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解決したコードの追記
test
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test
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@@ -184,198 +184,20 @@
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//不定休日の値を取得する処理
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$iholiday = [];
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$sql = "SELECT Iholiday FROM table3 WHERE Cusid=?";
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$sth = $dbh -> prepare($sql);
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$array = array($_SESSION['cusid']);
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$sth -> execute($array);
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$result = $sth->fetchall(PDO::FETCH_ASSOC);
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foreach($result as $row){
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$iholidays = new DateTime($row['Iholiday']);
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$iholiday[] = $iholidays -> format('Y/n/j');
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}
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function make_date_list($year,$month){
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//毎週の定休日
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global $weeks;
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$regular_holiday = explode(',', $weeks);
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//祝日
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$public_holidays = get_public_holidays();
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//不定休
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global $iholiday;
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$target = new \DateTime("{$year}-{$month}-01");
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$date = [];
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for($i = 0; $i < intval($target->format('t')); $i++){
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if($i > 0) $target->modify('+1day');
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//祝日
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if(isset($public_holidays[$target->format('Y/n/j')])){
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$holiday = true;
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//定休日
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}else if(in_array($target->format('w'),(array)$regular_holiday)){
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$holiday = true;
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//不定休
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}else if(in_array($target->format('Y/n/j'),(array)$iholiday)){
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$holiday = true;
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//今日
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}
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else{
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$holiday = false;
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}
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$dates[$i] = [
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'day' => $target->format('j'),
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'week' => $target->format('w'),
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'holiday' => $holiday,
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];
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}
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/*カレンダーの空欄箇所*/
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$first_date = current($dates);
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$end_date = end($dates);
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$damy_num = intval($first_date['week']);
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if($damy_num > 0){
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$damy = [];
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for($i = $damy_num - 1;0 <= $i;$i--){
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array_unshift($dates,['week'=>$i,'day'=>' ']);
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}
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}
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$damy_num = intval($end_date['week']);
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if($damy_num < 6){
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$damy = [];
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for($i = $damy_num; $i < 6;$i++){
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array_push($dates,['week'=>$i,'day'=>' ']);
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}
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}
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return $dates;
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}
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if(isset($_GET['y']) and $_GET['y']){
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$year = $_GET['y'];
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}else{
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$year = date('Y');
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}
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if(isset($_GET['m']) and $_GET['m']){
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$month = $_GET['m'];
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}else{
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$month = date('m');
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}
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$dates = make_date_list($year,$month);
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echo json_encode($dates);
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?>
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以下の年を跨ぐことを考慮したコードを追加して解決できました。
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188
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+
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189
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+
ありがとうございました。
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190
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+
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+
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```javascript
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if(nowDate.getFullYear() < year ){
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c += '<a href="" onClick="prev();return false;" class="cal_prev"></a>';
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c += '<p>'+year+'年'+month+'月</p>';
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}
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380
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381
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```
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