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我需要在按钮单击时向页面添加新的几个元素。我不明白 - 当我用 fireBug 一步一步地做所有事情时 - 一切都很好。如果我关闭 Firebug - 不会添加任何内容。这是js代码:

    function Appending()
{
    var id = document.getElementById("default-id").value;
    id++;
    var result = $.post("addForm.php",{idEl:id},"html");
    $('#vrWrapper').append(result.responseText);
    document.getElementById("default-id").value = id; 
}
function removeInput(id) {
    $("#" + id).remove();
}

这里是 PHP 代码(addForm.php):

<?php
$id=$_POST["idEl"];
echo '<div class="data" id="'.$id.'"><div class ="userdata"><div class = "input"> <label for="lastName">Last Name:</label> <input id="lastName" type="text"/></div><div class = "input"> <label for="firstName">First Name:</label> <input id="firstName" type="text" /></div><div class = "input"> <label for="city">City:</label> <input id="city" type="text"/></div><div class = "input"> <label for="state">State:</label> <select name="state"><option value="AL">Alabama</option><option value="AK">Alaska</option><option value="AZ">Arizona</option><option value="AR">Arkansas</option><option value="CA">California</option><option value="CO">Colorado</option><option value="CT">Connecticut</option><option value="DE">Delaware</option><option value="DC">District of Columbia</option><option value="FL">Florida</option><option value="GA">Georgia</option><option value="HI">Hawaii</option><option value="ID">Idaho</option><option value="IL">Illinois</option><option value="IN">Indiana</option><option value="IA">Iowa</option><option value="KS">Kansas</option><option value="KY">Kentucky</option><option value="LA">Louisiana</option><option value="ME">Maine</option><option value="MD">Maryland</option><option value="MA">Massachusetts</option><option value="MI">Michigan</option><option value="MN">Minnesota</option><option value="MS">Mississippi</option><option value="MO">Missouri</option><option value="MT">Montana</option><option value="NE">Nebraska</option><option value="NV">Nevada</option><option value="NH">New Hampshire</option><option value="NJ">New Jersey</option><option value="NM">New Mexico</option><option value="NY">New York</option><option value="NC">North Carolina</option><option value="ND">North Dakota</option><option value="OH">Ohio</option><option value="OK">Oklahoma</option><option value="OR">Oregon</option><option value="PA">Pennsylvania</option><option value="RI">Rhode Island</option><option value="SC">South Carolina</option><option value="SD">South Dakota</option><option value="TN">Tennessee</option><option value="TX">Texas</option><option value="UT">Utah</option><option value="VT">Vermont</option><option value="VA">Virginia</option><option value="WA">Washington</option><option value="WV">West Virginia</option><option value="WI">Wisconsin</option><option value="WY">Wyoming</option></select></div></div><div class ="workdata"><div class = "input"><input id="ch1" type="checkbox" class="check" /><label for="ch1">Немирович (бригадир)</label></div><div class = "input"><input id="ch2" type="checkbox" class="check" /><label for="ch1">Немирович (бригадир)</label></div><div class = "input"><input id="ch3" type="checkbox" class="check" /><label for="ch1">Немирович (бригадир)</label></div><div class = "input"><input id="ch4" type="checkbox" class="check" /><label for="ch1">Немирович (бригадир)</label></div><div class = "input"><input id="ch1" type="checkbox" class="check" /><label for="ch1">Немирович (бригадир)</label></div></div><div class ="buttons"><button type="button" id="addAdult" onClick="Appending()">Add Adult</button><button type="button" id="register">Register</button><button type="button" id="cancel" onClick = "removeInput('.$id.')">Cancel</button></div></div>';
?>

真的很奇怪,一切都是正确的 - FireBug 没有显示错误

在这里您可以看到它的实际效果:http ://tvorchestva.net/test/form/form.html

4

3 回答 3

4

将您的代码放在回调中post

function Appending()
{
    var el = document.getElementById("default-id");  // cache the selection
    var id = el.value;
    id++;
    el.value = id; 
    $.post("addForm.php",{idEl:id},function( htm ) {
        $('#vrWrapper').append( htm );  // <--  run after response is received
    },"html");
}

由于 Differeds,您可以在较新版本的 jQuery 中使用不同的表单。

基于文档中的示例:

$.post("example.php", function() {
      alert("success");
    })
    .success(function() { alert("second success"); })
    .error(function() { alert("error"); })
    .complete(function() { alert("complete"); });
于 2011-06-18T20:38:23.587 回答
0

试试这个

function Appending() {
    var id = document.getElementById("default-id").value;
    id++;
    $.post("addForm.php", {
        idEl: id
    }, function(html) {
        $('#vrWrapper').append(html);
    });

    document.getElementById("default-id").value = id;
}

function removeInput(id) {
    $("#" + id).remove();
}
于 2011-06-18T20:36:03.213 回答
0

问题是$.post期望函数调用成功作为第三个参数,但您传递的是“html”。

尝试更换

$.post("addForm.php",{idEl:id},"html");

$.post("addForm.php", {idEl:id}, function(d) {
    $('#vrWrapper').append( d );
}); );

正如 jQuery 文档所指出的,“大多数实现都会指定一个成功处理程序;” 在他们的示例中,他们“获取请求的 HTML 片段并将其插入页面”,这正是您想要做的。

于 2011-06-18T20:39:12.080 回答