Question

使用Jquery方法生成的按钮不起作用。生成的按钮的ID是gpaCalculate。我尝试了点击和方法，但它不起作用。我不知道哪里出错了。以下是我整个代码的链接：https://jsfiddle.net/4rxq49ko/

 $(document).on("click","#gpaCalculate",function(){
    displayDetails();
})

生成按钮JQuery

$("#DropDownList1").change(function () {
      myval = parseInt($(this).val());
      $(".tableAppend").empty();

      $(".tableAppend").append('<tr><th></th><th>Courses</th><th>Credit Hours</th><th>Grade</th></tr>');
      $(".tableAppend").append('<tr><input type="submit" name="calculateGPA" value="Click Me" id="calculateGPA" /></tr>');
      for (var i = 1 ; i <= myval ; i++) {
          $(".tableAppend").append('<tr><td>' + (i) + '.</td><td><input name="courseTextBox' + i + '" type="text" id="courseTextBox' + i + '"></td>' +
              '<td><input name="creditTextBox' + i + '" type="text" id="creditTextBox' + i + '">  </td>' +
              '<td><select id="gradeSelect' + i + '">' +
                                   '<option value ="A">A</option>' +
                                   '<option value ="A-">A-</option>' +
                                   '<option value ="B+">B+</option>' +
                                   '<option value ="B">B</option>' +
                                   '<option value ="B-">B-</option>' +
                                   '<option value ="C+">C+</option>' +
                                   '<option value ="C">C</option>' +
                                   '<option value ="C-">C-</option>' +
                                   '<option value ="D">D</option>' +
                                   '<option value ="F">F</option>' +
                 '</select></td>');

      }
      $(".tableAppend").append('<tr><td></td><td><button type="submit" id="gpaCalculate" >Calculate GPA</button></td>');
  });

Answer 1

试，

$("#gpaCalculate").onclick(function () {
    displayDetails();
})

Answer 2

试试这个：

$("#DropDownList1").change(function () {
myval = parseInt($(this).val());
$(".tableAppend").empty();

$(".tableAppend").append('<tr><th></th><th>Courses</th><th>Credit Hours</th><th>Grade</th></tr>');
$(".tableAppend").append('<tr><input type="submit" name="calculateGPA" value="Click Me" id="calculateGPA" /></tr>');
for (var i = 1 ; i <= myval ; i++) {
$(".tableAppend").append('<tr><td>' + i + '.</td><td><input name="courseTextBox' + i + '" type="text" id="courseTextBox' + i + '"></td>' +
  '<td><input name="creditTextBox' + i + '" type="text" id="creditTextBox' + i + '">  </td>' +
  '<td><select id="gradeSelect' + i + '">' +
     '<option value ="A">A</option>' +
     '<option value ="A-">A-</option>' +
     '<option value ="B+">B+</option>' +
     '<option value ="B">B</option>' +
     '<option value ="B-">B-</option>' +
     '<option value ="C+">C+</option>' +
     '<option value ="C">C</option>' +
     '<option value ="C-">C-</option>' +
     '<option value ="D">D</option>' +
     '<option value ="F">F</option>' +
     '</select></td>');

}
$(".tableAppend").append('<tr><td></td><td><button type="submit" id="gpaCalculate">Calculate GPA</button></td>');
$("#gpaCalculate").click(function (e) {
e.preventDefault();
displayDetails();
})
});

Answer 3

尝试使用此方法。

$( "#gpaCalculate" ).live( "click", function() {
    alert( "Goodbye!" );
});

生成的按钮不起作用-JQuery

3 个答案: