My database contains records that have:
1) First line delimited with double \\n\\n and the rest delimited with \\n:
Introduction
Line 1
Line 2
2) Everything delimited with \\n:
Line 1
Line 2
Line 3
3) Nothing delimited:
Introduction
My goal is to add ">>" before each line that's delimited with a single \\n. The spaghetti dinner that I cooked (and it works) looks as follows:
list($intro, $details) = split("\n\n", $dbInput);
if ($details) {
$temp = split("\n", $details);
array_walk($temp, create_function('&$v,$k', '$v = ">> $v";'));
$dbOutput = "$intro \n\n ".join("\n", $temp);
} else {
$temp = split("\n", $intro);
if (count($temp) > 1) {
array_walk($temp, create_function('&$v,$k', '$v = ">> $v";'));
$dbOutput = join("\n", $temp);
} else $dbOutput = $temp[0];
}
An example of what I want to achieve:
Introduction
>> Line 1
>> Line 2
>> Line 3
or
>> Line 1
>> Line 2
>> Line 3
or
Introduction
QUESTIONS: How would I optimize this code to combine array_walk with split and join in the same statement somehow along the following lines:
$a = join("\n" , array_walk(split("\n", $a), create_function(('&$v,$k', '$v = ">> $v";')) ) );
My personal method would be to use preg_prelace
to replace lines terminating in \\n
(and not \\n\\n
with >>
at the front.
My guess is you don't want the
Introduction
To have the >>
in front?
In that case:
$pattern = '/([^\n]+)(?=\n(?!\n))|(?<=(?<!\n)\n)([^\n]+)/';
$replacement = '>>$1$2';
$subject = 'Introduction
Line 1
Line 2
Line 3';
echo preg_replace ($pattern, $replacement, $subject);
/* echoes:
Introduction
>>Line 1
>>Line 2
>>Line 3
*/
Explanation of regex:
([^\n]+)(?=\n(?!\n)) # match a line followed by \n but not a \n\n
| # OR
(?<=(?<!\n)\n)([^\n]+) # match a line preceded by \n but not a \n\n
The first bit captures Line1
and Line2
, but not Line3
, as this is terminated by end-of-string and not a \\n
. The second bit of regex captures Line3
.
They also both exclude a single-line string like Introduction
.
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