Question

我试图编写一个从数据库中获取图像的功能，并根据图像大小将它们显示在不同大小的边框内的网页上。当我尝试在浏览器中解析此代码时，出现意外的T_else错误。任何人都可以解释我哪里出错了吗？感谢

<?
$cakepicsql = mysql_query("SELECT * from cakes WHERE category = '".$cake['category']."'      ORDER BY rand() LIMIT 1") or die(mysql_error()); 
while($cakepic = mysql_fetch_array( $cakepicsql ))  {

    $image_path = "http://WEBSITE/products/"; 
    $filename = $image_path + $_GET['id'] + "-1.jpg";
    $size = getimagesize($filename);
    if ($size[0] > $size[1])
        echo "<a href="http://www.WEBSITE.co.uk/shop/cakes/<?=$cake['category_p']?>"><img src="http://www.WEBSITE.co.uk/products/<?=$cakepic['id']?>-1.jpg" alt="<?=$cake['name']?>" width="280" height="274" border="0" class="rounded-image" /></a> ";
    else
        echo "<a href="http://www.WEBSITE.co.uk/shop/cakes/<?=$cake['category_p']?>"><img src="http://www.WEBSITE.co.uk/products/<?=$cakepic['id']?>-1.jpg" alt="<?=$cake['name']?>" width="202" height="274" border="0" class="rounded-image" /></a>";
    endif;
} ?>      
  </td>
</tr>

Answer 1

由于您正在使用备用控制结构，因此在每个命令后需要:，因此：

<? if ($size[0] > $size[1]) ?>

应该是

<? if ($size[0] > $size[1]): ?>

和

<? else ?>

应该是

 <? else: ?>

此外，这是非常混乱的，所有那些打开和关闭输出标签。只需在普通PHP括号（<?php / ?>）内完成整个操作并使用echo。

Answer 2

试试这个

   <?php
    $cakepicsql = mysql_query("SELECT * from cakes WHERE category = '".$cake['category']."'      ORDER BY rand() LIMIT 1") or die(mysql_error()); 
     while($cakepic = mysql_fetch_array( $cakepicsql ))  {
      $image_path = "http://WEBSITE/products/"; 
      $filename = $image_path + $_GET['id'] + "-1.jpg"; 
      $size = getimagesize($filename); 
   if ($size[0] > $size[1]) {
   <a href='http://www.WEBSITE.co.uk/shop/cakes/<?php echo $cake["category_p2"];?>'><img src="http://www.WEBSITE.co.uk/products/<?php echo $cakepic['id']; ?>-1.jpg" alt="<?php echo $cake['name'];?>" width="280" height="274" border="0" class="rounded-image" /></a>
 } else{ 
 <a href="http://www.WEBSITE.co.uk/shop/cakes/<?php echo $cake['category_p'];?>"><img src="http://www.WEBSITE.co.uk/products/<?php echo $cakepic['id'];?>-1.jpg" alt="<?php echo $cake['name'];?>" width="202" height="274" border="0" class="rounded-image" /></a>
   }
   } 
 ?>      
 </td>
</tr>

Answer 3

更好：

<?php
$cakepicsql = mysql_query("SELECT * from cakes WHERE category = '".$cake['category']."'      ORDER BY rand() LIMIT 1") or die(mysql_error()); 
while($cakepic = mysql_fetch_array( $cakepicsql ))  {

 $image_path = "http://WEBSITE/products/"; 
 $filename = $image_path + $_GET['id'] + "-1.jpg"; 
 $size = getimagesize($filename); 
 if ($size[0] > $size[1]):
?>
     <a href="http://www.WEBSITE.co.uk/shop/cakes/<?=$cake['category_p']?>"><img src="http://www.WEBSITE.co.uk/products/<?=$cakepic['id']?>-1.jpg" alt="<?=$cake['name']?>" width="280" height="274" border="0" class="rounded-image" /></a>
 <?php
  else:
?>
  <a href="http://www.WEBSITE.co.uk/shop/cakes/<?=$cake['category_p']?>"><img src="http://www.WEBSITE.co.uk/products/<?=$cakepic['id']?>-1.jpg" alt="<?=$cake['name']?>" width="202" height="274" border="0" class="rounded-image" /></a>
 <?php endif;
  }  
?> 
  </td>
</tr>

