Question

我有以下代码块实例化typeahead

$("#searchbox").typeahead({
    minLength: 3,
  }, {
    name: "Type0",
    displayKey: "name",
    source: bloodHound[0].ttAdapter(),
    templates: {
      header: "<h4 class='typeahead-header'><img src='Type0.png'>Type0</h4>",
      suggestion: Handlebars.compile(["{{name}}<br><small>{{address}}        </small>"].join(""))
    }
  }, {
    name: "Type1",
    displayKey: "name",
    source: bloodHound[1].ttAdapter(),
    templates: {
      header: "<h4 class='typeahead-header'><img src='Type1.png'>Type1</h4>",
      suggestion: Handlebars.compile(["{{name}}<br><small>{{address}}</small>"].join(""))
    }
  }
)

如果我想添加其他类型，例如Type2我必须复制并粘贴块，因此效率不高。我有一个名为types的数组中可用的类型。

如何使用一个循环数组的代码块并创建与上面相同的结果。所有建议都赞赏。

Answer 1

我不知道你的char sentence[100] = ""; cin.getline(sentence, 100); char letters[100] = ""; int counter = 0; for (int i = 0; i < strlen(sentence); i++) { cout << sentence[i] << endl; letters[i] = sentence[i]; } for (int i = 0; i < strlen(sentence); i++) { for (int j = 0; j < strlen(sentence); j++) { if (letters[j]==sentence[i]) { counter++; cout << letters[j] << " occurs " << counter << " in the sentence" << endl; } } }数组应该是什么样的，但这是我如何设置的：

types

如何重构js预先实例化

1 个答案: