PHP eco中的锚定标记内的图像标记
下面是我的php代码PHP eco中的锚定标记内的图像标记,php,html,mysql,sql,href,Php,Html,Mysql,Sql,Href,下面是我的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>"; $detail
while ($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>';
while($row=mysqli\u fetch\u数组($return\u data)){
$id=“id:”.$row['demo_id']。“
”;
$name=“name:”.$row['demo_name']。“
”;
$version=“version:”.$row['demo_version']。“
”;
$details=“details:”.$row['demo_details']。“
”;
$file=“文件链接:”.$row['file']。“
”;
$new=basename($row['file']);//只获取文件名,去掉路径。
'';
回声';
我希望“链接”将我带到我上传的图像“文件”
但是,由于我将锚定标记放在回音中,因此它假定为”。
$new'作为文本,而不是从$new变量中获取值
我能做些什么来避免这种情况呢?当我遇到这种情况时,我会把事情分解成这样的可管理的部分
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>';
while($row=mysqli\u fetch\u数组($return\u data)){
$id=“id:”.$row['demo_id']。“
”;
$name=“name:”.$row['demo_name']。“
”;
$version=“version:”.$row['demo_version']。“
”;
$details=“details:”.$row['demo_details']。“
”;
$file=“文件链接:”.$row['file']。“
”;
$new=basename($row['file']);//只获取文件名,去掉路径。
回声“;
您不应该需要此代码上的端口号或完整的域名,如果您确实使用它,那么当您移动到真正的实时服务器时,或者从一个域移动到另一个域时,您必须修改所有此代码,当然,您希望我们中的其他人会忘记这样做
所以试试这个吧
echo "<a href='demo_webpages_project/images/$new'>";
echo "<img src='../demo_webpages_project/images/$new'/>";
echo '</a>';
echo“;
当我遇到这种情况时,我会把事情分解成这样的可管理的部分
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>';
while($row=mysqli\u fetch\u数组($return\u data)){
$id=“id:”.$row['demo_id']。“
”;
$name=“name:”.$row['demo_name']。“
”;
$version=“version:”.$row['demo_version']。“
”;
$details=“details:”.$row['demo_details']。“
”;
$file=“文件链接:”.$row['file']。“
”;
$new=basename($row['file']);//只获取文件名,去掉路径。
回声“;
您不应该需要此代码上的端口号或完整的域名,如果您确实使用它,那么当您移动到真正的实时服务器时,或者从一个域移动到另一个域时,您必须修改所有此代码,当然,您希望我们中的其他人会忘记这样做
所以试试这个吧
echo "<a href='demo_webpages_project/images/$new'>";
echo "<img src='../demo_webpages_project/images/$new'/>";
echo '</a>';
echo“;
试试这句话
echo '<a href = "http://'.$_SERVER["SERVER_NAME"].':8080/demo_webpages_project/images/'.$new.'"> Link </a>';
echo';
试试这句话
echo '<a href = "http://'.$_SERVER["SERVER_NAME"].':8080/demo_webpages_project/images/'.$new.'"> Link </a>';
echo';
使用“
双引号如果你想解释变量,你只是在双引号/单引号的地狱里。简化它,它不在锚中,它挂在无人地带,这样代码就不应该compile@Ghost嗨,我应该在哪里使用双引号?如果我使用echo““它把我带到了影像folder@RiggsFolly我尝试了几乎所有的引号排列,但都是在vainuse“
双引号如果你想解释变量,你只是在双引号/单引号地狱里。简化它,它不在锚里,它挂在无人地带,所以代码不应该compile@Ghost嗨,我应该在哪里使用双引号。如果我使用echo“”;它带我去看电影folder@RiggsFolly我尝试了几乎所有的引号排列,但都没有成功