我正在制作一個像craigslist這樣的網站，人們可以在其中發布他們想要出售的東西，并帶有帖子標題，帖子描述和電子郵件。問題：我可以從數據庫中提取除映像之外的所有數據。上傳圖像時，它們存儲在 C：\xampp\htdocs\uploads 中，圖像名稱與其余信息一起保存在數據庫中。下面是我的php。我試圖從數據庫中提取圖像的名稱，并用它來像這樣抓住它echo "<img src='uploads/".$image."' width='200'> ";我是php的新手，所以任何提示都值得贊賞。這是存儲/檢索用戶上傳圖像的好方法嗎？謝謝<?php$host = "localhost"; /* Host name */$user = "root"; /* User */$password = ""; /* Password */$dbname = "mydb"; /* Database name */$con = mysqli_connect($host, $user, $password,$dbname);// Check connectionif (!$con) { die("Connection failed: " . mysqli_connect_error());}$sql= "SELECT * FROM products";$result = $con->query($sql);$image = "SELECT image FROM products";if ($result->num_rows > 0) { // output data of each row while($row = $result->fetch_assoc()) { echo "<img src='uploads/".$image."' width='200'> "; echo "Title: " . $row["title"]. " Price: $" . $row["price"]. " <br> " . $row["description"]. " <br> " . $row["contact"] . " <br><br>"; }} else { echo "0 results";}$con->close();?>
1 回答
智慧大石
TA貢獻1946條經驗 獲得超3個贊
您需要對圖像使用相同的行，例如：
$sql= "SELECT * FROM products";
$result = $con->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<img src='uploads/".$row["image"]."' width='200'> ";
echo "Title: " . $row["title"]. " Price: $" . $row["price"]. " <br> " . $row["description"]. " <br> " . $row["contact"] . " <br><br>";
}
} else {
echo "0 results";
}
$con->close();
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