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JavascriptappendChild(td)在一个地方工作,而不是在另一个地方工作

转载 作者:行者123 更新时间:2023-12-02 17:01:20 25 4
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考虑这个 HTML(表单中表格的一部分):

<tr id="EnterForRow">
<td>Entered for</td>
<td><input type="radio" name="enterfor" value "0" checked>Myself
<input type="radio" name="enterfor" value "1">Someone Else </td>
</tr>
<tr id="PrayerForRow">
<td>Prayer for </td>
<td> <input type="radio" name="prayerfor" value="0" checked>Myself
<input type="radio" name="prayerfor" value="1">Someone Else </td>
</tr>

当用户点击Someone Else时,我有Javascript来使一个新的文本输入框出现在该行上。 PrayerForRow 的 Javascript 可以工作,但 EnterForRow 的 Javascript 不起作用。我看不出任何明显的差异。我想我已经盯着它看了太久了..

这有效:

var prayforRad =  document.getElementsByName('prayerfor');

for(var i = 0; i < prayforRad.length; i++)
{
prayforRad[i].onclick = function()
{
var theValue = radioValue(document.getElementsByName('prayerfor'));
if (theValue == "1")
{
if (!document.getElementById("pfor"))
{
var newTd = document.createElement("td");
newTd.setAttribute("id", "pfor");
var pforRow = document.getElementById("PrayerForRow");
pforRow.appendChild(newTd);
newTd.innerHTML = '<td>For: <input type="text" name="PrayFor" id="PrayFor" size="25"></td>';
}
}
else
{
if (document.getElementById("pfor"))
{
var pforRow = document.getElementById("PrayerForRow");
var pf = document.getElementById("pfor");
pforRow.removeChild(pf);
}
}
}
}

这不会:

var enterforRad = document.getElementsByName('enterfor');
for(var j = 0; j < enterforRad.length; j++)
{
enterforRad[j].onclick = function()
{
var theValue2 = radioValue(document.getElementsByName('enterfor'));
if (theValue2 == "1")
{
if (!document.getElementById("efor"))
{
var newTD2 = document.createElement("td");
newTD2.setAttribute("id", "efor");
var eforRow = document.getElementById("EnterForRow");
eforRow.appendChild(newTD2);
newTD2.innerHTML = '<td>For: <input type="text" name="EntFor" id="EntFor" size="25"></td>';
}
}
else
{
if (document.getElementById("efor"))
{
var eforRow = document.getElementById("EnterForRow");
var ef = document.getElementById("efor");
eforRow.removeChild(ef);
}
}
}
}

感谢任何指点。

最佳答案

<td><input type="radio" name="enterfor" value "0" checked>Myself
<input type="radio" name="enterfor" value "1">Someone Else </td>

我认为你失去了两个“=”

关于JavascriptappendChild(td)在一个地方工作,而不是在另一个地方工作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25678220/

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