Question

早上好，我正在尝试执行以下操作，以便在类型不是'main'时在li中回显区域名称。然而，它只是输出字符串中的内容，有没有办法成功地执行此操作并保存我创建800页脚页面？

 $("#chk").prop("checked", "checked").trigger("change");

 $("#chk").change(function() {
    alert("triggered!"); 
 });

Answer 1

你可以用HTML编写你的PHP代码而不用双引号：

<?php if ($type == "main"): ?>
    <ul class="footer-nav">
         <h2>Links</h2>
         <li><a href="index.php">Home</a></li>
         <li><a href="services.php">Our Services</a></li>
         <li><a href="upvc.php">Upvc repairs</a></li>
         <li><a href="emergency.php">Emergency 24/7</a></li>
     </ul>
<?php else: ?>
    <ul class="footer-nav">
        <h2>Links</h2>
        <li><a href="locksmiths-<?= $areaname ?>.php">Home</a></li>
        <li><a href="services-<?= $areaname ?>.php">Our Services</a></li>
        <li><a href="services-upvc-repairs-<?= $area-name ?>.php">Upvc repairs</a></li>
        <li><a href="services-<?= $area-name ?>-24hour.php">Emergency 24/7</a></li>
    </ul>
<?php endif; ?>

Answer 2

$area-name 无效变量名称，可能是拼写错误。
看到您的HTML我建议您浏览Alternative syntax for control structures

以下是提高可读性的其他方法，首先删除echo。

<?php if ($type=="main" ): ?>
              <ul class="footer-nav">
                <h2>Links</h2>
                <li><a href="index.php">Home</a></li>
                <li><a href="services.php">Our Services</a></li>
                <li><a href="upvc.php">Upvc repairs</a></li>
                <li><a href="emergency.php">Emergency 24/7</a></li>
              </ul>
<?php else: ?>
              <ul class="footer-nav">
                <h2>Links</h2>
                <li><a href="locksmiths-<?php echo $areaname; ?>.php">Home</a></li>
                <li><a href="services-<?php echo $areaname; ?>.php">Our Services</a></li>
                <li><a href="services-upvc-repairs-<?php echo $areaname; ?>.php">Upvc repairs</a></li>
                <li><a href="services-<?php echo $areaname; ?>-24hour.php">Emergency 24/7</a></li>
              </ul>
<?php endif;?>

Answer 3

更好的方法是使用三元运算符，这样您就不需要重复HTML了，只需更改链接：

<ul class="footer-nav">
    <h2>Links</h2>
    <li><a href="<?= ($type == "main" ? 'index.php' :     'locksmiths-'.$areaname.'.php') ?>">Home</a></li>
    <li><a href="<?= ($type == "main" ? 'services.php' :  'services-'.$areaname.'.php') ?>">Our Services</a></li>
    <li><a href="<?= ($type == "main" ? 'upvc.php' :      'services-upvc-repairs-'.$areaname.'.php') ?>">Upvc repairs</a></li>
    <li><a href="<?= ($type == "main" ? 'emergency.php' : 'services-'.$areaname.'-24hour.php') ?>">Emergency 24/7</a></li>
</ul>

或者，您可能希望将HTML中的数据抽象为数组，这样可以更轻松地添加更多链接或使链接更加动态（在数据库中）。

<?php
$links = [
    [
        'name' => 'Home',
        'main' => 'index.php',
        'area' => 'locksmiths-'.$areaname.'.php'
    ], [
        'name' => 'Our Services',
        'main' => 'services.php',
        'area' => 'services-'.$areaname.'.php'
    ], [
        'name' => 'Upvc repairs',
        'main' => 'upvc.php',
        'area' => 'services-upvc-repairs-'.$areaname.'.php'
    ], [
        'name' => 'Emergency 24/7',
        'main' => 'emergency.php',
        'area' => 'services-'.$areaname.'-24hour.php'
    ],
];
?>
<ul class="footer-nav">
    <h2>Links</h2>
    <?php foreach ($links as $link): ?>
    <li><a href="<?= ($type == "main" ? $link['main'] : $link['area']) ?>"><?= $link['name'] ?></a></li>
    <?php endforeach ?>
</ul>

