[英]PHP: split string based on array
以下是我要解析的數據:
50‐59 1High300.00 Avg300.00
90‐99 11High222.00 Avg188.73
120‐1293High204.00 Avg169.33
第一部分是權重范圍,接下來是計數,然后是Highprice ,以Avgprice結尾。
例如,我需要將上面的數據解析成一個數組,看起來像
[0]50-59
[1]1
[2]High300.00
[3]Avg300.00
[0]90-99
[1]11
[2]High222.00
[3]Avg188.73
[0]120‐129
[1]3
[2]High204.00
[3]Avg169.33
我考慮過創建一個可能的權重范圍的數組,但我不知道如何使用數組的值來拆分字符串。
$arr = array("10-19","20-29","30-39","40-49","50-59","60-69","70-79","80-89","90-99","100-109","110-119","120-129","130-139","140-149","150-159","160-169","170-179","180-189","190-199","200-209","210-219","220-229","230-239","240-249","250-259","260-269","270-279","280-289","290-299","300-309");
任何想法將不勝感激。
希望這會奏效:
$string='50-59 1High300.00 Avg300.00
90-99 11High222.00 Avg188.73
120-129 3High204.00 Avg169.33';
$requiredData=array();
$dataArray=explode("\n",$string);
$counter=0;
foreach($dataArray as $data)
{
if(preg_match('#^([\d]+\-[\d]+) ([\d]+)([a-zA-Z]+[\d\.]+) ([a-zA-Z]+[\d\.]+)#', $data,$matches))
{
$requiredData[$counter][]=$matches[1];
$requiredData[$counter][]=$matches[2];
$requiredData[$counter][]=$matches[3];
$requiredData[$counter][]=$matches[4];
$counter++;
}
}
print_r($requiredData);
'#^([\d]+\-[\d]+) ([\d]+)([a-zA-Z]+[\d\.]+) ([a-zA-Z]+[\d\.]+)#'
我認為這不會起作用,因為您在正則表達式中的重量和計數之間有空間。 我正在努力解決的問題是像這樣沒有空間的一排。
120‐1293High204.00 Avg169.33
需要解析為[0]120‐129 [1]3 [2]High204.00 [3]Avg169.33
你說的對。 這可以通過將重量位數限制為三個並使空格可選來解決。
'#^(\d+-\d{1,3}) *…
$arr = array('50-59 1High300.00 Avg300.00',
'90-99 11High222.00 Avg188.73',
'120-129 3High204.00 Avg169.33');
foreach($arr as $str) {
if (preg_match('/^(\d+-\d{1,3})\s*(\d+)(High\d+\.\d\d) (Avg\d+\.\d\d)/i', $str, $m)) {
array_shift($m); //remove group 0 (ie. the whole match)
$result[] = $m;
}
}
print_r($result);
輸出:
Array
(
[0] => Array
(
[0] => 50-59
[1] => 1
[2] => High300.00
[3] => Avg300.00
)
[1] => Array
(
[0] => 90-99
[1] => 11
[2] => High222.00
[3] => Avg188.73
)
[2] => Array
(
[0] => 120-129
[1] => 3
[2] => High204.00
[3] => Avg169.33
)
)
解釋:
/ : regex delimiter
^ : begining of string
( : start group 1
\d+-\d{1,3} : 1 or more digits a dash and 1 upto 3 digits ie. weight range
) : end group 1
\s* : 0 or more space character
(\d+) : group 2 ie. count
(High\d+\.\d\d) : group 3 literal High followed by price
(Avg\d+\.\d\d) : Group 4 literal Avg followed by price
/i : regex delimiter and case Insensitive modifier.
為了更通用,您可以將High
和Avg
替換為[az]+
這是您可以信任的模式(模式演示):
/^((\d{0,2})0‐(?:\2)9) ?(\d{1,3})High(\d{1,3}\.\d{2}) ?Avg(\d{1,3}\.\d{2})/m
其他答案忽略了weight range
子字符串中的數字模式。 范圍起始整數始終以0
結尾,范圍結束整數始終以9
結尾; 范圍總是跨越十個整數。
我的模式將捕獲起始整數中0
之前的數字,並在破折號后立即引用它們,然后要求捕獲的數字后跟9
。
我想指出您的示例輸入有點棘手,因為您的‐
不是標准-
介於我鍵盤上的0
和=
之間。 這是我要解決的一個鬼鬼祟祟的小問題。
方法(演示):
$text = '50‐59 1High300.00 Avg300.00
90‐99 11High222.00Avg188.73
120‐1293High204.00 Avg169.33';
preg_match_all(
'/^((\d{0,2})0‐(?:\2)9) ?(\d{1,3})High(\d{1,3}\.\d{2}) ?Avg(\d{1,3}\.\d{2})/m',
$text,
$matches,
PREG_SET_ORDER
);
var_export(
array_map(
fn($captured) => [
'weight range' => $captured[1],
'count' => $captured[3],
'Highprice' => $captured[4],
'Avgprice' => $captured[5]
],
$matches
)
);
輸出:
array (
0 =>
array (
'weight range' => '50‐59',
'count' => '1',
'Highprice' => '300.00',
'Avgprice' => '300.00',
),
1 =>
array (
'weight range' => '50‐59',
'count' => '1',
'Highprice' => '300.00',
'Avgprice' => '300.00',
),
2 =>
array (
'weight range' => '50‐59',
'count' => '1',
'Highprice' => '300.00',
'Avgprice' => '300.00',
),
3 =>
array (
'weight range' => '50‐59',
'count' => '1',
'Highprice' => '300.00',
'Avgprice' => '300.00',
),
)
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