PHP匹配次数

Question

I have this PHP string :

$range = "['2018-03-17','2018-03-19','2018-03-18','2018-03-18']";

I'm wanting to loop through the entire string, and do something like..

if day == "Saturday" || day == "Sunday" and only one occurrence of this in string then { // Do something } else { // Do something else }

I've explored preg_match_all with a regex [0-9]{4}-[0-9]{2}-[0-9]{2} but I can't get my head around it, any help would be really appreciated.

Answer 1

Try this code.

 if (day == "Saturday" || day == "Sunday" && (substr_count($range,"2018-03-18")) == 1 ){ 
// Do something 
} else { 
// Do something else 
}

substr_count($range,"2018-03-18") will return 2 here. substr_count return the number of occurrence of a sub string in main string.

Hope it helps.

Thanks.

Answer 2

First, you need to decode your string to array:

$range = json_decode('["2018-03-17","2018-03-19","2018-03-18","2018-03-18"]');

And then loop:

$rangeCount = array_count_values($range);
forearch ($range as $d) {
    $date = new DateTime($d);
    $weekDay = (int) $date->format('N');
    $found = $rangeCount[$d];

    if ($weekDay === 6 || $weekDay === 7 && $found === 1) {
        // here is your code
    }
}

I haven't tested the code.

Answer 3

Try like this

    <?php
    $range_string = '["2018-03-17","2018-03-19","2018-03-18","2018-03-18"]';
    $range = json_decode($range_string);

    foreach($range as $date){
     $nameOfDay = date('l', strtotime($date));
     if (($nameOfDay == 'Saturday' || $nameOfDay =='Sunday') && count(array_keys($range, $date)) == 1) 
     {
        echo "$nameOfDay exits only once on $date"."<br/>";
     }else{
        echo "$nameOfDay exits multiple time on $date"."<br/>";
     }
}
    ?>

Output

Saturday exits only once on 2018-03-17
Monday exits multiple time on 2018-03-19
Sunday exits multiple time on 2018-03-18
Sunday exits multiple time on 2018-03-18

Demo : https://eval.in/970622