你只需要＆＃34;：＆＃34;在if / else语句中，您不需要在每一行上打开和关闭PHP标记。

Answer 4

<?php
$cakepicsql = mysql_query("SELECT * from cakes WHERE category = '".$cake['category']."'      ORDER BY rand() LIMIT 1") or die(mysql_error()); 
while($cakepic = mysql_fetch_array( $cakepicsql ))  {
    $image_path = "http://WEBSITE/products/";
    $filename = $image_path + $_GET['id'] + "-1.jpg"; 
    $size = getimagesize($filename);
    if ($size[0] > $size[1]) 
        echo "<a href='http://www.WEBSITE.co.uk/shop/cakes/".$cake['category_p']."'><img src='http://www.WEBSITE.co.uk/products/'".$cakepic['id']."-1.jpg' alt='".$cake['name']."' width='280' height='274' border='0' class='rounded-image' /></a>";
    else
        echo "<a href='http://www.WEBSITE.co.uk/shop/cakes/".$cake['category_p']."'><img src='http://www.WEBSITE.co.uk/products/".$cakepic['id']."-1.jpg' alt='".$cake['name']."' width='202' height='274' border='0' class='rounded-image' /></a>";
 } 
?>      
  </td>
</tr>

Answer 5

就我个人而言，我认为HEREDOC使您的代码看起来更整洁，更容易修复，并避免原始引用问题.......

<?php
$sql = <<<EOF
SELECT * from cakes
WHERE category = '{$cake['category']}'
ORDER BY rand() LIMIT 1
EOF;

$cakepicquery = mysql_query($sql) or die(mysql_error()); 
while($cakepic = mysql_fetch_array( $cakepicquery ))  {

    $image_path = "http://WEBSITE/products/"; 
    $filename = $image_path + $_GET['id'] + "-1.jpg";
    $size = getimagesize($filename);
    $height = 274;
    $width = ($size[0] > $size[1]) ? 280 : 202;
    echo <<<EOF
<a href="http://www.WEBSITE.co.uk/shop/cakes/{$cake['category_p']}">
    <img src="http://www.WEBSITE.co.uk/products/{$cakepic['id']}-1.jpg" alt="{$cake['name']}" width="{$width}" height="{$height}" border="0" class="rounded-image" />
</a>
EOF;
} ?>    
</td>
</tr>

顺便提一下，你的循环在变量使用方面似乎有点混乱。如果您正在编制URL一次，则无需再次执行此操作。为什么在一个位置使用_GET ['id']而在另一个位置使用数据库中的ID？

  echo <<<
<a href="http://www.WEBSITE.co.uk/shop/cakes/{$cake['category_p']}">
    <img src="{$filename}" alt="{$cake['name']}" width="{$width}" height="{$height}" border="0" class="rounded-image" />
</a>
EOF;

Answer 6

您需要在回声中转义双引号。要在字符串中包含引号，您需要使用不同的引号，例如：echo "width='280'";或转义引号，例如：echo "width=\"280\""否则您将结束字符串，使PHP期望代码遵循。

另外，您似乎正在尝试在PHP代码中嵌入PHP代码。

要在字符串中包含PHP数组值，您需要像这样编写它：

echo "<a href='http://www.WEBSITE.co.uk/shop/cakes/".$cake['category_p']."'>

因此，例如，此行应如下所示：

echo "<a href='http://www.WEBSITE.co.uk/shop/cakes/".$cake['category_p']."'><img src='http://www.WEBSITE.co.uk/products/".$cakepic['id']."-1.jpg' alt='".<?=$cake['name']."' width='280' height='274' border='0' class='rounded-image' /></a>";

编辑：只是添加我同意其他人，构建字符串的部分（例如$filename = "http://www.WEBSITE.co.uk/shop/cakes/<?=$cake['category_p']?>"）确实看起来更整洁，并且如果您以后必须更改某些内容，则会使代码更容易维护。

PHP If / Else Statement不使用嵌入式HTML代码

6 个答案: