PHP eco 内锚标记内的图像标记
Image tag inside anchor tag inside PHP eco
以下是我的php代码
while ($row = mysqli_fetch_array($return_data)) {
$id = "ID:".$row['demo_id']."<br>";
$name = "Name: ".$row['demo_name']."<br>";
$version = "Version: ".$row['demo_version']."<br>";
$details = "Details: ".$row['demo_details']."<br>";
$file = "File Link: ".$row['file']."<br>";
$new = basename( $row['file'] ); // GET FILE NAME ONLY, GET RID OF PATH.
'<img src = \"../demo_webpages_project/images/$new"/>';
echo '<a href = "http://'.$_SERVER["SERVER_NAME"].':8080/demo_webpages_project/images/".$new> Link </a>';
我希望 'link' 将我带到我上传的图片 'file'。
但是由于我将锚标记放在回声中,它假定为 '.
$new' 作为文字而不是从 $new 变量中获取值。
我该怎么做才能避免这种情况?
当我遇到这种情况时,我只是将事情分解成这样易于管理的部分。
while ($row = mysqli_fetch_array($return_data)) {
$id = "ID:".$row['demo_id']."<br>";
$name = "Name: ".$row['demo_name']."<br>";
$version = "Version: ".$row['demo_version']."<br>";
$details = "Details: ".$row['demo_details']."<br>";
$file = "File Link: ".$row['file']."<br>";
$new = basename( $row['file'] ); // GET FILE NAME ONLY, GET RID OF PATH.
echo "<a href='http://{$_SERVER['SERVER_NAME']}:8080/demo_webpages_project/images/$new'>";
echo "<img src='../demo_webpages_project/images/$new'/>";
echo '</a>';
您不需要此代码上的端口号或完整的域名,如果您确实使用它,则在移动到真实的实时服务器或从一个域到另一个,当然你喜欢我们其他人会忘记做的事情。
所以试试这个
echo "<a href='demo_webpages_project/images/$new'>";
echo "<img src='../demo_webpages_project/images/$new'/>";
echo '</a>';
试试这条线
echo '<a href = "http://'.$_SERVER["SERVER_NAME"].':8080/demo_webpages_project/images/'.$new.'"> Link </a>';
以下是我的php代码
while ($row = mysqli_fetch_array($return_data)) {
$id = "ID:".$row['demo_id']."<br>";
$name = "Name: ".$row['demo_name']."<br>";
$version = "Version: ".$row['demo_version']."<br>";
$details = "Details: ".$row['demo_details']."<br>";
$file = "File Link: ".$row['file']."<br>";
$new = basename( $row['file'] ); // GET FILE NAME ONLY, GET RID OF PATH.
'<img src = \"../demo_webpages_project/images/$new"/>';
echo '<a href = "http://'.$_SERVER["SERVER_NAME"].':8080/demo_webpages_project/images/".$new> Link </a>';
我希望 'link' 将我带到我上传的图片 'file'。
但是由于我将锚标记放在回声中,它假定为 '. $new' 作为文字而不是从 $new 变量中获取值。
我该怎么做才能避免这种情况?
当我遇到这种情况时,我只是将事情分解成这样易于管理的部分。
while ($row = mysqli_fetch_array($return_data)) {
$id = "ID:".$row['demo_id']."<br>";
$name = "Name: ".$row['demo_name']."<br>";
$version = "Version: ".$row['demo_version']."<br>";
$details = "Details: ".$row['demo_details']."<br>";
$file = "File Link: ".$row['file']."<br>";
$new = basename( $row['file'] ); // GET FILE NAME ONLY, GET RID OF PATH.
echo "<a href='http://{$_SERVER['SERVER_NAME']}:8080/demo_webpages_project/images/$new'>";
echo "<img src='../demo_webpages_project/images/$new'/>";
echo '</a>';
您不需要此代码上的端口号或完整的域名,如果您确实使用它,则在移动到真实的实时服务器或从一个域到另一个,当然你喜欢我们其他人会忘记做的事情。
所以试试这个
echo "<a href='demo_webpages_project/images/$new'>";
echo "<img src='../demo_webpages_project/images/$new'/>";
echo '</a>';
试试这条线
echo '<a href = "http://'.$_SERVER["SERVER_NAME"].':8080/demo_webpages_project/images/'.$new.'"> Link </a>';