将数据动态添加到现有表格时出现问题...

Melissa Mussitsch写道：

Melissa Mussitsch wrote:

我之前在创建一张全新的桌子时已经这样做了。
但是下面的代码不是工作和我一直收到错误：
Internet Explorer无法打开Internet站点.....
操作中止

这是代码：

myBody = document.getElementsByTagName（" body"）。item（0）;
mytablebody = document.getElementById（''product''）。getElementsByTa gName（" TBODY"）[0];

newrow = document.createElement（" TR"）;
newrow.setAttribute（" bgColor"，"＃EFECCB"）;

//系列
newcell = document.createElement（" TD"）;
currenttext = document.createTextNode（" Family"）;
newcell.appendChild（currenttext）;
newcell。的setAttribute（QUOT;宽度&QUOT ;，10）;
newcell.setAttribute（" align"，" center"）;
newrow.appendChild（newcell）;
//产品说明
newcell = document.createElement（" TD"）;
currenttext = document.createTextNode（" Product Description"）;
newcell.appendChild（currenttext）;
newcell.setAttribute（" align"，" center"）;
newrow.appendChild（newcell）;
mytablebody.appendChild（newrow）;

这是非常基本的代码。我手动放入TR，TD，TD，它工作正常。
我想确保它不是因为我有一些不匹配< TABLE>标签。

任何想法？

I''ve done this before while creating a brand new table.
But the code below is not working and I keep getting the error:
"Internet Explorer cannot open the Internet site .....
Operation aborted"

Here is the code:

myBody=document.getElementsByTagName("body").item( 0);
mytablebody = document.getElementById(''product'').getElementsByTa gName("TBODY" )[0];

newrow=document.createElement("TR");
newrow.setAttribute("bgColor", "#EFECCB");

//Family
newcell=document.createElement("TD");
currenttext=document.createTextNode("Family");
newcell.appendChild(currenttext);
newcell.setAttribute("width","10");
newcell.setAttribute("align", "center");
newrow.appendChild(newcell);
//Product Description
newcell=document.createElement("TD");
currenttext=document.createTextNode("Product Description");
newcell.appendChild(currenttext);
newcell.setAttribute("align", "center");
newrow.appendChild(newcell);
mytablebody.appendChild(newrow);

It''s pretty basic code. I put the TR, TD, TD in by hand and it worked fine.
I wanted to make sure it wasn''t because I had some mismatching <TABLE> tags.

Any ideas?

您确定上面发布的代码会导致错误吗？你怎么用b $ b来调用那段代码？你好像有一些链接导致

那条消息。


-


Martin Honnen
http://JavaScript.FAQTs.com/

经过深入研究后，我发现错误信息是一个非常标准的错误信息，可能意味着许多事情。我结束了

让代码成功写入一行。但是，在非常好的b $ b结束时，我仍然会收到此错误。在这些

语句之后留下的唯一内容是< / FORM>< / BODY>< / HTML>！

对于这个项目我试过添加这个代码到了一个已经很庞大的asp

页面。我想我将不得不将它分开以使其工作并且

然后尝试将其重新插入。当我将代码拉回时，我不会得到它

尽管如此。


我填写变量和数组之后调用此代码并且

已经绘制了一些Web表单对象页。我只需要我的

< SCRIPT>标记并调用函数来创建表并写一个

几行（现在）。当我从我的代码中删除这三行时我不会得到错误：

1）< SCRIPT language ="" JavaScript""> ; var

mybody = document.getElementsByTagName（" body"）。item（0）; mytable = document.cr

eateElement（" TABLE"）;< / SCRIPT>

2）< SCRIPT language ="" JavaScript""> WriteTable（mytable，<％= iarr Max％>，''''，>
''''）;< / SCRIPT>

3）< SCRIPT language ="" JavaScript""> mybody.appendChild（mytable）;< / SCRIPT>


当然，WriteTable完成大部分工作。


谢谢。

Melissa M


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Well after doing further research, I see that the error message is a
pretty standard one and could mean a number of things. I ended up
getting the code to write one row successfully. However, at the very
end, I still get this error. And the only thing left after these
statements is </FORM></BODY></HTML>!
For this project I''ve tried adding this code to an already enormous asp
page. I think I''m going to have to split it out to get it working and
then try to plug it back in. I don''t get it when I pull that code back
out though.

