getkey);
> 
> echo $key; //note this is returning the correct and expected value
> 
> echo "<br>"; 
> 
> // now that we have the key lets get the row
> 
> $query = "SELECT * FROM jobs WHERE pk_job = '$key'"; 
> 
> echo $query; //note this is apparently behaving correctly and gives me
a
> correct result when executed from the command line
> 
> echo "<br>";
> 
> $result = mysql_query($query);
> 
> echo $result;
> 
> echo "<br>";
> 
> //make array
> 
> $row = mysql_fetch_array ($result);
> 
> echo $row;
> 
>  
> 
> I'm pretty sure that this is a "DUH!" moment, but I'm not sure quite
> what it is that I've done.  I copied the code from another thing that I
> wrote a while back that's working fine and only changed SQL statements
> to reflect their new purpose in life.  
> 
>  
> 
> TIA,
> 
>  
> 
> Jimi
> 
>  
> 
>  
> 
> If computers get too powerful, we can organize them into a committee --
> that will do them in. -- Bradley's Bromide
> 
>  
> 
> 

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