如何使用Javascript提交POST变量？
[英]How do I submit a POST variable with Javascript?
问题描述
So I am trying to submit a variable and the name of the variable via a form. 
所以我试图通过表单提交变量和变量的名称。 I switched a button from submit to button because I need additional validation. 我从提交到按钮切换了一个按钮，因为我需要额外的验证。
Anyway, here's the button now: 无论如何，现在是按钮：
<button type="button" onclick="subForm()" name="del" id="deletebutton" value="'.$org.'">Delete</button>
Here's my current validation: 这是我目前的验证：
<script type="text/javascript">
function subForm()
{
if(confirm("Are you sure you want to delete this?"))
document.forms["addorg"].submit();
else
return false;
}
</script>
And here's my script on the other side: 这是另一方面的脚本：
if (isset($_POST["del"]) && ($_POST['del'] !== '')) {
$del = mysql_real_escape_string(html2txt($_POST['del']));
$resfile = mysql_query('SELECT file_loc from organization WHERE org_id = '.$del);
$org_name = mysql_real_escape_string(html2txt($_POST['orgname']));
if (!$resfile)
header('Location: '.$admin.'?error=query');
while ($filerow = mysql_fetch_array($resfile)) {
$fileplace = $filerow['file_loc'];
unlink(".".$fileplace);
rmdir($org_name);
}
mysql_query("DELETE from organization where org_id='".$del."'");
header('Location: '.$admin);
}
It is not currently deleting the records that I want. 它目前没有删除我想要的记录。 How do I pass along the "del" name to the other page? 如何将“del”名称传递给其他页面？
4 个解决方案
解决方案1
3 2012-01-31 18:26:06
You can use <input type="hidden"> : 您可以使用<input type="hidden"> ：
echo '<input type="hidden" name="org_id" value="'.$org_id.'" />'
This should render something like: 这应该呈现如下：
<input type="hidden" name="org_id" value="1" />
Using this code you can access the hidden field data using: 使用此代码，您可以使用以下方法访问隐藏的字段数据
$org_id = $_POST['org_id'];
解决方案2
1 2012-01-31 18:27:44
instead use onsubmit 而是使用onsubmit
<form method='POST' onsubmit='return subForm();'>
and 和
<script type="text/javascript">
function subForm()
{
if(confirm("Are you sure you want to delete this?"))
return true;
else
return false;
}
</script>
edit: you can also change 编辑：你也可以改变
if (isset($_POST["del"]) && ($_POST['del'] !== '')) {
to 至
if ( !empty($_POST['del']) ) {
but i think this line is your problem 但我认为这一行是你的问题
$resfile = mysql_query('SELECT file_loc from organization WHERE org_id = '.$del);
try 尝试
$resfile = mysql_query("SELECT file_loc from organization WHERE org_id = '".$del."' ");
解决方案3
0 2012-01-31 18:26:56
Looking at http://www.w3schools.com/tags/tag_button.asp suggests that the 'name' of a button isn't submitted as opposed to its 'value' or 'text'. 查看http://www.w3schools.com/tags/tag_button.asp表明，未提交按钮的“名称”而不是“值”或“文本”。 Why not use an input of type hidden as just suggested? 为什么不使用隐藏类型的输入？
解决方案4
0 2012-01-31 19:01:09
I suggest you rethink using a form at all and consider AJAX . 我建议你重新考虑使用表格并考虑AJAX 。 This would solve the problem of knowing which button was clicked and the page doesn't need to reload. 这将解决了解单击哪个按钮并且页面不需要重新加载的问题。
Here is a sample of what you're trying to do: 以下是您要尝试的示例：
<html>
<head>
<script type="text/JavaScript">
var xmlhttp;
if (window.XMLHttpRequest)
xmlhttp=new XMLHttpRequest();
else
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
function deleteOrganization(orgID)
{
xmlhttp.onreadystatechange = function()
{
if (xmlhttp.readyState == 4 && xmlhttp.status == 200)
{
// in your PHP file, have it echo something like "SUCCESS" or "FAIL", and the response will be alerted here
alert(xmlhttp.responseText);
refreshList(); // either call another function to update the list or return the new list after the success/fail response.
}
};
xmlhttp.open("GET", "delete.php?orgID="+ orgID, true);
// OR - xmlhttp.open("GET", "page.php?action=DELETE&orgID="+ orgID, true);
xmlhttp.send();
}
function refreshList()
{
// either delete the row with JavaScript or make another AJAX call to get the new list after the entry was deleted.
}
</script>
</head>
<body>
<button onclick="deleteOrganization('<?php echo $org; ?>')">Delete</button>
</body>
</html>
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