I have this code called after variables and arrays are filled in and
some web form objects are already drawn on the page. I simply have my
<SCRIPT> tag and call the function to create the table and write a
couple rows (for now). When I remove these three lines from my code I
don''t get the error:
1)<SCRIPT language=""JavaScript"">var
mybody=document.getElementsByTagName("body").item( 0);mytable=document.cr
eateElement("TABLE");</SCRIPT>
2)<SCRIPT language=""JavaScript"">WriteTable(mytable,<%=iarr Max%>, '''',
'''');</SCRIPT>
3)<SCRIPT language=""JavaScript"">mybody.appendChild(mytable );</SCRIPT>

Of course WriteTable does most of the work.

Thank you.
Melissa M

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Ok - 我已经能够简单了搞定这个并仍然得到错误。


我又有三行：

< SCRIPT language =" JavaScript"> var

mybody = document.getElementsByTagName（" body"）。item（0）; mytable = document.cr

eateElement（" TABLE"）;< / SCRIPT>

< SCRIPT language =" JavaScript"> WriteProductList（mytable，

<％= iarrMax％>）;< / SCRIPT>

< SCRIPT language =" JavaScript"> mybody.appendChild（mytable）; < / SCRIPT>


这是WriteProductList的代码：

函数WriteProductList（mytable，iarrMax）{


var mytablebody;

var newrow;

var newcell;

var currenttext;


mytable.setAttribute（" width"，" 100％"）;

mytable.setAttribute（" border"，" 1"）;

mytable。 setAttribute（" id"，" productselect"）;

mytablebody = document.createElement（" TBODY"）;


newrow = document.createElement （TR）;


//家庭

newcell = document.createElement（" TD"）;

currenttext = document.createTextNode（" Family"）;

newcell.appendChild（currenttext）;

newrow.appendChild（newcell）;

newrow.appendChild（newcell）;


mytablebody.appendChild（newrow）;

mytable.appendChild（mytablebody）;

}


它'非常简单，因为它在桌子上写了一行。

让我说我正在排桌子。

但是一旦它到达mybody.appendChild（mytable）;部分，出现

错误。如果没有它，我就不会排好。


有没有人看到明显错误的东西？


谢谢。>
Melissa M


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Ok - I''ve been able to simplify this and still get the error.

Again I have three lines:
<SCRIPT language="JavaScript">var
mybody=document.getElementsByTagName("body").item( 0);mytable=document.cr
eateElement("TABLE");</SCRIPT>
<SCRIPT language="JavaScript">WriteProductList(mytable,
<%=iarrMax%>);</SCRIPT>
<SCRIPT language="JavaScript">mybody.appendChild(mytable); </SCRIPT>

Here is the code for WriteProductList:
function WriteProductList(mytable, iarrMax) {

var mytablebody;
var newrow;
var newcell;
var currenttext;

mytable.setAttribute("width","100%");
mytable.setAttribute("border","1");
mytable.setAttribute("id","productselect");
mytablebody=document.createElement("TBODY");

newrow=document.createElement("TR");

//Family
newcell=document.createElement("TD");
currenttext=document.createTextNode("Family");
newcell.appendChild(currenttext);
newrow.appendChild(newcell);
newrow.appendChild(newcell);

mytablebody.appendChild(newrow);
mytable.appendChild(mytablebody);
}

It''s very simple in that it writes one row to the table.
Let me say that I am getting my row in the table.
But as soon as it gets to the "mybody.appendChild(mytable);" part, the
error appears. And without it, I don''t get the row.

Does anyone see something obviously wrong?

Thank you.
Melissa M

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