我是 PHP/JavaScript 的新手，我不确定我到底做错了什么。我正在尝试创建一个基于 Web 的表单，该表单具有复选框显示/隐藏选项。我的问题是我下面的代码给我一个解析错误。如果有人能帮助指出我的错误，我将不胜感激!
<?php
<html>
<head>
<title>Start</title>
<?PHP
$field1 = 'unchecked';
$field2 = 'unchecked';
$field3 = 'unchecked';
$field4 = 'unchecked';
$field5 = 'unchecked';
if (isset($_POST['Submit1'])) {
if (isset($_POST['field1'])) {
$field1 = $_POST['field1'];
if ($field1 == 'net') {
$field1 = 'checked';
}
}
if (isset($_POST['field2'])) {
$field2 = $_POST['field2'];
if ($field2 == 'word') {
$field2 = 'checked';
}
}
if (isset($_POST['field3'])) {
$field3 = $_POST['field3'];
if ($field3 == 'excel') {
$field3 = 'checked';
}
}
if (isset($_POST['field4'])) {
$field4 = $_POST['field4'];
if ($field4 == 'web') {
$field4 = 'checked';
}
}
if (isset($_POST['field5'])) {
$field5 = $_POST['field5'];
if ($field5 == 'php') {
$field5 = 'checked';
}
}
}
?>
</head>
<body>
<FORM NAME ="Sign-up" METHOD ="POST" ACTION ="presurvey.php">
<P>
<Input type = 'Checkbox' Name ='field1' value ="net"
<?PHP print $field1; ?>
>Mobility
<P>
<Input type = 'Checkbox' Name ='field2' value="value 1"
<?PHP print $field2; ?>
>Value 2
<P>
<Input type = 'Checkbox' Name ='field3' value="value 2"
<?PHP print $field3; ?>
>Value 3
<P>
<Input type = 'Checkbox' Name ='field4' value="value 3"
<?PHP print $field4; ?>
>Value 4
<P>
<Input type = 'Checkbox' Name ='field5' value="value 4"
<?PHP print $field5; ?>
>Name
<P>
<INPUT TYPE = "Submit" Name = "Submit1" VALUE = "Submit">
</FORM>
</body>
</html>
echo <script type="text.javascript">
// called onclick of checkbox
function toggleSub(box, id) {
// get reference to related content to display/hide
var el = document.getElementById(id);
if ( box.checked ) {
el.style.display = 'block';
} else {
el.style.display = 'none';
}
}
</script>;
?>
最佳答案
<html>
<head>
<title>Start</title>
<?php
$field1 = 'unchecked';
$field2 = 'unchecked';
$field3 = 'unchecked';
$field4 = 'unchecked';
$field5 = 'unchecked';
if (isset($_POST['Submit1'])) {
if (isset($_POST['field1'])) {
$field1 = $_POST['field1'];
if ($field1 == 'net') {
$field1 = 'checked';
}
}
if (isset($_POST['field2'])) {
$field2 = $_POST['field2'];
if ($field2 == 'word') {
$field2 = 'checked';
}
}
if (isset($_POST['field3'])) {
$field3 = $_POST['field3'];
if ($field3 == 'excel') {
$field3 = 'checked';
}
}
if (isset($_POST['field4'])) {
$field4 = $_POST['field4'];
if ($field4 == 'web') {
$field4 = 'checked';
}
}
if (isset($_POST['field5'])) {
$field5 = $_POST['field5'];
if ($field5 == 'php') {
$field5 = 'checked';
}
}
}
?>
</head>
<body>
<FORM NAME ="Sign-up" METHOD ="POST" ACTION ="presurvey.php">
<P>
<Input type = 'Checkbox' Name ='field1' value ="net"
<?php print $field1; ?>
>Mobility
<P>
<Input type = 'Checkbox' Name ='field2' value="value 1"
<?php print $field2; ?>
>Value 2
<P>
<Input type = 'Checkbox' Name ='field3' value="value 2"
<?php print $field3; ?>
>Value 3
<P>
<Input type = 'Checkbox' Name ='field4' value="value 3"
<?php print $field4; ?>
>Value 4
<P>
<Input type = 'Checkbox' Name ='field5' value="value 4"
<?php print $field5; ?>
>Name
<P>
<INPUT TYPE = "Submit" Name = "Submit1" VALUE = "Submit">
</FORM>
</body>
</html>
<script type="text/javascript">
function toggleSub(box, id) {
// get reference to related content to display/hide
var el = document.getElementById(id);
if ( box.checked ) {
el.style.display = 'block';
} else {
el.style.display = 'none';
}
}
</script>;
消除了您的错误，但不明白您到底想要实现什么？
关于javascript - 使用 PHP/JavaScript 显示/隐藏复选框，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/40189133/