[英]PHP MYSQL Display table column based on variable
Using the following variables 使用以下变量
I have the Following table named everyones_friends; 我有一个名为所有人的下表: id, john, jack, bill, dave 1, jack, john, dave, bill 2, bill, dave, john, jack 3 dave, bill, jack, john
id,约翰,杰克,条例草案,戴夫1,杰克,约翰,大卫,条例草案2,条例草案,戴夫,约翰,杰克3戴夫,条例草案,杰克,约翰
doing the following query; 进行以下查询;
$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']; 我的问题是,由于我知道$ display_column ['$ variable'],因此如何使用以前用户选择的变量显示“帐单”列; won't work.
将无法正常工作。 Based on their choice one of the columns above goes into a dropdown.
根据他们的选择,上面的列之一进入下拉列表。
update this line to be : 将此行更新为:
$display_column[$variable];
single quota in php doesn't parse php variables. php中的单个配额不会解析php变量。
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