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Re: [PHP] Catch line indetation




El 28/10/16 a les 15:19, Curtis Maurand ha escrit:
> Sorry for the top post.  In the regex you can use ^\s*[*]  which will catch any white space characters.
> 
> On October 28, 2016 8:51:35 AM EDT, Richard <inbound-lists-php@xxxxxxxxxxxxxxxxxxxxx> wrote:
>>
>>
>>> Date: Friday, October 28, 2016 12:09:31 +0100
>>> From: Ashley Sheridan <ash@xxxxxxxxxxxxxxxxxxxx>
>>>
>>> On 28 October 2016 12:01:16 BST, Narcis Garcia
>>> <informatica@xxxxxxxxx> wrote:
>>>> Hello, I have a string (I quote here only) as:
>>>>
>>>> '   <table>...</table>'
>>>>
>>>> As you can see there are 3 spaces at the beginning, but it could
>>>> be 0 or
>>>> 4 or any number of spaces.
>>>> How can I get a string with only the initial spaces part?
>>>>
>>>> '   <table>...</table>' -> '   '
>>>> 'hello' -> ''
>>>> ' hello' -> ' '
>>>>
>>>> Thanks.
>>>
>>> Have you tried regular expressions? Something like:
>>>
>>> ^( )*[^ ]
>>>
>>> The first captured match is the number of spaces, from 0 to any
>>> amount. Not the space between the brackets and before the closing
>>> square bracket
>>>
>>
>> You need to take into consideration that "whitespace" can be created
>> by more than the simple "space" (ascii 32) character. A "[horizontal]
>> tab" (ascii 9) is common, but also look at the top of php trim
>> function documentation:
>>
>>  <http://php.net/manual/en/function.trim.php>
>>
>> to see the characters that it handles as "whitespace". While "trim"
>> does the opposite of what you're after, you could likely use the
>> ctype_space function to accomplish your need:
>>
>>  <http://php.net/manual/en/function.ctype-space.php>
>>  
>> e.g., by iterating through the string until it returns "false". Also,
>> "strpos" could be useful.
>>
>>
>>
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> 

preg_replace('^\s*[*]', '${1}', '   <table>...</table>');
Warning: preg_replace(): No ending delimiter '^' found


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