Question

How do I echo a variable inside the tag? this code does not seem to work. Any help is appreciated.

echo "<a href='productdetails.php?image="$results['image']"'><img src='images/".$results["image"]."' class='img-responsive' /></a><br />";

Answer 1

喜欢这个

echo "<a href='productdetails.php?image="{$results['image']}"'>
    <img src='images/{$results["image"]}' class='img-responsive' />
</a><br />";

或

echo "<a href='productdetails.php?image="$results[image]"'>
    <img src='images/$results[image]' class='img-responsive' />
</a><br />";

或者喜欢你的原作，但使用正确的连接

echo "<a href='productdetails.php?image=" . $results['image'] . "'>
        <img src='images/".$results["image"]."' class='img-responsive' />
    </a><br />";

Answer 2

应该是这样的，你在.之前和之后错过了$results['image']，所以没有发生正确的连接：

echo "<a href='productdetails.php?image=" . $results['image'] . "'><img src='images/" . $results["image"] . "' class='img-responsive' /></a><br />";

方法2：把它放在变量中：

$image = $results['image'];
echo "<a href='productdetails.php?image=" . $image . "'><img src='images/" . $image . "' class='img-responsive' /></a><br />";

Answer 3

或者像这样：

echo 
    '<a href="productdetails.php?image=' . $results['image'] . '">
        <img src="images/' . $results['image'] . '" class="img-responsive" />
    </a><br />';

内部回声变化<a> tag

3 个答案: