我刚刚报名参加树屋训练,我还有这个小挑战要完成。我无法让它发挥作用。如果变量设置为"vanilla",我将获得成功的输出。然而,如果我使用"饼干面团",那么我会出现语法错误,我想这可能是因为饼干和面团之间的空格?
<?php
$flavor = cookie dough;
echo "<p> Your favourite flavor is ";
echo $flavor;
echo ".</P>";
if ($flavor == "cookie dough") {
echo "<p> Yeah I like cookie dough aswell! </p>";
}
?>
这里您忘记了""
$flavor = cookie dough;
应该是
$flavor = "cookie dough";
^ ^
用引号将$flavor
变量值括起来。
<?php
$flavor = "cookie dough";
echo "<p> Your favourite flavor is ";
echo $flavor;
echo ".</P>";
if ($flavor == "cookie dough") {
echo "<p> Yeah I like cookie dough aswell! </p>";
}
?>
简单地说,
$flavor = cookie dough;
不是正确的代码,请用引号括起来。您可以在这里看到定义字符串的可能方法。
$flavor = cookie dough;
应该是
$flavor = 'cookie dough';
或
$flavor = "cookie dough";
要定义字符串,应该使用"'"
或'"'
;
您忘记将cookie dough
括在引号中:
$flavor = "cookie dough";