我是新來的PHP的世界,並放在一起的表格,乘以一個輸入的值。但是,當我嘗試驗證某人是否未輸入任何值來返回錯誤消息時,它會顯示該消息。我的代碼如下。感謝您能否提出改進建議。這是爲什麼!isset似乎不起作用?
<?php
$counter = 0;
if(isset($_POST["submit"])) {
$start = $_POST["start"];
$end = $_POST["end"];
$multiply = $_POST["multiplication"];
// if($_POST["start"] == "" && $_POST["end"] == "" && $_POST["multiplication"] == "") {
// print "Please enter some values";
// }
if(!isset($_POST["start"], $_POST["end"], $_POST["multiplication"])) {
print "Please enter some values";
}
// for($start;$start<$end;$start++) {
// $counter = $counter +1;
// $multiplication = $counter * $multiply;
// print "$counter <br />";
// print "$counter multiplied by $multiply = $multiplication <br />";
// }
}
?>
<html>
<head>
<title>Sample Multiplication</title>
</head>
<body>
<form name="multiply" method="post" action="multiplication_sample.php">
<input type="text" name="start" value="<?php if(isset($_POST["start"])) { print $start; } ?>">
<input type="text" name="end" value="<?php if(isset($_POST["end"])) { print $end; } ?>">
<input type="text" name="multiplication" value="<?php if(isset($_POST["multiplication"])) { print $multiply; } ?>">
<input type="submit" name="submit" value="submit">
</form>
<?php
if(isset($_POST["submit"])) {
for($start;$start<$end;$start++) {
$counter = $counter + 1;
$multiplication = $counter * $multiply;
print "$counter multiplied by $multiply = $multiplication <br />";
}
}
?>
</body>
</html>
如果用戶沒有爲每個表單域指定任何值,我希望表單顯示消息。 – PeanutsMonkey 2011-04-10 22:16:59