<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Online PHP Script Execution</title>
</head>
<body>
<?php
$handle = fopen("http://data.kaohsiung.gov.tw/Opendata/DownLoad.aspx?Type=2&CaseNo1=AV&CaseNo2=1&FileType=1&Lang=C&FolderType=","rb");
$content = "";
while (!feof($handle)) {
$content .= fread($handle, 10000);
}
fclose($handle);
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "KHtravel";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$content = json_decode($content);
foreach ($content as $key => $value) {
$sql = "INSERT INTO travelstation (id, name, toldescribe, tel, address, picture1, px, py, opentime, changetime) VALUES ('$value->Id', '$value->Name', '$value->Toldescribe', '$value->Tel', '$value->Add', '$value->Picture1', '$value->Px', '$value->Py', '$value->Opentime', '$value->Changetime')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
}
mysqli_close($conn);
?>
</body>
</html>
- Sep 11 Fri 2015 12:16
[PHP] PHP讀JSON並將結果存進MySQL
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