[英]jquery function not calling on click
I have a function like so: 我有一个像这样的功能:
last_value = null;
$('td').click(function(e) {
console.log($(e.target).attr("id"));
if ($(e.target).is("td")) {
var the_form = $("#edit-form").detach();
if (last_value) {
$("#" + last_value.attr("id")).replaceWith(last_value);
}
last_value = $(e.target).clone();
$(the_form).removeClass("hidden");
var the_input = $(the_form).find("input.form-control");
$(the_input).attr("name", $(e.target).attr("id"));
$(the_input).attr("placeholder", $(e.target).text());
$(e.target).css('padding-top', 1);
$(e.target).css('padding-bottom', 1);
$(e.target).text("");
$(e.target).append(the_form);
}
});
This is supposed to take a table cell, and produce an inline form populated with the cell contents, which it replaces. 这应该采用一个表格单元格,并产生一个用单元格内容填充的内联表格,并将其替换。 Additionally, code has been added so that when a different cell is clicked, the contents revert to their original values.
此外,已添加代码,以便在单击其他单元格时,内容将恢复为其原始值。 However, the problem I'm running into is this: suppose I click one cell, A. The form appears the way it should.
但是,我遇到的问题是:假设我单击了一个单元格A。该表单以应有的方式显示。 Then suppose I click cell B. The form then "moves" to that cell, and the contents in cell A revert to their original values.
然后假设我单击单元格B。然后该窗体“移动”到该单元格,并且单元格A中的内容恢复为其原始值。 Now suppose I click on cell A again.
现在,假设我再次单击单元格A。 In this case, not only does the form not appear, it stays in cell B. In fact, the
console.log
doesn't even fire. 在这种情况下,不仅表单没有出现,而且还停留在单元格B中。实际上,
console.log
甚至不会触发。 What am I doing wrong here? 我在这里做错了什么?
Try with this 试试这个
$(document).find('td').on('click',function(e){
e.preventDefault();
console.log($(e.target).attr("id"));
if ($(e.target).is("td")) {
var the_form = $("#edit-form").detach();
if (last_value) {
$("#" + last_value.attr("id")).replaceWith(last_value);
}
last_value = $(e.target).clone();
$(the_form).removeClass("hidden");
var the_input = $(the_form).find("input.form-control");
$(the_input).attr("name", $(e.target).attr("id"));
$(the_input).attr("placeholder", $(e.target).text());
$(e.target).css('padding-top', 1);
$(e.target).css('padding-bottom', 1);
$(e.target).text("");
$(e.target).append(the_form);
}
});
Hope this will help you. 希望这会帮助你。
As was suggested in the comments under the OP, (thanks @Mohammad Akbari), changing the code to 如操作规范下的注释所建议,(感谢@Mohammad Akbari),将代码更改为
$('table').on('click', 'td', function(e){
console.log($(e.target).attr("id"));
if ($(e.target).is("td")) {
var the_form = $("#edit-form").detach();
if (last_value) {
$("#" + last_value.attr("id")).replaceWith(last_value);
}
last_value = $(e.target).clone();
$(the_form).removeClass("hidden");
var the_input = $(the_form).find("input.form-control");
$(the_input).attr("name", $(e.target).attr("id"));
$(the_input).attr("placeholder", $(e.target).text());
$(e.target).css('padding-top', 1);
$(e.target).css('padding-bottom', 1);
$(e.target).text("");
$(e.target).append(the_form);
}
});
fixes the problem. 解决问题。
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