返回指令在javascript中停止我的for循環

Question

當我使用for循環時，沒有任何問題

for (var h = 0; h < xt.length; h++) {
  var heart = '';
  console.log(xt[h]);
  if (row[0] == xt[h]) {
    heart = 'icon-heart';
  } else {
    heart = 'icon-    heart1';
  }
}

但是當我使用返回指令時，循環停止：

for (var h = 0; h < xt.length; h++) {
  var heart = '';
  console.log(xt[h]);
  if (row[0] == xt[h]) {
    heart = 'icon-heart';
  } else {
    heart = 'icon-heart1';
  }
  return '<i class="' + heart + '" onClick="addFAv(this)"  data-id="' + row[0] + '" data-name="' + row[1] + '"></i> <a href="#" class="icon-line-clipboard i-circled i-small"></a> <a href="#" class="icon-file-alt2 i-circled i-small"></a>';
}

你知道為什么嗎 ？

函數中的整個代碼是

$('#recevabilite').dataTable({
      "scrollY": "200px",
      "scrollCollapse": true,
      "paging": false,
      stateSave: true,
      "dom": '<"#recev.top">irt',
      "columnDefs": [{
            "visible": false,
            "targets": [3]
          },
          {
            "render": function(data, type, row) {
                //    var date1 = new Date()+10J;
                //affichage des icones  
                var x = localStorage.getItem("listeFavoris");
                xt = x.split(',');

                for (var h = 0; h < xt.length; h++) {
                  var heart = '';
                  console.log(xt[h]);
                  if (row[0] == xt[h]) {
                    heart = 'icon-heart';
                  } else {
                    heart = 'icon-heart1';
                  }
                  return '<i class="' + heart + '" onClick="addFAv(this)"  data-id="' + row[0] + '" data-name="' + row[1] + '"></i> <a href="#" class="icon-line-clipboard i-circled i-small"></a> <a href="#" class="icon-file-alt2 i-circled i-small"></a>';
                }

謝謝

Answer 1

當您執行return ，它會結束該函數，因此您不會執行循環的任何進一步迭代。

您需要連接每個循環迭代的值，然后在循環后返回最終結果：

            "render": function(data, type, row) {
                //    var date1 = new Date()+10J;
                //affichage des icones  
                var x = localStorage.getItem("listeFavoris");
                var xt = x.split(',');
                var html = '';
                for (var h = 0; h < xt.length; h++) {
                  var heart = '';
                  console.log(xt[h]);
                  if (row[0] == xt[h]) {
                    heart = 'icon-heart';
                  } else {
                    heart = 'icon-heart1';
                  }
                  html += '<i class="' + heart + '" onClick="addFAv(this)"  data-id="' + row[0] + '" data-name="' + row[1] + '"></i> <a href="#" class="icon-line-clipboard i-circled i-small"></a> <a href="#" class="icon-file-alt2 i-circled i-small"></a>';
                }
              return html;
            }

Answer 2

我找到了解決方案

  {"render": function(data, type, row) {
            //    var date1 = new Date()+10J;
            //affichage des icones  
            var x = localStorage.getItem("listeFavoris");
            var xt = x.split(',');
            var html = '<a href="#" class="icon-line-clipboard i-circled i-small"></a> <a href="#" class="icon-file-alt2 i-circled i-small"></a>';
            for (var h = 0; h < xt.length; h++) {
              var heart = '';
              console.log(xt[h]);
              if (row[0] == xt[h]) {
                heart = 'icon-heart';
                  h = xt.length; }
              html += '<i class="' + heart + '" onClick="addFAv(this)"  data-id="' + row[0] + '" data-name="' + row[1] + '"></i>';
            }
        if   (heart == '') {
        html += '<i class="icon-heart1" onClick="addFAv(this)"  data-id="' + row[0] + '" data-name="' + row[1] + '"></i>';}
          return html;

        }, "targets": 4}