我试图按顺序回显这些变量
$caption = "Some random caption";
$url = " https://example.com";
$tags = " tag1, tag2, tag3, tag4 ";
$message = $caption.$url.$tags;
并且输出不正常。就是这样
tag1, tag2, tag3, tag4 Some random caption https://example.com
实际代码如下:
``
function hashtag($string){
$newString = str_replace(' ', "", $string);
$nString = str_replace(',', " #", $newString);
echo $nString;
}
$name = "Full Name";
$hashName = " #".str_replace(' ', '', $name);
$string = ', tag1, tag2, tag3, tag4, tag5, tag6, tag7';
$tags = hashtag($string);
$url = ' http://example.com';
$caption = ' some random caption';
$message = $caption.$url.$tags;
echo $message;
您echo
您在函数中的主题标签,因此在您调用它时会显示该字符串。您的函数不返回任何内容,因此$tags
的值为NULL,并且标记不会出现在第二个echo
中。
在函数中使用return
语句解决您的问题:
function hashtag($string){
$newString = str_replace(' ', "", $string);
$nString = str_replace(',', " #", $newString);
return $nString;
}