Question

我试图返回字符串的某个部分。我看过substr，但我不相信这是我正在寻找的。

使用此字符串：

/text-goes-here/more-text-here/even-more-text-here/possibly-more-here

如何返回前两个//之间的所有内容，即text-goes-here

谢谢，

Answer 1

$str="/text-goes-here/more-text-here/even-more-text-here/possibly-more-here";
$x=explode('/',$str);
echo $x[1]; 
print_r($x);// to see all the string split by /

Answer 2

<?php
$String = '/text-goes-here/more-text-here/even-more-text-here/possibly-more-here';

$SplitUrl = explode('/', $String);

# First element
echo $SplitUrl[1];  // text-goes-here

# You can also use array_shift but need twice
$Split = array_shift($SplitUrl);
$Split = array_shift($SplitUrl);

echo $Split;  // text-goes-here
?>

Answer 3

上面的爆炸方法肯定有用。在第二个元素上匹配的原因是PHP在数组中插入空白元素，无论何时开始或在没有任何其他内容的情况下运行到分隔符。另一种可能的解决方案是使用正则表达式：

<?php
$str="/text-goes-here/more-text-here/even-more-text-here/possibly-more-here";

preg_match('#/(?P<match>[^/]+)/#', $str, $matches);

echo $matches['match'];

（？P＆lt; match＆gt; ...部分告诉它与命名的捕获组匹配。如果省略？P＆lt; match＆gt;部分，你最终会得到$ matches中的匹配部分[1] ]。$ matches [0]将包含带有正斜杠的部分，如“/ text-goes-here /".

Answer 4

只需使用preg_match：

preg_match('@/([^/]+)/@', $string, $match);
$firstSegment = $match[1]; // "text-goes-here"

，其中

@    - start of regex (can be any caracter)
/    - a litteral /
(    - beginning of a capturing group
[^/] - anything that isn't a litteral /
+    - one or more (more than one litteral /)
)    - end of capturing group
/    - a litteral /
@    - end of regex (must match first character of the regex)

返回字符串的一部分

4 个答案: