Question

我有一个带有此输入的表单：

<input id="cant_user" type="text" name="cant_user" value="1" onblur="addInputs()">

添加大于1的输入值时会克隆

#userData，我想将输入名称从user_age更改为age[n]

<div id="userData">
    <div class="panel-body" id="generated[0]">
        <div class="form-group">
            <label class="col-md-4 control-label">Age User [n]</label>
        <div class="col-md-6">
            <input type="number" class="form-control" name="user_age[]" value="0" required>
        </div>
    </div>
</div>

Javascript：

function addInputs() {
  var cantUsers = document.getElementById("cant_user").value;
  var countGenerated = $('[id^=generated]').length;
  if (countGenerated > 0) {
    $('[id^=generated]').not(':first').remove();
  }
  if (cantUsers != 1) {
    if (cantUsers > 1 && cantUsers < 10) {
      for (var i = 1; i < cantUsers; i++) {
        jQuery("#userData").append(jQuery("#userData").children().first().clone());
        jQuery("#userData").children().last().attr("id", "generated[" + i + "]");

        for (var n = 0; n <= i; m++) {
          var u = document.getElementsByName("user_age[]");
          $(u).removeAttr("name").attr({
            id: "generatedAge[" + n + "]",
            name: "age[" + n + "]"
          });
        }
      };
    }
  };
}

如何使每个标签和输入名称显示数组中的值，如：

<label>Age User 1<label>
<input name="age[0]">
<label>Age User 2<label>
<input name="age[1]">

Answer 1

我更改了一些代码，如果我理解你的想法正确可能会有所帮助

for ( var i = 1; i < cantUsers; i++) {
var userData = jQuery("#userData");
var childOfUserData = userData.children();
userData.append( childOfUserData.first().clone() ); 

childOfUserData = userData.children(); //because we just appen new data
childOfUserData.last().attr("id","generated[" + i + "]");
for ( var n = 0; n <= i; n++ ) {
    var child =  childOfUserData.get(i);
    var userAge = $("input", $(child));//because you only have one input
    userAge.attr("name","age[" + n + "]").attr("id","generatedAge[" + n + "]");
}

Answer 2

<input name="age[0]">

在上面的代码name中，属性值充当文本而不是数组值。

要在attribute中获取数组值，您需要更改JavaScript代码

$("#userData").children().last().attr("id", generated[i]);

$(u).removeAttr("name").attr({ id: generatedAge[n], name: age[n] });

现在，如果您检查属性值，您将收到数据。

希望这会对你有所帮助。

使用数组更改attrib名称

2 个答案: