如何 - 显示 属性 上的 If 语句
How to - If statement on Display Property
这会显示公司徽标
{display property='Logo' object_sid=$user_sid}
但是,如果不显示某些内容,我怎么知道他们是否有徽标显示它。
感谢您的帮助。
if-else
语句首先检查给定条件,如果满足则移至第一部分或 else
移至另一部分。
一个例子:
<?php
$display_property = "Logo";
if(isset($display_property)){
echo $display_property;
}else
{
echo "No display";
}
?>
Note: Given with the Information in your question, this example was
created. However this might not do the EXACT thing but would play a
vital role in understanding what you're struggling with.
这会显示公司徽标
{display property='Logo' object_sid=$user_sid}
但是,如果不显示某些内容,我怎么知道他们是否有徽标显示它。
感谢您的帮助。
if-else
语句首先检查给定条件,如果满足则移至第一部分或 else
移至另一部分。
一个例子:
<?php
$display_property = "Logo";
if(isset($display_property)){
echo $display_property;
}else
{
echo "No display";
}
?>
Note: Given with the Information in your question, this example was created. However this might not do the EXACT thing but would play a vital role in understanding what you're struggling with.