safari regExp error

lize·2022년 4월 21일
0

safari에서 에러 발생한 코드

for(let i=0; i < roomObj.length; i++) {
        $("table#ROOM_CODE tbody").append(
          `<tr class="tr_${i+1}">
            <th>${roomObj[i].TYPE_NAME}</th>
            <td>${roomObj[i].TYPE_NM}</td>

            <td>${roomObj[i].ADLT_BASE_PERS}명</td>
            <td>${roomObj[i].ADLT_MAX_PERS}명</td>
  
            <td>${roomObj[i].P01_WEEK_PRCE.replace(/\B(?<!\.\d*)(?=(\d{3})+(?!\d))/g, ",")}원</td>
            <td>${roomObj[i].P01_FRD_PRCE.replace(/\B(?<!\.\d*)(?=(\d{3})+(?!\d))/g, ",")}원</td>
            <td>${roomObj[i].P01_SAT_PRCE.replace(/\B(?<!\.\d*)(?=(\d{3})+(?!\d))/g, ",")}원</td>
          </tr>`
        );
      }

해결

for(let i=0; i < roomObj.length; i++) {
        $("table#ROOM_CODE tbody").append(
          `<tr class="tr_${i+1}">
            <th>${roomObj[i].TYPE_NAME}</th>
            <td>${roomObj[i].TYPE_NM}</td>
            <td>${roomObj[i].ADLT_BASE_PERS}명</td>
            <td>${roomObj[i].ADLT_MAX_PERS}명</td>
            <td>${Number(roomObj[i].P01_WEEK_PRCE).toLocaleString('en')}원</td>
            <td>${Number(roomObj[i].P01_FRD_PRCE).toLocaleString('en')}원</td>
            <td>${Number(roomObj[i].P01_SAT_PRCE).toLocaleString('en')}원</td>
          </tr>`
        );
      }

도움받은 링크: https://www.codeproject.com/Questions/5274806/Regex-works-in-chrome-but-breaks-in-safari-invalid
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