Question

如何回复表单输入而无需在验证后重新显示表单？我只能在表单后显示输入。这是我的代码

<?php
$postalCode = $_POST['postalCode'];
$postalCodeErr = "";
$postalCodeValidation = '/^[A-Z][0-9][A-Z][0-9][A-Z][0-9]$/';
$postalCodeIsValid = false;
?>
<html>
<body>
    INT322 Lab 3-1
    <br />
    <br />
    <form name="lab3form" action="index.php" method="post">
        Postal Code:
        <input type="text" name="postalCode" value="<?php if(isset($postalCode)) echo $postalCode; ?>" />
        <?php
        if(($postalCode != "") && preg_match($postalCodeValidation, $postalCode)) {
            $postalCodeIsValid = true;
        }
        else {
            $postalCodeErr = "Invalid Postal Code";
        }
        if(isset($postalCode)) echo " $postalCodeErr";
        ?>
        <br />
        <br />
        <input type="submit" name="submit" />
    </form>
</body>
</html>

<?php
if($_POST['submit'] && $postalCodeIsValid) {
    echo "Postal Code: $postalCode";
}
?>

Answer 1

如何将表单打包在else的{{1}}

中

if($_POST['submit'] && $postalCodeIsValid) { ... } else { ... }

完整代码的更新答案：

<?php
if($_POST['submit'] && $postalCodeIsValid) {
    echo "Postal Code: $postalCode";
}
else {
    <form name="lab3form" action="index.php" method="post">
        Postal Code:
        <input type="text" name="postalCode" value="<?php if(isset($postalCode)) echo $postalCode; ?>" />
        <?php
        if(($postalCode != "") && preg_match($postalCodeValidation, $postalCode)) {
            $postalCodeIsValid = true;
        }
        else {
            $postalCodeErr = "Invalid Postal Code";
        }
        if(isset($postalCode)) echo " $postalCodeErr";
        ?>
        <br />
        <br />
        <input type="submit" name="submit" />
    </form>
}
?>

请注意，您的正则表达式仅适用于<html> <body> INT322 Lab 3-1 <br /> <br /> <?php if(!empty($_POST['submit'])): $postalCode = $_POST['postalCode']; if(isValidPostalCode($postalCode)): echo "Postal Code: $postalCode"; else: form($postalCode, true); endif; else: form(); endif; ?> </body> </html> <?php function form($postalCode = null, $hasError = false) { ?> <form name="lab3form" action="postal.php" method="post"> Postal Code: <input type="text" name="postalCode" value="<?php if(isset($postalCode)) echo $postalCode; ?>" /> <?php if ($hasError): ?> <div class="error">Invalid Postal Code</div> <?php endif; ?> <br /> <br /> <input type="submit" name="submit" /> </form> <?php } function isValidPostalCode($postalCode) { $postalCodeValidation = '/^[A-Z][0-9][A-Z][0-9][A-Z][0-9]$/'; return !empty($postalCode) && preg_match($postalCodeValidation, $postalCode); } ?>等邮政编码 - 我不确定这是否是您想要的行为。

PHP - 显示表单输入而不重新显示表单

1 个答案: