我错误地购买了godaddy linux,现在我不知道如何做我以前用经典ASP做的简单照片库!
我创建了一个包含字段“image_path”和“no_of_images”等的MySQL表...我想要做的是连接到数据库表,将image_path导入img标记并循环直到存储在“no_of_images”中的数值满足了。
我尝试使用此代码执行此操作,但它无效。
<?php
$servername = "localhost";
$username="";
$password="";
$dbname="";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$myid = $_GET['id'];
$sql = "SELECT * FROM gallery where id='$myid'";
$result = $conn->query($sql);
$row = mysql_fetch_row($result);
$image_path = $row[3];
$no_of_images = $row[4];
$x = 1;
while($x <= $no_of_images ) {
echo "<img src="$image_path"/"$x".jpg><br>";
$x++;
}
$conn->close();
?>
答案 0 :(得分:4)
您在mysql函数中使用mysqli结果。您应该使用$result->fetch_row()
答案 1 :(得分:0)
如前所述,您需要切换到MySQL功能。使用此代码:
<?php
$servername = "localhost";
$username = "";
$password = "";
$dbname = "";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$myid = $_GET['id'];
$sql = "SELECT * FROM gallery where id='$myid'";
$result = $conn->query($sql);
$x = 1;
while ($row = $result -> fetch_row()){
$image_path = $row[3];
$no_of_images = $row[4];
echo "<img src=" .$image_path. "/" .$x. ".jpg><br>";
$x++;
}
$conn->close();