Php图库表
我需要做一个表,最好每行4-5张图片,但它会继续在我的页面上。任何帮助都将不胜感激Php图库表,php,html,gallery,Php,Html,Gallery,我需要做一个表,最好每行4-5张图片,但它会继续在我的页面上。任何帮助都将不胜感激 <?php $host = "localhost"; $user = "root"; $password = ""; $database = "imageupload"; $query = "SELECT id, url, name from images "; $connect = mysqli_connect($host,$user,$password,$database) or die("Pro
<?php
$host = "localhost";
$user = "root";
$password = "";
$database = "imageupload";
$query = "SELECT id, url, name from images ";
$connect = mysqli_connect($host,$user,$password,$database) or die("Problem connecting.");
$result = mysqli_query($connect,$query) or die("Bad Query.");
mysqli_close($connect);
while($row = $result->fetch_array())
{
echo "<td>";
echo "<td><h2><img src=" . $row['url'] . " width=150 height=150/></h2></td>";
echo "<td><h2>" .$row['name'] . "</h2></td>";
echo "</td>";
}
?>
<table>
您可以尝试以下方法:
echo "<table><tr>";
foreach($row = $result->fetch_array()) {
if($row['id'] % 4 == 0) { //Change it to 5 if you want 5 images per row
echo "</tr><tr>";
}
echo "<td><img src='" . $row['url'] . "' /></td>";
}
echo "</tr></table>";
echo”“;
foreach($row=$result->fetch\u array()){
如果($row['id']%4==0){//如果希望每行有5个图像,请将其更改为5
回声“;
}
回声“;
}
回声“;
您包含了一些代码,这些代码看起来好像是在尝试执行您需要执行的操作。它有什么作用?您过早地关闭了连接,并且您的
标签不正确。语法:…
。..
是你的代码。而且,我实际上没有看到任何
s。@don't也这样;语法错误层出不穷,大多数是HTML,其中一个是mysql。