Returning column names from a mySQL table using PHP

Question

Can't figure this out for the life of me. Trying to return the column names from the clients securities table, then return the result as an array. Can anybody point out where I'm getting off track?

mysql_select_db("HandlerProject", $con);        //Selects database
  $selectcols = "SELECT * FROM ".$clientname."securitiestable";     //selects all     columns from clients security table
  $tempcols = mysql_query($selectcols) or die(mysql_error());
  $returnedcols = $mysql_fetch_array($tempcols);
  $tempsymbol = mysql_query("SHOW COLUMNS FROM".$clientname."securitiestable");
  $symbol = $mysql_fetch_array($tempsymbol);

Throws a "Fatal error: Function name must be a string" for the $returnedcols and $symbol lines — JoeCortopassi
– JoeCortopassi, Commented Jan 17, 2010 at 1:40

martin clayton · Accepted Answer · 2011-02-09 19:33:13Z

5

Suggestions:

You've got $ signs prefixing the mysql_fetch_array() calls so you'd need to have assigned a value (function name you want to call) to $mysql_fetch_array (this is probably why you're seeing the error you mention in your comment).

Also you have a missing space after FROM in the second query

//                                          v
$tempsymbol = mysql_query("SHOW COLUMNS FROM ".$clientname."securitiestable");

Last thing to check - is $clientname set?

Having said that - take Bill Karwin's advice!

edited Feb 9, 2011 at 19:33

answered Jan 17, 2010 at 1:29

martin clayton

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Bill Karwin · Accepted Answer · 2010-01-17 02:19:34Z

3

I would use mysql_fetch_assoc() for the SELECT query, and then call array_keys() on any row of the result.

$selectcols = "SELECT * FROM ".$clientname."securitiestable";
$tempcols = mysql_query($selectcols) or die(mysql_error());
$returnedcols = mysql_fetch_assoc($tempcols);
$colnames = array_keys($returnedcols);

Your fatal error is because of a separate issue: you have a $ symbol at the start of your function call. This is legal PHP syntax, because you can put the name of a function in a variable and call it indirectly:

function foo($arg)
{
  echo $arg . "!\n";
}

$bar = "foo";

$bar("hello world");

But in your case, it's probably not what you intended. If you want to call a function by its literal name, don't put a $ in front of it. If you have a string variable that contains the name of a function, then you can use the variable as I show above.

edited Jan 17, 2010 at 2:19

answered Jan 17, 2010 at 1:35

Bill Karwin

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2 Comments

user849137 Over a year ago

I know this post is old, but I'm just curious; Your method above - Will that not select EVERY row in the DB? When the SHOW COLUMNS FROM would just grab the column names? Excuse my dumbness.

Bill Karwin Over a year ago

If you use mysql_unbuffered_query() you can fetch just the first row.

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