Question

Not sure but this may be a bug in php. Here is my dummy code.

<script language="javascript">
function fb_logout() {
document.getElementById("hidden_logout_id").value
alert(document.getElementById("hidden_logout_id").value);
document.getElementById("form_1").submit();
}
</script>

echo "<form id="form_1" action="" . $_SERVER[ PHP_SELF ] . "" method="post">";
echo "<input type="hidden" id="hidden_logout_id" name="hidden_logout_name" value="1"/>";
echo "<input type="text" id="h_logout2" name="h_logout2" value="1"/>";    
echo "</form>";
var_dump($_POST);
echo "<span onclick="fb_logout();">Logout</span>";

When doing a post trough the click on the span the values of the hidden and the text inside the form do not get posted: var_dump($_POST) shows an empty variable. But it s strange that if I remove the hidden from the form (or I just place it outside the form) it works and it passes in the $_POST the value of the input text remaining in the form. Does this has to do with that the hidden is modified from an event outside the form?

感谢任何帮助。

Answer 1

该法典可能有所帮助。

<?php
    if(isset($_POST[ hidden_logout_name ])) var_dump($_POST);   
?>
<script language="javascript">
function fb_logout() {    
    alert(document.getElementById("hidden_logout_id").value);
    document.getElementById("hidden_logout_id").value = "hidden_logout_value";
    document.getElementById("form_1").submit();
}
</script>

<form id="form_1" action="" method="post">
    <input type="hidden" id="hidden_logout_id" name="hidden_logout_name" value=""/>
    <input type="text" id="h_logout2" name="h_logout2" value="1"/>
</form>
<button onclick="fb_logout();">Logout</button>

Answer 2

document.getElementById("hidden_logout_id").value

您没有就此一线采取任何行动,或是把其价值分配给变数。

var logout_value=document.getElementById("hidden_logout_id");
alert(logout_value.value);
document.getElementById("form_1").submit();

愿你帮助。

Answer 3

展望

hidden_logout_name

而不是

hidden_logout_id

因此,你的价值在于

$_POST[ hidden_logout_name ]

友情链接