此代碼回顯指定目錄中的所有文件名，但僅將文件夾中最后一個文件的名稱插入數據庫。我必須怎么做才能將目錄中的所有文件名導入數據庫？非常感謝<?php$servername = "localhost";$username = "root";$password = "mypswrd";$dbname = "mydb";// Create connection$conn = new mysqli($servername, $username, $password, $dbname);// Check connectionif ($conn->connect_error) { die("Connection failed: " . $conn->connect_error);}$dir = "myFiles"; // directory path to files// Open a directory, and read its contentsif (is_dir($dir)) { if ($dh = opendir($dir)) { while (($file = readdir($dh)) !== false) { echo $file . "</br>"; $sql = "INSERT INTO myTable (fileName) VALUES ('$file')"; } closedir($dh); }}if ($conn->query($sql) === true) { echo "New record created successfully";} else { echo "Error: " . $sql . "<br>" . $conn->error;}$conn->close();?>
1 回答
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TA貢獻1845條經驗 獲得超8個贊
這是因為您的“if ($conn->query($sql) === TRUE”）在 while 循環之外。
你在 while 循環中構建你的 $sql，但你只在 while 循環之后執行查詢，因此你只插入最后一條記錄。
//建議
<?php
$servername = "localhost";
$username = "root";
$password = "mypswrd";
$dbname = "mydb";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$dir = "myFiles"; // directory path to files
// Open a directory, and read its contents
if (is_dir($dir)){
if ($dh = opendir($dir)){
while (($file = readdir($dh)) !== false){
echo $file . "</br>";
$sql = "INSERT INTO myTable (fileName) VALUES ('$file')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
}
else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
closedir($dh);
}
}
$conn->close();
?>
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