I have a php page that on load shows a table by default, and that table is:
我有一个php页面,加载时默认显示一个表,该表是:
echo "<table class='tftable' border='1' id='table_L'>";
There is another table:
还有另一张表:
echo "<table class='tftable' border='1' id='table_P' style='display:none'>";
which I toggle on the page by the following javascript function: function toggleTables(table_id)
我通过以下javascript函数在页面上切换:function toggleTables(table_id)
{
if(table_id == "table_L") {
document.getElementById('table_L').style.display = "table";
document.getElementById('table_P').style.display = "none";
}
else if(table_id == "table_P") {
document.getElementById('table_P').style.display = "table";
document.getElementById('table_L').style.display = "none";
}
}
So depending on which link user clicks, it displays the appropriate table. But then I have a form that based on selected values feeds the table data upon Submit. And the problem is, if user selects the table_P, which by default is hidden, then selects some dropdown values and submits the form, the page refreshes and table_L is loaded back again. How can I keep the same table selected even after the page reload? Is there some values I need to store to remember upon form submit?
因此,根据用户点击的链接,它会显示相应的表格。但是我有一个基于选定值的表单在提交时提供表数据。问题是,如果用户选择默认隐藏的table_P,然后选择一些下拉值并提交表单,页面刷新并再次加载table_L。如何在页面重新加载后保持选择相同的表格?在表单提交时是否需要存储一些值以便记住?
<form method="post" action="">
selections...
<INPUT TYPE="submit" name="submit" title="Run It!" value="SUBMIT" />
</form>
1 个解决方案
#1
0
You could have a hidden field in the form that knows which table is currently visible. On submit, use that form data form the client to echo the proper table.
您可以在表单中有一个隐藏字段,该字段知道当前可见的表。在提交时,从客户端使用该表单数据来回显正确的表。
So your form would look like this:
所以你的表单看起来像这样:
<form method="post" action="">
...
<input type="hidden" name="tableid" value="table_P">
<button type="submit">Submit</button>
</form>
And serverside, the post body would contain the name of the table, and you would print the correct table.
而在服务器端,帖子正文将包含表的名称,您将打印正确的表。
#1
0
You could have a hidden field in the form that knows which table is currently visible. On submit, use that form data form the client to echo the proper table.
您可以在表单中有一个隐藏字段,该字段知道当前可见的表。在提交时,从客户端使用该表单数据来回显正确的表。
So your form would look like this:
所以你的表单看起来像这样:
<form method="post" action="">
...
<input type="hidden" name="tableid" value="table_P">
<button type="submit">Submit</button>
</form>
And serverside, the post body would contain the name of the table, and you would print the correct table.
而在服务器端,帖子正文将包含表的名称,您将打印正确的表。