bob*_*bob 0 php scope function
我有一个简单的Vars.php页面:
<?php
//Vars.php
$WeekOfDateSelected = date('l, m/d/Y', strtotime($MonthYear));
$NextSundayOfDateSelected = date('l, m/d/Y', strtotime('this Sunday', strtotime($WeekOfDateSelected)));
?>
我有另一个包含Vars.php的PHP并构建一个表:
<html>
<?php
//AnalyticsTest.php
include($_SERVER['DOCUMENT_ROOT']."/~/~/~/~/~/Vars.php");
function WeekTable() {
echo "<table id=\"a\">
<tr>
<th style=\"text-align: center;\"><a href=\"#\">< previous week</a></th>
<th colspan=\"4\" style=\"text-align: center;\"><h2>Week of ";
echo $WeekOfDateSelected;
echo " - ";
echo $NextSundayOfDateSelected;
echo "</th>
<th style=\"text-align: center;\"><a href=\"#\">next week ></a></th>
</tr>
</table>";
}
?>
</html>
基本上,当我WeekTable()正确调用所有输出时,除了PHP变量$WeekOfDateSelected和$NextSundayOfDateSelected输出空白.
您需要将这些变量作为参数传递给该函数.否则他们超出了范围.
function WeekTable($WeekOfDateSelected,$NextSundayOfDateSelected) {
// ...
}
确保在调用函数时传递它们:
WeekTable($WeekOfDateSelected,$NextSundayOfDateSelected);
你也可以使用global关键字,但这是一个糟糕的编程习惯,所以我不会在这里显示.