Answer 4

这应该可以解决您的问题。你忘了把别人说成是正确的陈述

        <?php if ($type=="main" ) {
            echo '
            <ul class="footer-nav">
            <h2>Links</h2>
            <li><a href="index.php">Home</a></li>
            <li><a href="services.php">Our Services</a></li>
            <li><a href="upvc.php">Upvc repairs</a></li>
            <li><a href="emergency.php">Emergency 24/7</a></li>
          </ul>';}
          else {
            echo '
            <ul class="footer-nav">
            <h2>Links</h2>
            <li><a href="locksmiths-'.$areaname.'.php">Home</a></li>
            <li><a href="services-'.$areaname.'.php">Our Services</a></li>
            <li><a href="services-upvc-repairs-'.$areaname.'.php">Upvc repairs</a></li>
            <li><a href="services-'.$areaname.'-24hour.php">Emergency 24/7</a></li>
          </ul>';}
?>

Answer 5

您有一些语法错误，请像这样使用

<?php if ($type=="main" ) {
                echo '
                <ul class="footer-nav">
                <h2>Links</h2>
                <li><a href="index.php">Home</a></li>
                <li><a href="services.php">Our Services</a></li>
                <li><a href="upvc.php">Upvc repairs</a></li>
                <li><a href="emergency.php">Emergency 24/7</a></li>
              </ul>';
            }else{
                echo '
                <ul class="footer-nav">
                <h2>Links</h2>
                <li><a href="locksmiths-'.$areaname.'.php">Home</a></li>
                <li><a href="services-'.$areaname.'.php">Our Services</a></li>
                <li><a href="services-upvc-repairs-'.$areaname.'.php">Upvc repairs</a></li>
                <li><a href="services-'.$areaname.'-24hour.php">Emergency 24/7</a></li>
              </ul>';
            }
  ?>

确定变量名称，例如我将$area-name替换为$areaname

Answer 6

无需将回声置于回声中..请遵循以下更改..

<?php if ($type=="main" ) :
            echo '
            <ul class="footer-nav">
            <h2>Links</h2>
            <li><a href="index.php">Home</a></li>
            <li><a href="services.php">Our Services</a></li>
            <li><a href="upvc.php">Upvc repairs</a></li>
            <li><a href="emergency.php">Emergency 24/7</a></li>
          </ul>';
          else:
            echo '
            <ul class="footer-nav">
            <h2>Links</h2>
            <li><a href="locksmiths-'.$areaname.'.php">Home</a></li>
            <li><a href="services-.'$areaname.'.php">Our Services</a></li>
            <li><a href="services-upvc-repairs-.'$area-name.'.php">Upvc repairs</a></li>
            <li><a href="services-'.$area-name.'-24hour.php">Emergency 24/7</a></li>
          </ul>'
    endif;?>

Answer 7

最短的代码

您只能echo一次公共数据，就像我做的那样..

变量$area-name无效......我将其转换为$ area_name。

<?php 
    echo '<ul class="footer-nav"><h2>Links</h2>'; //common data
if ($type=="main")
    echo '
        <li><a href="index.php">Home</a></li>
        <li><a href="services.php">Our Services</a></li>
        <li><a href="upvc.php">Upvc repairs</a></li>
        <li><a href="emergency.php">Emergency 24/7</a></li>';
else echo '
        <li><a href="locksmiths-'.$areaname.'.php">Home</a></li>
        <li><a href="services-'.$areaname.'.php">Our Services</a></li>
        <li><a href="services-upvc-repairs-'.$area_name.'.php">Upvc repairs</a></li>
        <li><a href="services-'.$area_name.'-24hour.php">Emergency 24/7</a></li>';
echo '</ul>';//common data
?>

Answer 8

$areaname = 'wales';
$pages = [
    'index.php' => 'Home',
    'services.php' => 'Our Services',
    'upvc.php' => 'Upvc repairs',
    'emergency.php' => 'Emergency 24/7'
];
$area_urls = [
    "locksmiths-$areaname.php",
    "services-$areaname.php",
    "services-upvc-repairs-$areaname.php",
    "services-$areaname-24hour.php"
];
if($type != 'main')
    $pages = array_combine($area_urls, $pages);
// Output menu
echo "<h2>Links</h2>\n<ul class='footer-nav'>\n";
    foreach($pages as $link => $page)
        printf("\t<li><a href='%s'>%s</a></li>\n", $link, $page);
echo "</ul>\n";

在echo中使用echo建立一个php if satement

8 个答案: