if else 语句不发布变量
[英]if else statement not posting variable
问题描述
在这段代码中，我试图制作一个简单的表单，询问用户他们想要哪种信封。 在用户选择表单中的选项后，我希望它显示总数。 $shipment 变量应该是价格，问题是当我回显总数时它没有打印到屏幕上。 除了 $shipment 之外，此代码中的所有其他内容都按预期显示在屏幕上。 任何关于错误的提示都值得赞赏！
<html>
<body>
<form action="" method="post">
Package Type :
<select name="package" id="package" value="type">
<option value="Flate Rate Envelope">Flate Rate Envelope</option>}
<option value="Small Rate Envelope">Small Rate Envelope</option>
<option value="Medium Rate Envelope">Medium Rate Envelope</option>
<option value="Large Rate Envelope">Large Rate Envelope </option>
</select>
<input type="submit" value="submit" name="submit">
</form>
</body>
</html>
<?php
echo "<br>";
echo $_POST["package"];
echo "<br>";
$shipment = 0;
if ($_POST["package"] == 'Flate Rate Envelope'){
$shipment = 6.70;
echo "Total Shipping Cost: $".$_POST[$shipment];
}
else if ($_POST["package"] == 'Small Rate Envelope'){
$shipment = 7.20;
echo "Total Shipping Cost: $".$_POST["$shipment"];
}
else if ($_POST["package"] == 'Medium Rate Envelope'){
$shipment = 13.65;
echo "Total Shipping Cost: $".$_POST["$shipment"];
}
else if ($_POST["package"] == 'Large Rate Envelope'){
$shipment = 18.90;
echo "Total Shipping Cost: $".$_POST["$shipment"];
}
?>
1 个解决方案
解决方案1
0 已采纳 2022-09-26 15:06:12
您正在回显不存在的变量$_POST[$shipment] 。 将其更改为$shipment 。
echo "<br>";
echo $_POST["package"];
echo "<br>";
$shipment = 0;
if ($_POST["package"] == 'Flate Rate Envelope'){
$shipment = 6.70;
echo "Total Shipping Cost: $".$shipment;
}
else if ($_POST["package"] == 'Small Rate Envelope'){
$shipment = 7.20;
echo "Total Shipping Cost: $".$shipment;
}
else if ($_POST["package"] == 'Medium Rate Envelope'){
$shipment = 13.65;
echo "Total Shipping Cost: $".$shipment;
}
else if ($_POST["package"] == 'Large Rate Envelope'){
$shipment = 18.90;
echo "Total Shipping Cost: $".$shipment;
}
?>
PS不要这样做。 它既不安全，又是一段错误的代码。 此外，检查您的 $POST 变量是否与 SQL 注入有关。
PHP 变量 if else 语句
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