Question

我正在运行在每个页面上加载main.php文件的网站，它有数千页。每个页面网址都不同，例如

mywebsite.com/United-Kingdom.php

和

mywebsite.com/United-Stats-of-america.php.

每个页面都有唯一的页面标题，由PHP从数据库加载，（<h4 style="font-size:20px;color:#148ADB;"><?php echo $cityname['country']; ?>Information</h4>）。

我想用网址或页面标题自动更改标题jpg图像，例如，如果用户使用联合王国网址，则应在标题上加载/images/country/united-kingdom.jpg并且当用户打开美国统计数据网页时应该在标题上加载图像/images/country/united-stats.jpg。

我怎样才能实现这个目标？

= now（）按票价排序＆＃34;）; $ ARR1 = mysql_fetch_array（$ SEL1）; ？＆GT;

<div id="content-top-txt" style="height:auto">
<div id="content-top-inner1">

<div id="content-top-txt-inner1" style="height:auto">       

        <div class="content">

        <h4 style="font-size:20px;color:#148ADB;"><?php echo $cityname['country']; ?>specials</h4>

        <p style='text-align:justify; font-size:13px; color:#3c3c3c;'>

        <strong style="text-transform:uppercase;color:#148ADB;"><?php echo $cityname['country']; ?></strong> with choice of range prices think of comfort for lesser cost<strong style="text-transform:uppercase"><?php echo $cityname['country']; ?></strong> starts from <?php

                    $tempcode=$arr1['des_code'];

                    $stemp=mysql_query("SELECT * FROM cities_var WHERE code='$tempcode'");

                    $temp=mysql_fetch_array($stemp);

                    if($temp['code']=='')

                    {                                               

                     ?>

                     <strong><?php echo $arr1['des_name'];?> (<?php echo $arr1['des_code']; ?>) </strong>

                    <?php

                    }

                    else

                    {

                    ?>

                    <a href="<?php echo $temp['url'] ?>" target="_blank"><strong><?php echo $arr1['des_name'];?> (<?php echo $arr1['des_code']; ?>) </strong></a>

                    <?php

                    }

                    ?>

                    @ <strong>&pound;<?php echo $arr1['fare']; ?></strong> with & table below<?php echo $arr1['des_name'];?> & other cities in <strong style="text-transform:uppercase"><?php echo $cityname['country']; ?></strong> About<strong>UK</strong>.General information About <strong style="text-transform:uppercase"><?php echo $cityname['country']; ?></strong>call @ 123456789 or use <a href="../../form.php">Form</a> or email:- <strong>Email@address.com</strong></p>

        <p><?php    echo $cvar['overview2']; ?></p>

        <p>Note:-Prices given are excluding taxes</p>

        </div>

<div id="content-top-inner2">

<div id="content-top-txt-inner2" style="height:auto">      

        <!-- slider start here -->

        <div class="slider-main">



        </div>   <!-- slider end here-->

        </div><!-- Content-top-txt-inner2 end-->

        </div><!-- content-top-inner2 end here-->

</div><!--contaent-top-txt end here -->

</div><!--content-top end here -->



<div id="content-middle">

<div id="content-middle-txt">   

        <?php

        include("../../files/lb1.php");

        ?>

    <div class="content-middle-txt-right">

        <?php       

        include("../../files/d-search-compare.php");            

        ?>

     </div>

     <div style="clear:both"></div>

  <div class="more"><p><span>Note :&nbsp;</span><?php //echo $sql_rec['note']; ?></p><p>paragraph text.</p>

</div> <!--content-middle-txt end -->

</div>  <!--content-middle end -->

<div id="content-bottom">

<div id="content-bottom-txt">                   

    <div id="bottom-all-box">

    <?php                                       

        include("../../files/lb2.php");

    ?>

    <div id="content-bottom-right">

    <?php

        include("../../files/rb1.php");

    ?> 

    </div>

    </div>

    <?php

    /*  

        include("../../files/country-left-box3.php");

        include("../../files/country-right-box1.php");

    */

    ?>

Answer 1

我会将所有图像名称更改为PHP文件的确切页面名称。因此，例如，/united-kingdom.php您的图片应命名为united-kingdom.jpg。

使用相同的内容，您可以使用$_SERVER['REQUEST_URI']获取您请求的当前页面网址。

现在，您可以使用当前页面名称定义图像。

$url= $_SERVER['REQUEST_URI'];
$imageName = substr($url, 0, strpos($imageName, "."));

变量$imageName将包含united-kingdom。因此，您可以将图像定义为：

<img src="<?php echo $imageName; ?>.jpg" alt="" />

此代码未经过完整测试，只是为了让您了解如何处理它。

使用URL或页面标题在每个页面上更改图像源的PHP代码

1 个答案: