PHP MYSQL Display table column based on variable

Question

Using the following variables

I have the Following table named everyones_friends; id, john, jack, bill, dave 1, jack, john, dave, bill 2, bill, dave, john, jack 3 dave, bill, jack, john

doing the following query;

$sql = "SELECT * FROM everyones_friends";
$mydata = mysql_query($sql);
while($record = mysql_fetch_array($mydata));
$display_column['$variable'];

My question is how do I display the "bill" column using a previously user selected variable since I know the $display_column['$variable']; won't work. Based on their choice one of the columns above goes into a dropdown.

Possible duplicate of stackoverflow.com/questions/20019484/mysql-how-to-use-a-variable-as-column-name — Woody
– Woody, Commented Jan 6, 2016 at 19:28

Gouda Elalfy · Accepted Answer · 2016-01-06 19:28:38Z

1

update this line to be :

$display_column[$variable];

single quota in php doesn't parse php variables.

answered Jan 6, 2016 at 19:28

Gouda Elalfy

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