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Php 提交表格的问题_Php_Javascript - Fatal编程技术网

Php 提交表格的问题

Php 提交表格的问题,php,javascript,Php,Javascript,有这样一种形式: <form name="myform" action="All.php" method="get"> <input type='text' id='tx' name='tx' value=''> <td height="8" align="right" valign="middle"><a href="javascript: submitform('All')">All Cities</a> | <a hr

有这样一种形式:

 <form name="myform" action="All.php" method="get">
 <input type='text' id='tx' name='tx' value=''>
 <td height="8" align="right" valign="middle"><a href="javascript: submitform('All')">All Cities</a> | <a href="javascript: submitform('Newyork')">Newyork</a> | <a href="javascript: submitform('Police')">Police</a></td></form>

<script type="text/javascript">
function submitform(val)
{
$("#tx").val(val);
document.myform.submit();
}
</script>
在All.php中:

 <form method="get" action="index1.php" id="frm1">
 <select id="category" onchange="showSelected1();" style="width:150px">
 <option></option>
 <option value="1" <?php if($_GET["categoryText"] == "Marriages") echo "selected"; ?> >Marriages</option>
 <option value="2" <?php if($_GET["categoryText"] == "Birthdays") echo "selected"; ?> >Birthdays</option></select>
 <input type="hidden" id="categoryText" name="categoryText" value='<?= $_GET['categoryText']; ?>'/>

<select id="city" onchange="showSelected();" style="width:150px">
<option><?= $_GET['tx']; ?></option>
<option></option>
<option value="1" <?php if($_GET["cityText"] == "Newyork") echo "selected"; ?> >Newyork</option>
<option value="2" <?php if($_GET["cityText"] == "Police") echo "selected"; ?> >Police</option></select>
<input type="hidden" id="cityText" name="cityText" value='<?= $_GET['cityText']; ?>'/>
<input type="button" value="Submit" onclick="formSubmit()"/></form>
<script language="JavaScript" type="text/javascript">
<!--
function showSelected()
{
var selObj = document.getElementById('city');
var cityTextObj = document.getElementById('cityText');

var selIndex = selObj.selectedIndex;
cityTextObj.value = selObj.options[selIndex].text;
}
//-->
</script>
<script language="JavaScript" type="text/javascript">
<!--
function showSelected1()
{
var selObj = document.getElementById('category');
var categoryTextObj = document.getElementById('categoryText');

var selIndex = selObj.selectedIndex;
categoryTextObj.value = selObj.options[selIndex].text;

}
//-->
</script>
表示从第一页内容填充到All.php中的表单本身。如果有人单击submit按钮而不更改选择城市,则在All.php中,它不会显示,也不会将城市值带到index1.php。为什么?你可以试试

document.getElementById("myform").submit();

您在下面的代码中写入了错误的变量名

<script type="text/javascript">
function submitform(val)
{
$("#hx").val(val);
document.myform.submit();
}
</script> 
您使用的是hx而不是tx,因此您的代码可能是这样的

<script type="text/javascript">
function submitform(val)
{
$("#tx").val(val);
document.myform.submit();
}
</script>
更改此选项,然后重试您的代码

我在代码中看到了许多其他问题,请学习如何使用单引号和双引号。这是你写的

value='<?= $_GET['categoryText']; ?>'  which is wrong.
正确的方法就是价值=


根据此更新您的代码,然后再次运行,如果有任何问题,请返回此处。

好的……我已经更正了。这个问题仍然存在。选择“categoryText”和“cityText”,给定submit,它将转到index1.php。如果在'index1.php'中点击submit,则不更改任何文本,即不提交表单。这是echo选择的问题吗?