Question

我试图从Pokemon API中获取名称并将其放入html中的div字段，但jQuery中的.text（）函数似乎并没有将文本放入html中把它放在.getJSON函数中。只是好奇这个问题会是什么？感谢。

<div class="container">
    <div class="row">
        <div class="col-sm-3" id="1">
            <div id="name1"></div>
        </div>
        <div class="col-sm-3" id="2">
            <div id="name2"></div>
        </div>
        <div class="col-sm-3" id="3">
            <div id="name3"></div>
        </div>
        <div class="col-sm-3" id="4">
            <div id="name4"></div>
        </div>
     </div>
  </div>

Javascript代码

$(document).ready(function() {
/*Works*/
for(var j = 1; j < 5; j++){
    $("#name" + j).text("HELLO");
}
/*Doesn't work*/
for(var j = 1; j < 5; j++){
    var webAddress2 = "http://pokeapi.co/api/v1/pokemon/" + j;
    $.getJSON(webAddress2, function(data) {
        console.log("test");
        $("#name" + j).text("SOME TEXT");
    });
}
});

Answer 1

由于getJSON在成功时是异步的，因此执行j的值将是5。没有标识为name5的元素，因此不会发生任何事情。要使其有效，请使用 closure 功能。

$(document).ready(function() {
  /*Works*/
  for (var j = 1; j < 5; j++) {
    $("#name" + j).text("HELLO");
  }
  /*Doesn't work*/
  for (var j = 1; j < 5; j++) {
    var webAddress2 = "http://pokeapi.co/api/v1/pokemon/" + j;
    (function(j) {
      $.getJSON(webAddress2, function(data) {
        console.log("test");
        $("#name" + j).text("SOME TEXT");
      });
    })(j);
  }
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="container">
  <div class="row">
    <div class="col-sm-3" id="1">
      <div id="name1"></div>
    </div>
    <div class="col-sm-3" id="2">
      <div id="name2"></div>
    </div>
    <div class="col-sm-3" id="3">
      <div id="name3"></div>
    </div>
    <div class="col-sm-3" id="4">
      <div id="name4"></div>
    </div>
  </div>
</div>

或使用 let 作为块级范围变量。

$(document).ready(function() {
  /*Works*/
  for (var j = 1; j < 5; j++) {
    $("#name" + j).text("HELLO");
  }
  /*Doesn't work*/
  for (let j = 1; j < 5; j++) {
    var webAddress2 = "http://pokeapi.co/api/v1/pokemon/" + j;
    $.getJSON(webAddress2, function(data) {
      console.log("test");
      $("#name" + j).text("SOME TEXT");
    });
  }
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="container">
  <div class="row">
    <div class="col-sm-3" id="1">
      <div id="name1"></div>
    </div>
    <div class="col-sm-3" id="2">
      <div id="name2"></div>
    </div>
    <div class="col-sm-3" id="3">
      <div id="name3"></div>
    </div>
    <div class="col-sm-3" id="4">
      <div id="name4"></div>
    </div>
  </div>
</div>

jQuery text（）里面.getJSON（）

1 个答案: