donglianglu8136 2016-01-06 19:21
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PHP MYSQL显示基于变量的表列

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.

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  • drg17404 2016-01-06 19:28
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    update this line to be :

    $display_column[$variable];
    

    single quota in php doesn't parse php variables.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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