我需要帮助使用复选框和PHP。我只是想确定PHP是否选中复选框的值。
示例：
<?php
include ("inc/conf.php");
$id = $_SESSION['id'];
if(isset($_POST['subfrm'])){
$gtid = $_REQUEST['tid'];
$ch1 = $_REQUEST['ch1'];
if($ch1 == "ON"){
$gch1 = "Y";
} else {
$gch1 = "N";
}
$ch2 = $_REQUEST['ch2'];
if($ch2 == "ON"){
$gch2 = "Y";
} else {
$gch2 = "N";
}
mysql_query("UPDATE SET ctable ch1='$gch1', ch2='$gch2' WHERE id='$gtid'");
}
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input type="hidden" name="tid" value="<?php echo $id; ?>" />
<input type="checkbox" name="ch1" />Hats
<input type="checkbox" name="ch2" />Watches
<textarea name="thetext"></textarea>
<input type="submit" name="subfrm" value="PUNCH ME" />
</form>
if(isset($_REQUEST["ch1"])){
$gch1 = "Y";
} else {
$gch1 = "N";
}
if(isset($_REQUEST["ch2"])){
$gch2 = "Y";
} else {
$gch2 = "N";
}
您不需要检查值是什么，因为如果不检查，它将不会提交任何数据，如果是，它将提交on的值。
试试这个：
<?php
$ch1 = isset($_REQUEST['ch1']);
如果未选中复选框，则其相应的变量将不会显示在请求中。
一起通过这个。它会让您知道选中了哪个复选框，并在表单提交时保留该复选框。
<?php
$message = '';
$ch1_checked = false;
$ch2_checked = false;
if(isset($_POST['submit_button'])) {
// Form was submitted
$ch1_checked = isset($_POST['ch1']);
$ch2_checked = isset($_POST['ch2']);
if($ch1_checked && $ch2_checked) {
$message .= 'Both were checked.';
} else if($ch1_checked) {
$message .= 'Checkbox 1 was checked.';
} else if($ch2_checked) {
$message .= 'Checkbox 2 was checked.';
} else {
$message .= 'Neither were checked.';
}
}
?>
<?php echo $message; ?>
<form id="my_form" action="test.php" method="post">
<input type="checkbox" name="ch1" value="ch1" <?php if($ch1_checked) echo 'checked'; ?> />Checkbox 1<br />
<input type="checkbox" name="ch2" value="ch2" <?php if($ch2_checked) echo 'checked'; ?> />Checkbox 2<br />
<input type="submit" name="submit_button" value="Go!" />
</form>